Chapter 3 Square-Square Root & Cube-Cube Root

Concept: The square of an odd number is always an odd number, and the square of an even number is always an even number.

We are looking for a number among the options that is the square of an odd number.

Based on the concept, the square of an odd number must be an odd number itself.

Let's examine the given options:

(a) 256: This number is even (it ends in 6). Therefore, it cannot be the square of an odd number. We can confirm that $\sqrt{256} = 16$, which is an even number.

(b) 361: This number is odd (it ends in 1). This could potentially be the square of an odd number. Let's find its square root.

We find that $\sqrt{361} = 19$.

Since $19$ is an odd number, $361$ is the square of an odd number.

(c) 144: This number is even (it ends in 4). Therefore, it cannot be the square of an odd number. We can confirm that $\sqrt{144} = 12$, which is an even number.

(d) 400: This number is even (it ends in 0). Therefore, it cannot be the square of an odd number. We can confirm that $\sqrt{400} = 20$, which is an even number.

From the analysis, only 361 is an odd number among the options, and its square root (19) is also an odd number.

Therefore, the square of an odd number among the given choices is 361.

The correct answer is (b) 361.

Concept: The unit digit of the square of a number is determined by the unit digit of the original number. To find the unit digit of a square, find the square of the unit digit of the original number and take its unit digit.

We need to find which of the given numbers, when squared, results in a number with 1 at its units place.

Let's examine the unit digit of each number and the unit digit of its square:

(a) For $19^2$: The unit digit of 19 is 9. The square of the unit digit is $9^2 = 81$. The unit digit of 81 is 1.

(b) For $17^2$: The unit digit of 17 is 7. The square of the unit digit is $7^2 = 49$. The unit digit of 49 is 9.

(c) For $18^2$: The unit digit of 18 is 8. The square of the unit digit is $8^2 = 64$. The unit digit of 64 is 4.

(d) For $16^2$: The unit digit of 16 is 6. The square of the unit digit is $6^2 = 36$. The unit digit of 36 is 6.

Only in option (a), where we square 19, does the resulting number ($19^2 = 361$) have a unit digit of 1.

Therefore, $19^2$ will have 1 at its units place.

The correct answer is (a) $19^2$.

Concept: The number of natural numbers lying between the squares of two consecutive natural numbers, $n^2$ and $(n+1)^2$, is always equal to $2n$.

Here, the two consecutive natural numbers are 18 and 19.

We can consider $n = 18$. The next consecutive natural number is $n+1 = 19$.

We need to find the number of natural numbers between $18^2$ and $19^2$.

Using the concept, the number of natural numbers between $n^2$ and $(n+1)^2$ is given by $2n$.

Substituting $n = 18$, the number of natural numbers is $2 \times 18 = 36$.

Alternate Method:

Calculate the squares of the numbers:

$18^2 = 324$

$19^2 = 361$

We need to find the number of natural numbers that are greater than 324 and less than 361.

These numbers are $325, 326, ..., 360$.

To find the count of these numbers, we can subtract the smaller number (just above 324, which is 325) from the larger number (just below 361, which is 360) and add 1 (because we are counting inclusively from the number after the first square up to the number before the second square).

Number of natural numbers = (Last number in the range) - (First number in the range) + 1

Number of natural numbers = $360 - 325 + 1$

$360 - 325 = 35$

$35 + 1 = 36$

Both methods confirm that there are 36 natural numbers between $18^2$ and $19^2$.

Comparing this result with the given options:

(a) 30

(b) 37

(c) 35

(d) 36

The correct answer is (d) 36.

Concept: A perfect square is a number obtained by squaring a natural number. The unit digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. A number whose unit digit is 2, 3, 7, or 8 cannot be a perfect square.

We need to identify which of the given numbers is not a perfect square.

Let's look at the unit digit of each option:

(a) 361: The unit digit is 1. Numbers ending in 1 can be perfect squares (e.g., $1^2=1$, $9^2=81$, $11^2=121$, $19^2=361$). $\sqrt{361} = 19$, which is a natural number. So, 361 is a perfect square.

(b) 1156: The unit digit is 6. Numbers ending in 6 can be perfect squares (e.g., $4^2=16$, $6^2=36$, $14^2=196$, $16^2=256$, $24^2=576$, $26^2=676$, $34^2=1156$). $\sqrt{1156} = 34$, which is a natural number. So, 1156 is a perfect square.

(c) 1128: The unit digit is 8. According to the concept, a number ending in 8 cannot be a perfect square.

(d) 1681: The unit digit is 1. Numbers ending in 1 can be perfect squares (e.g., $1^2=1$, $9^2=81$, $11^2=121$, $19^2=361$, $21^2=441$, $29^2=841$, $31^2=961$, $39^2=1521$, $41^2=1681$). $\sqrt{1681} = 41$, which is a natural number. So, 1681 is a perfect square.

Based on the unit digit rule, the number 1128, which ends in 8, cannot be a perfect square.

Therefore, the number that is not a perfect square is 1128.

The correct answer is (c) 1128.

Concept: The unit digit of a perfect square is determined by the unit digit of the original number being squared. The unit digits of the squares of the digits 0 through 9 are as follows:

• $0^2 = 0$ (Unit digit is 0)
• $1^2 = 1$ (Unit digit is 1)
• $2^2 = 4$ (Unit digit is 4)
• $3^2 = 9$ (Unit digit is 9)
• $4^2 = 16$ (Unit digit is 6)
• $5^2 = 25$ (Unit digit is 5)
• $6^2 = 36$ (Unit digit is 6)
• $7^2 = 49$ (Unit digit is 9)
• $8^2 = 64$ (Unit digit is 4)
• $9^2 = 81$ (Unit digit is 1)

The possible unit digits of a perfect square are the unit digits found above: 0, 1, 4, 5, 6, and 9.

The digits that are not possible unit digits of a perfect square are the remaining digits: 2, 3, 7, and 8.

We need to find which digit from the given options cannot be the unit digit of a perfect square.

Let's check each option against the list of possible and impossible unit digits:

• (a) 1: 1 is a possible unit digit of a perfect square ($1^2 = 1$, $11^2=121$, etc.).
• (b) 6: 6 is a possible unit digit of a perfect square ($4^2 = 16$, $6^2 = 36$, etc.).
• (c) 5: 5 is a possible unit digit of a perfect square ($5^2 = 25$, $15^2 = 225$, etc.).
• (d) 3: 3 is not a possible unit digit of a perfect square. It is in the list of digits (2, 3, 7, 8) that cannot be the unit digit of a perfect square.

Therefore, a perfect square can never have the digit 3 at its ones place.

The correct answer is (d) 3.

We need to evaluate the expression $\sqrt{176 + \sqrt{2401}}$.

First, let's evaluate the inner square root, $\sqrt{2401}$.

We need to find a number whose square is 2401.

Let's estimate. $40^2 = 1600$ and $50^2 = 2500$. So the number is between 40 and 50.

The unit digit of 2401 is 1. The unit digit of the square root must be either 1 (since $1^2=1$) or 9 (since $9^2=81$, which ends in 1).

Let's try 41 and 49.

$41^2 = 41 \times 41 = 1681$.

$49^2 = 49 \times 49$.

$\begin{array}{cc}& & 4 & 9 \\ \times & & 4 & 9 \\ \hline && 4 & 4 & 1 \\ & 1 & 9 & 6 & \times \\ \hline 2 & 4 & 0 & 1 \\ \hline \end{array}$

So, $\sqrt{2401} = 49$.

Now substitute the value of $\sqrt{2401}$ back into the original expression:

$\sqrt{176 + \sqrt{2401}} = \sqrt{176 + 49}$

Next, perform the addition inside the square root:

$176 + 49 = 225$

So the expression becomes $\sqrt{225}$.

Finally, evaluate $\sqrt{225}$.

We need to find a number whose square is 225.

We know that $10^2 = 100$ and $20^2 = 400$. So the number is between 10 and 20.

The unit digit of 225 is 5. The unit digit of the square root must be 5 (since $5^2=25$).

Let's try 15.

$15^2 = 15 \times 15 = 225$.

So, $\sqrt{225} = 15$.

Thus, the value of $\sqrt{176 + \sqrt{2401}}$ is 15.

Comparing this result with the given options:

(a) 14

(b) 15

(c) 16

(d) 17

The correct answer is (b) 15.

Given:

$\sqrt{5625} = 75$

To Find:

The value of $\sqrt{0.5625} + \sqrt{56.25}$.

Solution:

We are given that $\sqrt{5625} = 75$. We need to find the values of $\sqrt{0.5625}$ and $\sqrt{56.25}$ using this information.

Consider the term $\sqrt{0.5625}$.

We can write $0.5625$ as $\frac{5625}{10000}$.

So, $\sqrt{0.5625} = \sqrt{\frac{5625}{10000}}$.

Using the property $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, we get:

$\sqrt{0.5625} = \frac{\sqrt{5625}}{\sqrt{10000}}$.

We are given $\sqrt{5625} = 75$.

We know that $\sqrt{10000} = \sqrt{100^2} = 100$.

Therefore, $\sqrt{0.5625} = \frac{75}{100} = 0.75$.

Alternatively, when taking the square root of a decimal number, if there are $n$ digits after the decimal point in the original number, there will be $n/2$ digits after the decimal point in its square root (assuming $n$ is even). Here, $0.5625$ has 4 decimal places, so its square root will have $4/2 = 2$ decimal places. Since $\sqrt{5625} = 75$, $\sqrt{0.5625}$ must be $0.75$.

Consider the term $\sqrt{56.25}$.

We can write $56.25$ as $\frac{5625}{100}$.

So, $\sqrt{56.25} = \sqrt{\frac{5625}{100}}$.

Using the property $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, we get:

$\sqrt{56.25} = \frac{\sqrt{5625}}{\sqrt{100}}$.

We are given $\sqrt{5625} = 75$.

We know that $\sqrt{100} = \sqrt{10^2} = 10$.

Therefore, $\sqrt{56.25} = \frac{75}{10} = 7.5$.

Alternatively, $56.25$ has 2 decimal places, so its square root will have $2/2 = 1$ decimal place. Since $\sqrt{5625} = 75$, $\sqrt{56.25}$ must be $7.5$.

Now we need to find the sum of these two values:

$\sqrt{0.5625} + \sqrt{56.25} = 0.75 + 7.5$

Adding the two decimal numbers:

$\begin{array}{cc} & 0 & . & 7 & 5 \\ + & 7 & . & 5 & 0 \\ \hline & 8 & . & 2 & 5 \\ \hline \end{array}$

So, $0.75 + 7.5 = 8.25$.

Thus, the value of $\sqrt{0.5625} + \sqrt{56.25}$ is 8.25.

Comparing this result with the given options:

(a) 82.5

(b) 0.75

(c) 8.25

(d) 75.05

The correct answer is (c) 8.25.

Concept: A perfect square is the square of a natural number. We need to find the natural numbers whose squares are greater than 1 and less than 50.

We list the squares of natural numbers:

• $1^2 = 1$
• $2^2 = 4$
• $3^2 = 9$
• $4^2 = 16$
• $5^2 = 25$
• $6^2 = 36$
• $7^2 = 49$
• $8^2 = 64$

We are looking for perfect squares that lie between 1 and 50. This means the perfect square must be strictly greater than 1 and strictly less than 50.

From the list above, the perfect squares that satisfy this condition are:

• $4$ (since $1 < 4 < 50$)
• $9$ (since $1 < 9 < 50$)
• $16$ (since $1 < 16 < 50$)
• $25$ (since $1 < 25 < 50$)
• $36$ (since $1 < 36 < 50$)
• $49$ (since $1 < 49 < 50$)

The perfect squares 1 and 64 are not between 1 and 50.

Counting these perfect squares, we find there are 6 such numbers (4, 9, 16, 25, 36, and 49).

Therefore, there are 6 perfect squares between 1 and 50.

The final answer is 6.

Concept: When a number ending in zeroes is raised to a power, the number of zeroes in the result is the number of zeroes in the original number multiplied by the power.

We need to find the number of zeroes in the cube of 100.

The number is 100.

The number of zeroes in 100 is 2.

We are calculating the cube, which means raising to the power of 3.

Using the concept, the number of zeroes in $100^3$ is the number of zeroes in 100 multiplied by 3.

Number of zeroes $= (\text{Number of zeroes in 100}) \times 3$

Number of zeroes $= 2 \times 3 = 6$.

Alternatively, we can calculate $100^3$ directly:

$100^3 = 100 \times 100 \times 100$

$100 \times 100 = 10000$

$10000 \times 100 = 1000000$

The number 1,000,000 has 6 zeroes.

Therefore, the cube of 100 will have 6 zeroes.

The final answer is 6.

Concept: To find the square of a decimal number, we multiply the number by itself. We can ignore the decimal point during multiplication and then place the decimal point in the result. The number of decimal places in the product is the sum of the number of decimal places in the numbers being multiplied.

We need to find the square of 6.1.

The square of 6.1 is $6.1 \times 6.1$ or $(6.1)^2$.

Let's multiply 61 by 61, ignoring the decimal point for now:

$\begin{array}{cc}& & 6 & 1 \\ \times & & 6 & 1 \\ \hline && & 6 & 1 \\ & 3 & 6 & 6 & \times \\ \hline 3 & 7 & 2 & 1 \\ \hline \end{array}$

So, $61 \times 61 = 3721$.

Now, let's consider the decimal places. The number 6.1 has 1 decimal place.

When we multiply 6.1 by 6.1, the total number of decimal places in the product will be the sum of the decimal places in each number, which is $1 + 1 = 2$.

So, we need to place the decimal point in 3721 such that there are 2 digits after the decimal point.

Starting from the right, we move the decimal point 2 places to the left: 37.21.

Therefore, $(6.1)^2 = 37.21$.

The final answer is 37.21.

Concept: To find the cube of a decimal number, we multiply the number by itself three times. We can ignore the decimal point during multiplication and then place the decimal point in the result. The number of decimal places in the product is the sum of the number of decimal places in the numbers being multiplied.

We need to find the cube of 0.3.

The cube of 0.3 is $0.3 \times 0.3 \times 0.3$ or $(0.3)^3$.

Let's multiply 3 by 3 by 3, ignoring the decimal point for now:

$3 \times 3 = 9$

$9 \times 3 = 27$

So, $3 \times 3 \times 3 = 27$.

Now, let's consider the decimal places. The number 0.3 has 1 decimal place.

When we multiply 0.3 by 0.3 by 0.3, the total number of decimal places in the product will be the sum of the decimal places in each number, which is $1 + 1 + 1 = 3$.

So, we need to place the decimal point in 27 such that there are 3 digits after the decimal point. We can add leading zeros if needed.

Starting from the right of 27, we move the decimal point 3 places to the left: 0.027.

Therefore, $(0.3)^3 = 0.027$.

The final answer is 0.027.

Concept: The unit digit of the square of a number is determined by the unit digit of the original number. To find the unit digit of a square, find the square of the unit digit of the original number and take its unit digit.

We need to find the digit at the units place of $68^2$.

The original number is 68.

The unit digit of 68 is 8.

Now, we find the square of the unit digit:

$8^2 = 64$.

The unit digit of $8^2 = 64$ is 4.

Therefore, the unit digit of $68^2$ will be 4.

We can verify this by calculating $68^2$:

$68 \times 68$

$\begin{array}{cc}& & 6 & 8 \\ \times & & 6 & 8 \\ \hline && 5 & 4 & 4 \\ & 4 & 0 & 8 & \times \\ \hline 4 & 6 & 2 & 4 \\ \hline \end{array}$

$68^2 = 4624$. The unit digit is indeed 4.

The final answer is 4.

Concept: The square root of a number $x$ is a number $y$ such that $y^2 = x$. Every positive number $x$ has two square roots: one positive and one negative. The positive square root is often referred to as the principal square root.

The standard mathematical symbol used to denote the positive square root (or principal square root) of a number $x$ is the radical sign, $\sqrt{\phantom{x}}$, placed over the number.

Thus, the positive square root of a number $x$ is denoted by $\sqrt{x}$.

The final answer is $\sqrt{x}$.

Concept: A perfect cube is a number that is the cube of a natural number. In the prime factorization of a perfect cube, the exponent of each prime factor is a multiple of 3.

We need to find the least number that should be multiplied with 9 to make it a perfect cube.

First, find the prime factorization of 9.

$9 = 3 \times 3 = 3^2$.

For a number to be a perfect cube, the exponent of each prime factor in its prime factorization must be a multiple of 3 (i.e., 3, 6, 9, etc.).

The prime factorization of 9 is $3^2$. The exponent of the prime factor 3 is 2.

To make the exponent of 3 a multiple of 3, the smallest multiple of 3 that is greater than or equal to 2 is 3.

We need the power of 3 to be $3^3$. Currently, it is $3^2$.

To get $3^3$ from $3^2$, we need to multiply by $3^{(3-2)} = 3^1 = 3$.

So, we need to multiply 9 by 3 to make it a perfect cube.

Let's check the result:

$9 \times 3 = 27$.

The prime factorization of 27 is $3 \times 3 \times 3 = 3^3$. The exponent of the prime factor 3 is 3, which is a multiple of 3. Therefore, 27 is a perfect cube ($3^3$).

The least number to be multiplied with 9 to make it a perfect cube is 3.

The final answer is 3.

We need to determine if the square of 0.4 is equal to 0.16.

To find the square of 0.4, we calculate $(0.4)^2$.

$(0.4)^2 = 0.4 \times 0.4$

We can multiply 4 by 4, ignoring the decimal point for now:

$4 \times 4 = 16$.

The number 0.4 has 1 decimal place. When multiplying 0.4 by 0.4, the total number of decimal places in the product will be $1 + 1 = 2$.

So, we place the decimal point in 16 such that there are 2 digits after the decimal point.

Moving the decimal point 2 places to the left from the right of 16 gives 0.16.

Thus, $(0.4)^2 = 0.16$.

The statement says that the square of 0.4 is 0.16, which matches our calculation.

Therefore, the statement is true.

The final answer is T.

We need to determine if the cube root of 729 is equal to 8.

The cube root of a number $x$, denoted by $\sqrt[3]{x}$, is a number $y$ such that $y^3 = x$.

If the cube root of 729 is 8, then $8^3$ should be equal to 729.

Let's calculate $8^3$:

$8^3 = 8 \times 8 \times 8$

$8 \times 8 = 64$

$64 \times 8$

$\begin{array}{cc}& & 6 & 4 \\ \times & & & 8 \\ \hline & 5 & 1 & 2 \\ \hline \end{array}$

So, $8^3 = 512$.

Since $512 \neq 729$, the cube root of 729 is not 8.

Let's find the actual cube root of 729 by prime factorization:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 = (3^2)^3 = 9^3$.

Therefore, $\sqrt[3]{729} = 9$.

The statement claims that the cube root of 729 is 8, which is false.

The final answer is F.

Concept: The number of natural numbers lying between the squares of two consecutive natural numbers, $n^2$ and $(n+1)^2$, is always equal to $2n$.

Here, the two consecutive natural numbers are 10 and 11.

We can consider $n = 10$. The next consecutive natural number is $n+1 = 11$.

We need to find the number of natural numbers between $10^2$ and $11^2$.

Using the concept, the number of natural numbers between $n^2$ and $(n+1)^2$ is given by $2n$.

Substituting $n = 10$, the number of natural numbers is $2 \times 10 = 20$.

Alternate Method:

Calculate the squares of the numbers:

$10^2 = 100$

$11^2 = 121$

We need to find the number of natural numbers that are greater than 100 and less than 121.

These numbers are $101, 102, ..., 120$.

To find the count of these numbers, we can subtract the smaller number (just above 100, which is 101) from the larger number (just below 121, which is 120) and add 1.

Number of natural numbers = $120 - 101 + 1$

$120 - 101 = 19$

$19 + 1 = 20$

Both methods show that there are 20 natural numbers between $10^2$ and $11^2$.

The statement claims that there are 21 natural numbers between $10^2$ and $11^2$, which is false.

The final answer is F.

Concept: The sum of the first $n$ odd natural numbers is equal to $n^2$.

We are asked about the sum of the first 7 odd natural numbers.

Here, the number of odd natural numbers is $n=7$.

Using the concept, the sum of the first $n$ odd natural numbers is $n^2$.

For $n=7$, the sum is $7^2 = 49$.

Alternate Method:

List the first 7 odd natural numbers and find their sum:

The first 7 odd natural numbers are 1, 3, 5, 7, 9, 11, and 13.

Sum = $1 + 3 + 5 + 7 + 9 + 11 + 13$

Sum = $(1+13) + (3+11) + (5+9) + 7$

Sum = $14 + 14 + 14 + 7$

Sum = $3 \times 14 + 7$

Sum = $42 + 7 = 49$.

Both methods show that the sum of the first 7 odd natural numbers is 49.

The statement claims that the sum of the first 7 odd natural numbers is 49, which is true.

The final answer is T.

We need to determine if the statement "The square root of a perfect square of n digits will have $\frac{n}{2}$ digits if n is even" is true or false.

Let $N$ be a perfect square with $n$ digits.

Let $\sqrt{N}$ be its positive square root, and let $\sqrt{N}$ have $k$ digits.

The number of digits $k$ in the positive square root of a perfect square with $n$ digits depends on whether $n$ is even or odd.

The rule is:

• If $n$ is an even number, then the number of digits in its square root is $k = \frac{n}{2}$.
• If $n$ is an odd number, then the number of digits in its square root is $k = \frac{n+1}{2}$.

The statement specifically talks about the case where $n$ is even.

According to the rule, when the number of digits $n$ is even, the number of digits in the square root is indeed $\frac{n}{2}$.

Let's consider a few examples where $n$ is even:

• If $n=2$ (e.g., 64), $k = \frac{2}{2} = 1$. $\sqrt{64}=8$, which has 1 digit.
• If $n=4$ (e.g., 1600), $k = \frac{4}{2} = 2$. $\sqrt{1600}=40$, which has 2 digits.
• If $n=6$ (e.g., 400000 not perfect square, $400081 = 633.9^2$ not perfect square. $400000$ to $999999$. smallest 6 digit perfect square $317^2 = 100489$. largest $999^2=998001$), $k = \frac{6}{2} = 3$. $\sqrt{100489} = 317$, which has 3 digits. $\sqrt{998001}=999$, which has 3 digits.

The statement accurately describes the rule for the number of digits in the square root when the original number of digits is even.

Therefore, the statement is true.

The final answer is T.

Concept: A property of square numbers is that they can be expressed as the sum of consecutive odd natural numbers starting from 1. The square of a natural number $n$ is equal to the sum of the first $n$ odd natural numbers.

That is, $n^2 = 1 + 3 + 5 + ... + (2n-1)$.

We need to express the number 36 as a sum of successive odd natural numbers.

First, recognize that 36 is a perfect square.

$36 = 6^2$.

Since $36 = 6^2$, we know that 36 can be expressed as the sum of the first 6 odd natural numbers.

The first 6 odd natural numbers are 1, 3, 5, 7, 9, and 11.

Let's find their sum:

Sum = $1 + 3 + 5 + 7 + 9 + 11$

Sum = $4 + 5 + 7 + 9 + 11$

Sum = $9 + 7 + 9 + 11$

Sum = $16 + 9 + 11$

Sum = $25 + 11$

Sum = $36$.

Thus, 36 can be expressed as the sum of the first 6 successive odd natural numbers.

The expression is $1 + 3 + 5 + 7 + 9 + 11$.

The final answer is $1 + 3 + 5 + 7 + 9 + 11$.

Concept: A number is a perfect square if and only if in its prime factorization, the exponent of each prime factor is an even number.

We need to check if 90 is a perfect square using prime factorization.

First, find the prime factorization of 90.

$\begin{array}{c|cc} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, the prime factorization of 90 is $2 \times 3 \times 3 \times 5 = 2^1 \times 3^2 \times 5^1$.

Now, examine the exponents of each prime factor in the factorization:

• The exponent of the prime factor 2 is 1.
• The exponent of the prime factor 3 is 2.
• The exponent of the prime factor 5 is 1.

For a number to be a perfect square, the exponent of each prime factor must be an even number.

In the prime factorization of 90, the exponents of 2 (which is 1) and 5 (which is 1) are odd numbers.

Since the exponents of all prime factors are not even, 90 is not a perfect square.

The final answer is 90 is not a perfect square because the exponents of its prime factors (2 and 5) are odd (1).

Concept: A number is a perfect cube if and only if in its prime factorization, the exponent of each prime factor is a multiple of 3 (i.e., 3, 6, 9, etc.). This means the prime factors can be grouped into triplets.

We need to check if 1728 is a perfect cube using prime factorization.

First, find the prime factorization of 1728.

$\begin{array}{c|cc} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 1728 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^6 \times 3^3$.

Now, examine the exponents of each prime factor in the factorization:

• The exponent of the prime factor 2 is 6.
• The exponent of the prime factor 3 is 3.

For a number to be a perfect cube, the exponent of each prime factor must be a multiple of 3.

In the prime factorization of 1728, the exponent of 2 (which is 6) is a multiple of 3 ($6 = 3 \times 2$), and the exponent of 3 (which is 3) is a multiple of 3 ($3 = 3 \times 1$).

Since the exponents of all prime factors are multiples of 3, 1728 is a perfect cube.

We can group the factors into triplets: $(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) = 2^3 \times 2^3 \times 3^3 = (2 \times 2 \times 3)^3 = 12^3$.

So, $\sqrt[3]{1728} = 12$.

The final answer is 1728 is a perfect cube because the exponents of its prime factors (2 and 3) are multiples of 3 (6 and 3 respectively).

Concept: The distributive law states that for any numbers a, b, and c, $a \times (b+c) = a \times b + a \times c$. We can use this law, often combined with the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$ or $(a-b)^2 = a^2 - 2ab + b^2$, to find the square of a number by expressing it as a sum or difference of two numbers.

We need to find the square of 43 using the distributive law.

We can write 43 as a sum of two numbers, for example, $40 + 3$.

The square of 43 is $(43)^2 = (40 + 3)^2$.

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, where $a=40$ and $b=3$:

$(40 + 3)^2 = (40)^2 + 2 \times 40 \times 3 + (3)^2$

Calculate each term:

$(40)^2 = 40 \times 40 = 1600$.

$2 \times 40 \times 3 = 80 \times 3 = 240$.

$(3)^2 = 3 \times 3 = 9$.

Now, add the terms:

$(40 + 3)^2 = 1600 + 240 + 9$

$1600 + 240 = 1840$

$1840 + 9 = 1849$.

Thus, the square of 43 is 1849.

We can also use the distributive law step-by-step without explicitly using the identity:

$(40 + 3)^2 = (40 + 3) \times (40 + 3)$

Using the distributive law, we multiply each term in the first parenthesis by each term in the second parenthesis:

$(40 + 3) \times (40 + 3) = 40 \times (40 + 3) + 3 \times (40 + 3)$

Apply the distributive law again:

$= (40 \times 40) + (40 \times 3) + (3 \times 40) + (3 \times 3)$

$= 1600 + 120 + 120 + 9$

$= 1600 + 240 + 9$

$= 1840 + 9$

$= 1849$.

Both methods give the same result.

The final answer is 1849.

Concept: A Pythagorean triplet is a set of three positive integers $a$, $b$, and $c$, such that $a^2 + b^2 = c^2$. For any natural number $m > 1$, the numbers $2m$, $m^2-1$, and $m^2+1$ form a Pythagorean triplet.

We need to find a Pythagorean triplet whose smallest number is 6.

We can use the formula for Pythagorean triplets based on a natural number $m > 1$: $2m$, $m^2-1$, and $m^2+1$.

The smallest number in the triplet is 6. We can set the smallest expression from the formula equal to 6 and solve for $m$. The expressions are $2m$ and $m^2-1$. Since $m > 1$, $2m$ grows linearly with $m$, while $m^2-1$ grows quadratically. For small values of $m > 1$, $m^2-1$ might be smaller than $2m$.

Let's check for which $m > 1$ the value $m^2-1$ equals 6.

$m^2 - 1 = 6$

$m^2 = 6 + 1$

$m^2 = 7$

Since 7 is not a perfect square, $m$ is not a natural number in this case.

Now, let's assume that the smallest number, 6, corresponds to $2m$.

$2m = 6$

$m = \frac{6}{2}$

$m = 3$.

Since $m=3$ is a natural number and $3 > 1$, we can use this value of $m$ to generate a Pythagorean triplet.

The triplet members are $2m$, $m^2-1$, and $m^2+1$. Substitute $m=3$:

$2m = 2 \times 3 = 6$

$m^2 - 1 = 3^2 - 1 = 9 - 1 = 8$

$m^2 + 1 = 3^2 + 1 = 9 + 1 = 10$

The Pythagorean triplet is (6, 8, 10).

Let's check if this is a Pythagorean triplet:

$6^2 + 8^2 = 36 + 64 = 100$

$10^2 = 100$

Since $6^2 + 8^2 = 10^2$, (6, 8, 10) is a Pythagorean triplet.

The smallest number in this triplet is 6.

The final answer is (6, 8, 10).

Concept: To find the cube root of a number using prime factorization, first find the prime factorization of the number. Then, group the prime factors into triplets. For each triplet of a prime factor, take one factor. The product of these single factors is the cube root of the number.

We need to find the cube root of 5832 using prime factorization.

First, find the prime factorization of 5832.

$\begin{array}{c|cc} 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 5832 is $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2^3 \times 3^6$.

Now, group the prime factors into triplets:

$5832 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)$

$5832 = 2^3 \times 3^3 \times 3^3$

To find the cube root, take one factor from each triplet:

$\sqrt[3]{5832} = \sqrt[3]{2^3 \times 3^3 \times 3^3}$

Using the property $\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$ and $\sqrt[3]{a^3} = a$, we get:

$\sqrt[3]{5832} = \sqrt[3]{2^3} \times \sqrt[3]{3^3} \times \sqrt[3]{3^3}$

$\sqrt[3]{5832} = 2 \times 3 \times 3$

$\sqrt[3]{5832} = 6 \times 3$

$\sqrt[3]{5832} = 18$.

We can verify this by cubing 18:

$18^3 = 18 \times 18 \times 18$

$18 \times 18 = 324$

$324 \times 18$

$\begin{array}{cc}& & 3 & 2 & 4 \\ \times & & & 1 & 8 \\ \hline & 2 & 5 & 9 & 2 \\ & 3 & 2 & 4 & \times \\ \hline 5 & 8 & 3 & 2 \\ \hline \end{array}$

$18^3 = 5832$.

Thus, the cube root of 5832 is 18.

The final answer is 18.

We need to find the square root of 22.09 using the long division method.

Steps for Long Division Method for Square Root:

1. Place bars over every pair of digits starting from the unit digit. For the decimal part, place bars over every pair of digits starting from the first decimal place.

2. Find the largest number whose square is less than or equal to the first period (the leftmost pair or single digit).

3. Subtract the square from the first period and bring down the next period to the right of the remainder to make the new dividend.

4. Double the quotient and write it with a blank digit on its right. Find a suitable digit for the blank space such that the product of the new number and this digit is less than or equal to the new dividend.

5. Write the digit in the quotient and the new number. Subtract the product.

6. Repeat steps 3, 4, and 5 until the remainder is 0 or the desired number of decimal places is reached.

7. Place the decimal point in the quotient above the decimal point in the number.

Applying the long division method to 22.09:

Pair the digits: 22.09 becomes $\overline{22}.\overline{09}$.

The first period is 22.

The largest square less than or equal to 22 is $4^2 = 16$.

Subtract 16 from 22: $22 - 16 = 6$.

Bring down the next period (09). The new dividend is 609.

Double the quotient (4) to get 8. Place a blank digit next to it: 8_.

We need to find a digit 'x' such that $8x \times x \leq 609$.

Trying digits: $87 \times 7 = 609$.

Write 7 in the quotient after the decimal point and next to 8.

Subtract 609 from 609: $609 - 609 = 0$.

The remainder is 0.

The long division process is shown below:

$\begin{array}{r} 4\ . \ 7\phantom{)} \\ 4{\overline{\smash{\big)}\,22.09\phantom{)}}} \\ \underline{-~\phantom{(}16\phantom{.09)}} \\ 87{\overline{\smash{\big)}\,609\phantom{)}}}\\ \underline{-~\phantom{()}(609)}\\ 0\phantom{)} \end{array}$

The quotient is 4.7.

Therefore, the square root of 22.09 is 4.7.

The final answer is 4.7.

Concept: A number is divisible by several numbers if it is divisible by their Least Common Multiple (LCM). A number is a perfect square if, in its prime factorization, the exponent of each prime factor is an even number.

We need to find the smallest perfect square that is divisible by 3, 4, 5, and 6. This number must be a multiple of 3, 4, 5, and 6.

The smallest number that is a multiple of 3, 4, 5, and 6 is their LCM.

Let's find the prime factorization of each number:

• $3 = 3^1$
• $4 = 2 \times 2 = 2^2$
• $5 = 5^1$
• $6 = 2 \times 3 = 2^1 \times 3^1$

To find the LCM of 3, 4, 5, and 6, we take the highest power of each prime factor present in the factorizations:

LCM$(3, 4, 5, 6) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$.

So, any number divisible by 3, 4, 5, and 6 must be a multiple of 60.

Now, we need this multiple of 60 to be a perfect square.

Let's look at the prime factorization of 60: $60 = 2^2 \times 3^1 \times 5^1$.

For a number to be a perfect square, the exponent of each prime factor in its prime factorization must be even.

In the prime factorization of 60:

• The exponent of 2 is 2 (which is even).
• The exponent of 3 is 1 (which is odd).
• The exponent of 5 is 1 (which is odd).

To make 60 a perfect square, we need to multiply it by the prime factors that have odd exponents, raised to a power that makes their exponent even. The smallest such exponent is 2.

We need to increase the exponent of 3 from 1 to 2, so we multiply by $3^{2-1} = 3^1 = 3$.

We need to increase the exponent of 5 from 1 to 2, so we multiply by $5^{2-1} = 5^1 = 5$.

The required multiplier is $3 \times 5 = 15$.

The smallest perfect square divisible by 3, 4, 5, and 6 is the LCM multiplied by the necessary factors to make the exponents even:

Smallest perfect square $= 60 \times (3 \times 5)$

Smallest perfect square $= 60 \times 15 = 900$.

Let's check the prime factorization of 900:

$900 = 60 \times 15 = (2^2 \times 3^1 \times 5^1) \times (3^1 \times 5^1) = 2^{2+0} \times 3^{1+1} \times 5^{1+1} = 2^2 \times 3^2 \times 5^2$.

Since the exponents of all prime factors (2, 3, and 5) are even (which is 2), 900 is a perfect square.

Also, since 900 is a multiple of LCM(3, 4, 5, 6), it is divisible by 3, 4, 5, and 6.

The final answer is 900.

Given:

Length of the ladder = 10 m

Distance of the foot of the ladder from the wall = 6 m

To Find:

The height of the wall.

Solution:

Let the wall be represented by a vertical line segment, the ground by a horizontal line segment, and the ladder by a hypotenuse connecting the top of the wall to the foot of the ladder on the ground.

Since the wall is vertical to the ground, the wall, the ground, and the ladder form a right-angled triangle.

In this right-angled triangle:

• The length of the ladder is the hypotenuse (10 m).
• The distance of the foot of the ladder from the wall is one leg (6 m).
• The height of the wall is the other leg (let's call it $h$ meters), which we need to find.

We can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let $h$ be the height of the wall.

According to the Pythagorean theorem:

(Distance from wall)$^2$ + (Height of wall)$^2$ = (Length of ladder)$^2$

$6^2 + h^2 = 10^2$

Now, solve for $h$:

$6^2 = 36$

$10^2 = 100$

So the equation becomes:

$36 + h^2 = 100$

Subtract 36 from both sides of the equation:

$h^2 = 100 - 36$

$h^2 = 64$

To find $h$, take the square root of both sides:

$h = \sqrt{64}$

Since the height must be a positive value, we take the positive square root:

$h = 8$

The height of the wall is 8 meters.

The final answer is 8 m.

Given:

Length of the rectangle = 20 m

Width of the rectangle = 15 m

To Find:

The length of a diagonal of the rectangle.

Solution:

Consider a rectangle. The diagonals of a rectangle divide it into two right-angled triangles. The sides of the rectangle form the two legs of the right-angled triangle, and the diagonal forms the hypotenuse.

Let the length of the rectangle be $l = 20$ m and the width be $w = 15$ m.

Let the length of the diagonal be $d$ m.

In the right-angled triangle formed by the length, the width, and the diagonal, the sides are the legs and the diagonal is the hypotenuse.

We can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

According to the Pythagorean theorem:

(Length)$^2$ + (Width)$^2$ = (Diagonal)$^2$

$l^2 + w^2 = d^2$

$20^2 + 15^2 = d^2$

Now, calculate the squares and solve for $d$:

$20^2 = 20 \times 20 = 400$

$15^2 = 15 \times 15 = 225$

So the equation becomes:

$400 + 225 = d^2$

$625 = d^2$

To find $d$, take the positive square root of both sides (since length must be positive):

$d = \sqrt{625}$

We can find the square root of 625:

We know $20^2 = 400$ and $30^2 = 900$. The number is between 20 and 30.

The unit digit of 625 is 5, so the unit digit of its square root must be 5.

Let's try 25.

$25^2 = 25 \times 25 = 625$.

So, $\sqrt{625} = 25$.

$d = 25$

The length of the diagonal of the rectangle is 25 meters.

The final answer is 25 m.

Given:

Area of the rectangular field = 2450 m$^2$

Length of the field is twice its breadth.

To Find:

The perimeter of the rectangular field.

Solution:

Let the breadth of the rectangular field be $b$ meters.

According to the given information, the length $l$ is twice the breadth.

So, $l = 2b$ meters.

The area of a rectangle is given by the formula:

Area = Length $\times$ Breadth

Given that the area is 2450 m$^2$, we have:

$l \times b = 2450$

Substitute the expression for length ($l = 2b$) into the area equation:

$(2b) \times b = 2450$

$2b^2 = 2450$

To find $b^2$, divide both sides of the equation by 2:

$b^2 = \frac{2450}{2}$

$b^2 = 1225$

To find the breadth $b$, take the positive square root of both sides (since breadth must be a positive value):

$b = \sqrt{1225}$

We can find the square root of 1225. We know that $30^2 = 900$ and $40^2 = 1600$. The unit digit of 1225 is 5, so the unit digit of its square root must be 5. Let's try 35:

$35^2 = 35 \times 35 = 1225$

So, $b = 35$ meters.

Now, find the length $l$ using the relationship $l = 2b$:

$l = 2 \times 35$

$l = 70$ meters.

The perimeter of a rectangle is given by the formula:

Perimeter = $2 \times (\text{Length} + \text{Breadth})$

Perimeter = $2 \times (l + b)$

Substitute the values of $l$ and $b$:

Perimeter = $2 \times (70 + 35)$

Perimeter = $2 \times (105)$

Perimeter = $210$ meters.

The perimeter of the field is 210 meters.

The final answer is 210 m.

Given:

Total number of students = 6250

Number of children left out = 9

The remaining students are arranged in rows such that the number of students in each row is equal to the number of rows.

To Find:

The number of children in each row of the square arrangement.

Solution:

The total number of students is 6250.

When the students were arranged, 9 children were left out.

The number of students who were arranged in the square formation is the total number of students minus the number of students left out.

Number of students arranged = $6250 - 9 = 6241$.

These 6241 students were arranged in rows such that the number of students in each row is equal to the number of rows.

Let $x$ be the number of rows.

Let $y$ be the number of students in each row.

According to the problem, $x = y$.

The total number of students arranged is the product of the number of rows and the number of students in each row.

Number of students arranged = Number of rows $\times$ Number of students in each row

$6241 = x \times y$

Since $x = y$, we have:

$6241 = x \times x$

$6241 = x^2$

To find the number of students in each row (which is $x$), we need to find the square root of 6241.

$x = \sqrt{6241}$

Let's find the square root of 6241 using the long division method.

Pair the digits: $\overline{62}\overline{41}$.

The first period is 62.

The largest square less than or equal to 62 is $7^2 = 49$.

Subtract 49 from 62: $62 - 49 = 13$.

Bring down the next period (41). The new dividend is 1341.

Double the quotient (7) to get 14. Place a blank digit next to it: 14_.

We need to find a digit 'x' such that $14x \times x \leq 1341$.

The unit digit of 6241 is 1. The unit digit of the square root must be either 1 (since $1^2=1$) or 9 (since $9^2=81$, which ends in 1).

Let's try 141 $\times$ 1 = 141 (too small).

Let's try 149 $\times$ 9.

$\begin{array}{cc}& & 1 & 4 & 9 \\ \times & & & & 9 \\ \hline & 1 & 3 & 4 & 1 \\ \hline \end{array}$

$149 \times 9 = 1341$. This matches the new dividend.

Write 9 in the quotient.

Subtract 1341 from 1341: $1341 - 1341 = 0$.

The remainder is 0.

The long division process is shown below:

$\begin{array}{r} 79\phantom{)} \\ 7{\overline{\smash{\big)}\,6241\phantom{)}}} \\ \underline{-~\phantom{(}49\phantom{41)}} \\ 149{\overline{\smash{\big)}\,1341\phantom{)}}}\\ \underline{-~\phantom{()}(1341)}\\ 0\phantom{)} \end{array}$

The square root of 6241 is 79.

So, $x = 79$.

The number of rows is 79, and the number of students in each row is 79.

The final answer is 79.

Given:

The number is 1500.

To Find:

The least number that must be added to 1500 to make it a perfect square, and the square root of that perfect square.

Solution:

To find the least number that must be added to 1500 to make it a perfect square, we should find the smallest perfect square that is greater than 1500.

We can do this by finding the square root of 1500 using the long division method.

Applying the long division method to 1500:

Pair the digits: $\overline{15}\overline{00}$.

The first period is 15.

The largest square less than or equal to 15 is $3^2 = 9$.

Subtract 9 from 15: $15 - 9 = 6$.

Bring down the next period (00). The new dividend is 600.

Double the quotient (3) to get 6. Place a blank digit next to it: 6_.

We need to find a digit 'x' such that $6x \times x \leq 600$.

Try some digits: $68 \times 8 = 544$. $69 \times 9 = 621$.

The largest product less than or equal to 600 is 544, using the digit 8.

Write 8 in the quotient.

Subtract 544 from 600: $600 - 544 = 56$.

The long division process is shown below:

$\begin{array}{r} 38\phantom{)} \\ 3{\overline{\smash{\big)}\,1500\phantom{)}}} \\ \underline{-~\phantom{(}9\phantom{00)}} \\ 68{\overline{\smash{\big)}\,600\phantom{)}}}\\ \underline{-~\phantom{()}(544)}\\ 56\phantom{)} \end{array}$

The square root of 1500 is approximately 38 with a remainder of 56.

This means that $38^2 < 1500$. We found that $38^2 = 1444$.

The next consecutive integer after 38 is 39.

The square of 39 will be the smallest perfect square greater than 1500.

Let's calculate $39^2$:

$39^2 = 39 \times 39$

$\begin{array}{cc}& & 3 & 9 \\ \times & & 3 & 9 \\ \hline && 3 & 5 & 1 \\ & 1 & 1 & 7 & \times \\ \hline 1 & 5 & 2 & 1 \\ \hline \end{array}$

$39^2 = 1521$.

The smallest perfect square greater than 1500 is 1521.

To find the least number that must be added to 1500 to get 1521, we subtract 1500 from 1521.

Least number to be added = $1521 - 1500 = 21$.

The perfect square obtained by adding 21 to 1500 is $1500 + 21 = 1521$.

The square root of this perfect square is $\sqrt{1521}$.

From our calculation, $\sqrt{1521} = 39$.

The least number to be added is 21.

The perfect square is 1521.

The square root of the perfect square is 39.

The final answer is Least number to be added: 21, Square root of the perfect square: 39.

Concept: A number is a perfect square if and only if in its prime factorization, the exponent of each prime factor is an even number. To find the smallest number by which a given number must be divided to get a perfect square, we find the prime factorization of the number and identify the prime factors that have odd exponents. The smallest number to divide by is the product of these prime factors raised to the power of their odd exponents.

We need to find the smallest number by which 1620 must be divided to get a perfect square.

First, find the prime factorization of 1620.

$\begin{array}{c|cc} 2 & 1620 \\ \hline 2 & 810 \\ \hline 3 & 405 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, the prime factorization of 1620 is $2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 = 2^2 \times 3^4 \times 5^1$.

Now, examine the exponents of each prime factor in the factorization:

• The exponent of the prime factor 2 is 2 (which is even).
• The exponent of the prime factor 3 is 4 (which is even).
• The exponent of the prime factor 5 is 1 (which is odd).

For a number to be a perfect square, the exponent of each prime factor must be an even number.

In the prime factorization of 1620, the prime factor 5 has an odd exponent (1).

To make the resulting number a perfect square by division, we need to divide by the prime factors with odd exponents, raised to the power of their odd exponents. This makes their new exponents even (specifically, zero in this case, which is an even number).

The prime factor with an odd exponent is 5, and its exponent is 1.

The smallest number to divide by is $5^1 = 5$.

Let's verify by dividing 1620 by 5:

$\frac{1620}{5} = 324$.

The prime factorization of 324 is obtained by dividing the prime factorization of 1620 by 5:

$1620 \div 5 = (2^2 \times 3^4 \times 5^1) \div 5^1 = 2^2 \times 3^4 \times 5^{(1-1)} = 2^2 \times 3^4 \times 5^0 = 2^2 \times 3^4$.

In the prime factorization of 324 ($2^2 \times 3^4$), the exponents of both prime factors (2 and 4) are even. Therefore, 324 is a perfect square ($\sqrt{324} = \sqrt{2^2 \times 3^4} = 2^{2/2} \times 3^{4/2} = 2^1 \times 3^2 = 2 \times 9 = 18$).

The smallest number by which 1620 must be divided to get a perfect square is 5.

The final answer is 5.

Solution:

To find which number's square is 196, we can square each of the given options:

(a) $11^2 = 11 \times 11 = 121$

(b) $12^2 = 12 \times 12 = 144$

(c) $14^2 = 14 \times 14 = 196$

(d) $16^2 = 16 \times 16 = 256$

Comparing the results with 196, we see that $14^2 = 196$.

Therefore, 196 is the square of 14.

The correct option is (c) 14.

Solution:

We know that the square of an even number is always an even number, and the square of an odd number is always an odd number.

Let's examine the given options:

(a) 144 is an even number.

(b) 169 is an odd number.

(c) 441 is an odd number.

(d) 625 is an odd number.

Since the square of an even number must be even, only option (a) 144 can be the square of an even number among the given options.

Let's verify if 144 is a perfect square and if its square root is even.

We find the square root of 144:

$\sqrt{144} = 12$

The number 12 is an even number.

Therefore, 144 is the square of the even number 12.

The correct option is (a) 144.

Solution:

To find the units place of the square of a number, we only need to consider the units place of the number itself.

If a number ends in 9, its units digit is 9.

We find the square of the units digit:

$9^2 = 81$

The units digit of $9^2$ is 1.

Therefore, the units place of the square of any number ending in 9 will be 1.

For example:

$9^2 = 81$ (units digit is 1)

$19^2 = 361$ (units digit is 1)

$29^2 = 841$ (units digit is 1)

The correct option is (c) 1.

Solution:

To find the units place of the square of a number, we only need to consider the units place of the number itself.

Let's check the units digit of the square for each option:

(a) $14^2$: The units digit of 14 is 4. The units digit of $4^2 = 16$ is 6.

(b) $62^2$: The units digit of 62 is 2. The units digit of $2^2 = 4$ is 4.

(c) $27^2$: The units digit of 27 is 7. The units digit of $7^2 = 49$ is 9.

(d) $35^2$: The units digit of 35 is 5. The units digit of $5^2 = 25$ is 5.

We are looking for the number whose square has 4 at the units place. From the above calculations, only $62^2$ has 4 at the units place.

The correct option is (b) 62².

Solution:

We are asked to find the number of natural numbers that lie between $5^2$ and $6^2$.

First, let's calculate the squares of the two numbers:

$5^2 = 5 \times 5 = 25$

$6^2 = 6 \times 6 = 36$

We need to find the number of natural numbers that are greater than 25 and less than 36.

These numbers are 26, 27, 28, 29, 30, 31, 32, 33, 34, and 35.

Counting these numbers, we find there are 10 natural numbers between 25 and 36.

Alternatively:

There is a general formula for the number of non-perfect square natural numbers between the squares of two consecutive natural numbers, $n^2$ and $(n+1)^2$. The number of such numbers is $2n$.

In this question, the consecutive natural numbers are 5 and 6. So, $n=5$.

The number of natural numbers between $5^2$ and $6^2$ is $2 \times 5 = 10$.

The correct option is (b) 10.

Solution:

A perfect square is a number obtained by squaring an integer. The units digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. A number whose units digit is 2, 3, 7, or 8 cannot be a perfect square.

Let's look at the units digit of each option:

(a) 841: The units digit is 1. A number ending in 1 can be a perfect square (e.g., $1^2=1$, $9^2=81$). In fact, $29^2 = 841$, so 841 is a perfect square.

(b) 529: The units digit is 9. A number ending in 9 can be a perfect square (e.g., $3^2=9$, $7^2=49$). In fact, $23^2 = 529$, so 529 is a perfect square.

(c) 198: The units digit is 8. A number ending in 8 cannot be a perfect square.

(d) All of the above: Since (a) and (b) are perfect squares, this option is incorrect.

Based on the units digit rule, 198 cannot be a perfect square because its units digit is 8.

The correct option is (c) 198.

Solution:

To find the one's digit (units digit) of the cube of a number, we only need to find the one's digit of the cube of the one's digit of the original number.

The given number is 23.

The one's digit of 23 is 3.

Now, we cube the one's digit:

$3^3 = 3 \times 3 \times 3 = 9 \times 3 = 27$

The one's digit of 27 is 7.

Therefore, the one's digit of the cube of 23 is 7.

The correct option is (b) 7.

Given:

Area of the square board = 144 square units.

To Find:

The length of each side of the board.

Solution:

The area of a square is calculated by squaring the length of its side.

Area = $(\text{side})^2$

We are given that the area is 144 square units. So,

$(\text{side})^2 = 144$

To find the length of the side, we need to calculate the square root of the area.

$\text{side} = \sqrt{144}$

We know that $12 \times 12 = 144$.

Therefore, $\sqrt{144} = 12$.

The length of each side of the board is 12 units.

The correct option is (b) 12 units.

Solution:

We need to find the location of $\sqrt{25}$ on the given number line.

First, calculate the value of $\sqrt{25}$.

$\sqrt{25} = 5$

Now, we examine the number line provided in the image. The points A, B, C, and D are marked on the number line at specific integer positions.

Point A is located at 3.

Point B is located at 4.

Point C is located at 5.

Point D is located at 6.

Since $\sqrt{25} = 5$, the location of $\sqrt{25}$ on the number line is at the point marked C.

The correct option is (c) C.

Solution:

A Pythagorean triplet consists of three positive integers $a$, $b$, and $c$, such that $a^2 + b^2 = c^2$.

For any natural number $m > 1$, the triplet $(2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet because:

$(2m)^2 + (m^2 - 1)^2 = 4m^2 + (m^4 - 2m^2 + 1) = m^4 + 2m^2 + 1 = (m^2 + 1)^2$

The members of this triplet are $2m$, $m^2 - 1$, and $m^2 + 1$.

If one member of a Pythagorean triplet is given as $2m$, then the other two members are $m^2 - 1$ and $m^2 + 1$. Note that the order of the last two members can be switched, as seen in option (b).

Comparing this with the given options, option (b) lists $m^2 + 1$ and $m^2 - 1$ as the other two members.

The correct option is (b) m² + 1 , m² – 1.

Solution:

We need to find the sum of the given successive odd numbers: 1, 3, 5, 7, 9, 11, 13, and 15.

We can find the sum by adding the numbers directly:

$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15$

$= 4 + 5 + 7 + 9 + 11 + 13 + 15$

$= 9 + 7 + 9 + 11 + 13 + 15$

$= 16 + 9 + 11 + 13 + 15$

$= 25 + 11 + 13 + 15$

$= 36 + 13 + 15$

$= 49 + 15$

$= 64$

Alternatively:

The sum of the first $n$ successive odd natural numbers is equal to $n^2$.

Let's count the number of terms in the given sum: 1, 3, 5, 7, 9, 11, 13, 15.

There are 8 terms in the sequence. So, $n=8$.

The sum is equal to $n^2 = 8^2$.

$8^2 = 8 \times 8 = 64$

The sum of the successive odd numbers 1, 3, 5, 7, 9, 11, 13, and 15 is 64.

The correct option is (b) 64.

Solution:

We are asked to find the sum of the first n odd natural numbers.

Let's look at the sum of the first few odd natural numbers:

Sum of the first 1 odd number = $1 = 1^2$

Sum of the first 2 odd numbers = $1 + 3 = 4 = 2^2$

Sum of the first 3 odd numbers = $1 + 3 + 5 = 9 = 3^2$

Sum of the first 4 odd numbers = $1 + 3 + 5 + 7 = 16 = 4^2$

From this pattern, we can see that the sum of the first $n$ odd natural numbers is equal to $n^2$.

The formula for the sum of the first $n$ odd natural numbers is $\sum\limits_{i=1}^{n} (2i-1) = n^2$.

The correct option is (b) $n^2$.

Solution:

A perfect cube is a number that can be obtained by multiplying an integer by itself three times (cubing it). To check if a number is a perfect cube, we can use prime factorization. If the prime factors of a number can be grouped into triplets, then the number is a perfect cube.

Let's find the prime factorization of each option:

(a) 243:

$\begin{array}{c|cc} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$. The prime factor 3 appears 5 times, which is not a multiple of 3. Thus, 243 is not a perfect cube.

(b) 216:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3$. The prime factors 2 and 3 appear in groups of three. Thus, 216 is a perfect cube.

(c) 392:

$\begin{array}{c|cc} 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$392 = 2 \times 2 \times 2 \times 7 \times 7 = 2^3 \times 7^2$. The prime factor 7 appears 2 times, which is not a multiple of 3. Thus, 392 is not a perfect cube.

(d) 8640:

$\begin{array}{c|cc} 2 & 8640 \\ \hline 2 & 4320 \\ \hline 2 & 2160 \\ \hline 2 & 1080 \\ \hline 2 & 540 \\ \hline 2 & 270 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$8640 = 2^6 \times 3^3 \times 5^1$. The prime factor 5 appears once, which is not a multiple of 3. Thus, 8640 is not a perfect cube.

From the prime factorizations, only 216 has all prime factors appearing in groups of three.

The correct option is (b) 216.

Given:

Length of one leg of the right triangle = $3x$

Length of the other leg of the right triangle = $4x$

To Find:

The length of the hypotenuse.

Solution:

In a right-angled triangle, the lengths of the sides are related by the Pythagorean theorem, which states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides (the legs).

Let the lengths of the legs be $a$ and $b$, and the length of the hypotenuse be $c$.

According to the Pythagorean theorem:

$a^2 + b^2 = c^2$

Given the lengths of the legs are $3x$ and $4x$, we can substitute these values into the formula:

$(3x)^2 + (4x)^2 = c^2$

Calculate the squares of the terms:

$(3x)^2 = 3^2 \times x^2 = 9x^2$

$(4x)^2 = 4^2 \times x^2 = 16x^2$

Substitute these back into the equation:

$9x^2 + 16x^2 = c^2$

Combine the terms on the left side:

$(9+16)x^2 = c^2$

$25x^2 = c^2$

To find the length of the hypotenuse $c$, take the square root of both sides of the equation:

$c = \sqrt{25x^2}$

$c = \sqrt{25} \times \sqrt{x^2}$

Since the length of a side must be positive, we consider the positive square root. $\sqrt{25} = 5$ and $\sqrt{x^2} = x$ (assuming $x$ represents a positive length).

$c = 5 \times x$

$c = 5x$

Thus, the length of the hypotenuse is $5x$.

The correct option is (a) 5x.

Solution:

Let's examine the given number pattern: 1, 4, 9, 16, 25, ...

We can observe the relationship between each number and its position in the sequence:

The first term is 1, which is $1^2$.

The second term is 4, which is $2^2$.

The third term is 9, which is $3^2$.

The fourth term is 16, which is $4^2$.

The fifth term is 25, which is $5^2$.

The pattern shows that each number is the square of the corresponding natural number representing its position in the sequence.

Therefore, the next two numbers will be the square of the 6th and 7th natural numbers.

The 6th number is $6^2 = 6 \times 6 = 36$.

The 7th number is $7^2 = 7 \times 7 = 49$.

So, the next two numbers in the pattern are 36 and 49.

The correct option is (b) 36, 49.

Solution:

The units digit of the square of a number is determined by the units digit of the number itself.

We need to find which of the given squares ends with the digit 1.

Let's examine the units digit of each number and its square:

(a) For $43^2$, the units digit of 43 is 3. The units digit of $3^2 = 9$ is 9.

(b) For $67^2$, the units digit of 67 is 7. The units digit of $7^2 = 49$ is 9.

(c) For $52^2$, the units digit of 52 is 2. The units digit of $2^2 = 4$ is 4.

(d) For $59^2$, the units digit of 59 is 9. The units digit of $9^2 = 81$ is 1.

The square of a number ends with the digit 1 if the number itself ends with 1 or 9.

Among the given numbers, only 59 ends with 9.

Therefore, $59^2$ will end with the digit 1.

The correct option is (d) 59².

Solution:

The units digit of a perfect square is determined by the units digit of the number being squared.

Let's look at the units digits of the squares of the digits 0 through 9:

$0^2 = 0$ (units digit 0)

$1^2 = 1$ (units digit 1)

$2^2 = 4$ (units digit 4)

$3^2 = 9$ (units digit 9)

$4^2 = 16$ (units digit 6)

$5^2 = 25$ (units digit 5)

$6^2 = 36$ (units digit 6)

$7^2 = 49$ (units digit 9)

$8^2 = 64$ (units digit 4)

$9^2 = 81$ (units digit 1)

The possible units digits of a perfect square are 0, 1, 4, 5, 6, and 9.

This means that a number ending in 2, 3, 7, or 8 can never be a perfect square.

Let's check the given options:

(a) 1 is a possible units digit of a perfect square.

(b) 8 is not a possible units digit of a perfect square.

(c) 0 is a possible units digit of a perfect square.

(d) 6 is a possible units digit of a perfect square.

Therefore, a perfect square can never have the digit 8 in its ones place.

The correct option is (b) 8.

Solution:

A perfect cube is a number that can be obtained by cubing an integer. To determine if a number is a perfect cube, we can find its prime factorization. A number is a perfect cube if and only if the exponents of all prime factors in its prime factorization are multiples of 3.

Let's examine the prime factorization of each option:

(a) 216:

$216 = 6^3 = (2 \times 3)^3 = 2^3 \times 3^3$. The exponents (3 and 3) are multiples of 3. So, 216 is a perfect cube.

(b) 567:

Let's find the prime factorization of 567:

$\begin{array}{c|cc} 3 & 567 \\ \hline 3 & 189 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$567 = 3 \times 3 \times 3 \times 3 \times 7 = 3^4 \times 7^1$. The exponents (4 and 1) are not multiples of 3. So, 567 is not a perfect cube.

(c) 125:

$125 = 5^3$. The exponent (3) is a multiple of 3. So, 125 is a perfect cube.

(d) 343:

$343 = 7^3$. The exponent (3) is a multiple of 3. So, 343 is a perfect cube.

From the analysis, only 567 is not a perfect cube because the exponents in its prime factorization are not multiples of 3.

The correct option is (b) 567.

Solution:

We are asked to find the value of $\sqrt[3]{1000}$.

The cube root of a number is the value that, when multiplied by itself three times, gives the original number.

We need to find a number $x$ such that $x \times x \times x = 1000$, or $x^3 = 1000$.

Let's check the options:

(a) $10^3 = 10 \times 10 \times 10 = 100 \times 10 = 1000$.

(b) $100^3 = 100 \times 100 \times 100 = 10000 \times 100 = 1000000$.

(c) $1^3 = 1 \times 1 \times 1 = 1$.

Since $10^3 = 1000$, the cube root of 1000 is 10.

So, $\sqrt[3]{1000} = 10$.

The correct option is (a) 10.

Solution:

We are given that $m$ is the square of a natural number $n$.

This can be written mathematically as:

$m = n^2$

We need to express $n$ in terms of $m$. To do this, we can take the square root of both sides of the equation:

$\sqrt{m} = \sqrt{n^2}$

Since $n$ is a natural number, $n$ is positive. The square root of $n^2$ for a positive number $n$ is $n$.

So, $\sqrt{n^2} = n$.

Substituting this back into the equation, we get:

$n = \sqrt{m}$

Thus, if $m$ is the square of a natural number $n$, then $n$ is the square root of $m$.

The correct option is (d) $\sqrt{m}$.

Solution:

Let a perfect square number have $N$ digits.

If $N$ is even, let $N = n$. We are given that $n$ is even.

The number of digits in the square root of a perfect square number with $n$ digits (where $n$ is even) is $\frac{n}{2}$.

Let's verify this with some examples:

Consider a 2-digit perfect square (n=2): 16, 25, 36, ..., 81, 100 (Note: 100 is 3 digits, so 2-digit perfect squares are up to 81).

$\sqrt{16} = 4$ (1 digit)

$\sqrt{25} = 5$ (1 digit)

$\sqrt{81} = 9$ (1 digit)

Here, $n=2$. The number of digits in the square root is 1. Using the formula $\frac{n}{2} = \frac{2}{2} = 1$. This matches.

Consider a 4-digit perfect square (n=4): 1024, 1225, ..., 9801.

$\sqrt{1024} = 32$ (2 digits)

$\sqrt{1225} = 35$ (2 digits)

$\sqrt{9801} = 99$ (2 digits)

Here, $n=4$. The number of digits in the square root is 2. Using the formula $\frac{n}{2} = \frac{4}{2} = 2$. This matches.

The rule is that if a perfect square has $n$ digits and $n$ is even, its square root has $\frac{n}{2}$ digits.

The correct option is (b) $\frac{n}{2}$ digit.

Solution:

We are given that $m$ is the cube root of $n$.

This relationship can be written mathematically as:

$m = \sqrt[3]{n}$

To find $n$ in terms of $m$, we need to remove the cube root from $n$. We can do this by cubing both sides of the equation.

$(m)^3 = (\sqrt[3]{n})^3$

When we cube a cube root, the operations cancel each other out:

$m^3 = n$

So, $n$ is equal to $m^3$.

The correct option is (a) m³.

Solution:

We need to find the value of the expression $\sqrt{248 + \sqrt{52+\sqrt{144}}}$.

We evaluate the expression from the innermost square root outwards.

First, evaluate the innermost square root:

$\sqrt{144}$

Since $12^2 = 144$, we have $\sqrt{144} = 12$.

Substitute this value back into the expression:

$\sqrt{248 + \sqrt{52+12}}$

Now, evaluate the expression inside the next square root:

$52 + 12 = 64$

So, the expression becomes:

$\sqrt{248 + \sqrt{64}}$

Now, evaluate the square root of 64:

$\sqrt{64}$

Since $8^2 = 64$, we have $\sqrt{64} = 8$.

Substitute this value back into the outermost square root:

$\sqrt{248 + 8}$

Finally, evaluate the expression inside the outermost square root:

$248 + 8 = 256$

The expression simplifies to:

$\sqrt{256}$

We need to find the square root of 256. We know that $16^2 = 16 \times 16 = 256$.

Therefore, $\sqrt{256} = 16$.

The value of the expression is 16.

The correct option is (c) 16.

Given:

$\sqrt{4096} = 64$

To Find:

The value of $\sqrt{4096}$ + $\sqrt{40.96}$

Solution:

We are given the value of $\sqrt{4096} = 64$.

Now, we need to find the value of $\sqrt{40.96}$.

We can write 40.96 as a fraction:

$40.96 = \frac{4096}{100}$

Using the property of square roots, $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, we have:

$\sqrt{40.96} = \sqrt{\frac{4096}{100}} = \frac{\sqrt{4096}}{\sqrt{100}}$

We know that $\sqrt{4096} = 64$ (given) and $\sqrt{100} = 10$ (since $10 \times 10 = 100$).

Substitute these values:

$\sqrt{40.96} = \frac{64}{10}$

$\sqrt{40.96} = 6.4$

Now, we need to calculate the sum $\sqrt{4096}$ + $\sqrt{40.96}$.

$\sqrt{4096}$ + $\sqrt{40.96}$ = $64 + 6.4$

Perform the addition:

$64.0 + 6.4 = 70.4$

Thus, the value of $\sqrt{4096}$ + $\sqrt{40.96}$ is 70.4.

The correct option is (d) 70.4.

Solution:

We are looking for the number of perfect squares that lie between 1 and 100. This means the perfect squares must be greater than 1 and less than 100.

Let's list the perfect squares of natural numbers starting from 1:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

$9^2 = 81$

$10^2 = 100$

The perfect squares that are strictly between 1 and 100 are the numbers that are greater than 1 and less than 100.

These numbers are: 4, 9, 16, 25, 36, 49, 64, and 81.

Let's count these numbers:

4 is the 1st number.

9 is the 2nd number.

16 is the 3rd number.

25 is the 4th number.

36 is the 5th number.

49 is the 6th number.

64 is the 7th number.

81 is the 8th number.

There are 8 perfect squares between 1 and 100.

There are 8 perfect squares between 1 and 100.

Solution:

We need to find the number of perfect cubes that lie between 1 and 1000 (strictly greater than 1 and strictly less than 1000).

A perfect cube is a number that is the cube of an integer.

Let's list the cubes of natural numbers starting from 1:

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$

$4^3 = 64$

$5^3 = 125$

$6^3 = 216$

$7^3 = 343$

$8^3 = 512$

$9^3 = 729$

$10^3 = 1000$

We are looking for perfect cubes greater than 1 and less than 1000.

From the list, the numbers that satisfy this condition are 8, 27, 64, 125, 216, 343, 512, and 729.

Counting these numbers, we find there are 8 perfect cubes between 1 and 1000.

There are 8 perfect cubes between 1 and 1000.

Solution:

To find the units digit of the square of a number, we only need to consider the units digit of the original number.

The units digit of the number 1294 is 4.

Now, we find the square of this units digit:

$4^2 = 4 \times 4 = 16$

The units digit of 16 is 6.

Therefore, the units digit in the square of 1294 is 6.

The units digit in the square of 1294 is 6.

Solution:

We need to find the number of zeroes in the square of 500.

The number 500 can be written as $5 \times 100$, or $5 \times 10^2$.

The square of 500 is $(500)^2$.

$(500)^2 = (5 \times 100)^2 = 5^2 \times (100)^2$

$5^2 = 25$

$(100)^2 = 100 \times 100 = 10000$

So, $(500)^2 = 25 \times 10000 = 250000$

The number 250000 has 4 zeroes at the end.

Alternatively:

A property of squaring numbers ending in zeroes is that the number of zeroes in the square is twice the number of zeroes in the original number.

The number 500 has 2 zeroes.

So, the square of 500 will have $2 \times 2 = 4$ zeroes.

The square of 500 will have 4 zeroes.

Solution:

We need to find the number of natural numbers that lie strictly between $n^2$ and $(n+1)^2$.

First, let's expand $(n+1)^2$:

$(n+1)^2 = n^2 + 2n + 1$

The natural numbers that are between $n^2$ and $(n+1)^2$ are those greater than $n^2$ and less than $n^2 + 2n + 1$.

These numbers are $n^2 + 1, n^2 + 2, n^2 + 3, \dots, n^2 + 2n$.

To find the count of these numbers, we can subtract the first number from the last number and add 1 (inclusive count).

Number of natural numbers = (Last number) - (First number) + 1

= $(n^2 + 2n) - (n^2 + 1) + 1$

= $n^2 + 2n - n^2 - 1 + 1$

= $2n$

Thus, there are $2n$ natural numbers between $n^2$ and $(n+1)^2$.

There are $2n$ natural numbers between $n^2$ and $(n + 1)^2$.

Solution:

We need to find the number of digits in the square root of 24025.

The number 24025 has 5 digits.

Let $n$ be the number of digits in the perfect square. Here, $n = 5$.

The rule for finding the number of digits in the square root of a perfect square with $n$ digits is:

• If $n$ is even, the number of digits in the square root is $\frac{n}{2}$.
• If $n$ is odd, the number of digits in the square root is $\frac{n+1}{2}$.

In this case, $n=5$, which is an odd number.

So, the number of digits in the square root of 24025 is $\frac{5+1}{2}$.

$\frac{5+1}{2} = \frac{6}{2} = 3$

The square root of 24025 will have 3 digits.

Let's verify this by finding the square root:

The number 24025 ends in 25, so its square root must end in 5.

Consider numbers ending in 5 whose squares might be close to 24025. $100^2 = 10000$, $200^2 = 40000$. The square root must be between 100 and 200.

Let's try 150. $150^2 = (15 \times 10)^2 = 15^2 \times 10^2 = 225 \times 100 = 22500$. This is close to 24025.

Let's try 155. $155^2 = (150+5)^2 = 150^2 + 2 \times 150 \times 5 + 5^2 = 22500 + 1500 + 25 = 24025$.

So, $\sqrt{24025} = 155$. The number 155 has 3 digits, which confirms our rule.

The square root of 24025 will have 3 digits.

Solution:

We need to find the square of 5.5.

The square of a number is the result of multiplying the number by itself.

$(5.5)^2 = 5.5 \times 5.5$

We can perform the multiplication:

$5.5 \times 5.5 = 30.25$

Alternatively, we can think of $5.5 = \frac{11}{2}$.

$(5.5)^2 = \left(\frac{11}{2}\right)^2 = \frac{11^2}{2^2} = \frac{121}{4}$

Converting the fraction to a decimal:

$\frac{121}{4} = 30.25$

The square of 5.5 is 30.25.

The square of 5.5 is 30.25.

Solution:

We are asked to find the value of $\sqrt{5.3 \times 5.3}$.

The expression inside the square root is $5.3 \times 5.3$, which is the same as $(5.3)^2$.

So, we need to find the value of $\sqrt{(5.3)^2}$.

The square root operation is the inverse of squaring (for non-negative numbers).

For any non-negative number $a$, $\sqrt{a^2} = a$.

In this case, $a = 5.3$, which is a positive number.

Therefore, $\sqrt{(5.3)^2} = 5.3$.

The value of $\sqrt{5.3 \times 5.3}$ is 5.3.

The square root of 5.3 × 5.3 is 5.3.

Solution:

We need to find the number of zeroes in the cube of 100.

The number 100 can be written as $10^2$.

The cube of 100 is $(100)^3$.

$(100)^3 = (10^2)^3$

Using the property of exponents $(a^m)^n = a^{m \times n}$, we get:

$(10^2)^3 = 10^{2 \times 3} = 10^6$

$10^6$ is equal to 1 followed by 6 zeroes, which is 1,000,000.

The number 1,000,000 has 6 zeroes.

Alternatively:

A property of cubing numbers ending in zeroes is that the number of zeroes in the cube is three times the number of zeroes in the original number.

The number 100 has 2 zeroes.

So, the cube of 100 will have $3 \times 2 = 6$ zeroes.

The cube of 100 will have 6 zeroes.

Solution:

We are asked to convert 1 square meter ($1\text{ m}^2$) to square centimeters ($\text{cm}^2$).

We know the relationship between meters and centimeters for linear measurement:

So, $1\text{ m} = 100\text{ cm}$.

To convert square units, we square the linear conversion factor:

$1\text{ m}^2 = (1\text{ m}) \times (1\text{ m})$

Substitute the equivalent value in centimeters for each meter:

$1\text{ m}^2 = (100\text{ cm}) \times (100\text{ cm})$

Multiply the numbers and the units:

$1\text{ m}^2 = (100 \times 100) \text{ cm} \times \text{cm}$

$1\text{ m}^2 = 10000 \text{ cm}^2$

Thus, 1 square meter is equal to 10000 square centimeters.

1m² = 10000 cm².

Solution:

We are asked to convert 1 cubic meter ($1\text{ m}^3$) to cubic centimeters ($\text{cm}^3$).

We know the basic linear conversion between meters and centimeters:

So, $1\text{ m} = 100\text{ cm}$.

To convert cubic units, we need to cube the linear conversion factor. This means we multiply the conversion factor by itself three times.

$1\text{ m}^3 = (1\text{ m}) \times (1\text{ m}) \times (1\text{ m})$

Substitute the equivalent value in centimeters for each meter:

$1\text{ m}^3 = (100\text{ cm}) \times (100\text{ cm}) \times (100\text{ cm})$

Now, multiply the numbers and the units:

$1\text{ m}^3 = (100 \times 100 \times 100) \text{ cm} \times \text{cm} \times \text{cm}$

$100 \times 100 \times 100 = 10000 \times 100 = 1,000,000$

So, $1\text{ m}^3 = 1,000,000 \text{ cm}^3$

Thus, 1 cubic meter is equal to 1,000,000 cubic centimeters.

1m³ = 1000000 cm³.

Solution:

To find the one's digit (units digit) of the cube of a number, we only need to consider the one's digit of the original number and find its cube's one's digit.

The given number is 38.

The one's digit of 38 is 8.

Now, we find the cube of the one's digit:

$8^3 = 8 \times 8 \times 8 = 64 \times 8 = 512$

The one's digit of 512 is 2.

Therefore, the one's digit in the cube of 38 is 2.

Ones digit in the cube of 38 is 2.

Solution:

We need to find the square of 0.7.

The square of a number is the result of multiplying the number by itself.

$(0.7)^2 = 0.7 \times 0.7$

To multiply decimals, we can multiply them as if they were whole numbers and then place the decimal point in the product.

$7 \times 7 = 49$

The number 0.7 has one decimal place. When we multiply 0.7 by 0.7, the total number of decimal places in the product will be the sum of the decimal places in the numbers being multiplied ($1 + 1 = 2$).

So, we place the decimal point in 49 such that there are two decimal places.

$0.7 \times 0.7 = 0.49$

The square of 0.7 is 0.49.

The square of 0.7 is 0.49.

Solution:

We need to find the sum of the first six odd natural numbers.

The first six odd natural numbers are 1, 3, 5, 7, 9, and 11.

We can add these numbers:

$1 + 3 + 5 + 7 + 9 + 11$

$= 4 + 5 + 7 + 9 + 11$

$= 9 + 7 + 9 + 11$

$= 16 + 9 + 11$

$= 25 + 11$

$= 36$

Alternatively:

The sum of the first $n$ odd natural numbers is given by the formula $n^2$.

Here, we are asked for the sum of the first six odd natural numbers, so $n=6$.

Sum $= 6^2 = 6 \times 6 = 36$

The sum of the first six odd natural numbers is 36.

The sum of first six odd natural numbers is 36.

Solution:

To find the digit at the ones place (units digit) of the square of a number, we only need to consider the units digit of the original number.

The units digit of the number 57 is 7.

Now, we find the square of this units digit:

$7^2 = 7 \times 7 = 49$

The units digit of 49 is 9.

Therefore, the digit at the ones place of $57^2$ is 9.

The digit at the ones place of 57² is 9.

Given:

The hypotenuse of a right triangle is 17 cm.

To Find:

The lengths of the other two sides (legs) of the right triangle.

Solution:

In a right-angled triangle, the lengths of the sides are related by the Pythagorean theorem. If the lengths of the two legs are $a$ and $b$, and the length of the hypotenuse is $c$, then $a^2 + b^2 = c^2$.

We are given that the hypotenuse $c = 17$ cm.

So, we need to find two numbers $a$ and $b$ such that $a^2 + b^2 = 17^2$.

$17^2 = 17 \times 17 = 289$

We need to find two perfect squares that add up to 289. Let's list some perfect squares:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

$9^2 = 81$

$10^2 = 100$

$11^2 = 121$

$12^2 = 144$

$13^2 = 169$

$14^2 = 196$

$15^2 = 225$

$16^2 = 256$

We are looking for two numbers $a$ and $b$ from the list such that $a^2 + b^2 = 289$. Let's try combining squares:

If we take $a^2 = 64$ (which means $a = 8$), then $b^2 = 289 - 64 = 225$. We see from the list that $225 = 15^2$, so $b = 15$.

Thus, $8^2 + 15^2 = 64 + 225 = 289 = 17^2$.

The lengths of the legs are 8 cm and 15 cm.

Alternate Solution:

We know that for any natural number $m > 1$, the triplet $(2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

We are given that the hypotenuse is 17 cm. In this triplet, the hypotenuse is $m^2 + 1$.

Subtract 1 from both sides:

Take the square root of both sides (since $m$ is a natural number):

Now, substitute $m=4$ into the formulas for the other two members of the triplet (the legs):

Leg 1 = $2m = 2 \times 4 = 8$

Leg 2 = $m^2 - 1 = 4^2 - 1 = 16 - 1 = 15$

The sides (legs) of the right triangle are 8 cm and 15 cm.

The sides of a right triangle whose hypotenuse is 17 cm are 8 cm and 15 cm.

Solution:

We need to find the value of $\sqrt{1.96}$.

We can convert the decimal number 1.96 into a fraction:

$1.96 = \frac{196}{100}$

Now, we need to find the square root of this fraction:

$\sqrt{1.96} = \sqrt{\frac{196}{100}}$

Using the property of square roots, $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, we can write:

$\sqrt{\frac{196}{100}} = \frac{\sqrt{196}}{\sqrt{100}}$

We know that $14^2 = 196$, so $\sqrt{196} = 14$.

We also know that $10^2 = 100$, so $\sqrt{100} = 10$.

Substitute these values into the expression:

$\frac{\sqrt{196}}{\sqrt{100}} = \frac{14}{10}$

Convert the fraction back to a decimal:

$\frac{14}{10} = 1.4$

So, $\sqrt{1.96} = 1.4$.

$\sqrt{1.96}$ = 1.4.

Solution:

We need to find the value of $(1.2)^3$.

Cubing a number means multiplying the number by itself three times.

$(1.2)^3 = 1.2 \times 1.2 \times 1.2$

First, let's calculate $1.2 \times 1.2$:

Multiply 12 by 12: $12 \times 12 = 144$.

Since there is one decimal place in 1.2, the product $1.2 \times 1.2$ will have $1 + 1 = 2$ decimal places.

$1.2 \times 1.2 = 1.44$

Now, we multiply this result by 1.2 again:

$1.44 \times 1.2$

Multiply 144 by 12:

$144 \times 12 = 1728$.

The number 1.44 has 2 decimal places, and 1.2 has 1 decimal place. The product $1.44 \times 1.2$ will have $2 + 1 = 3$ decimal places.

So, $1.44 \times 1.2 = 1.728$.

Therefore, $(1.2)^3 = 1.728$.

(1.2)³ = 1.728.

Solution:

We need to determine whether the cube of an odd number is odd or even.

Recall the properties of multiplying odd and even numbers:

Odd $\times$ Odd = Odd

Odd $\times$ Even = Even

Even $\times$ Even = Even

The cube of a number is the number multiplied by itself three times. If the number is odd, let's call it $O$.

$O^3 = O \times O \times O$

First, $O \times O = \text{Odd} \times \text{Odd} = \text{Odd}$.

Then, the cube is $(\text{Odd} \times \text{Odd}) \times O = \text{Odd} \times \text{Odd}$.

Odd $\times$ Odd = Odd.

Let's check with some examples of odd numbers:

$1^3 = 1$ (Odd)

$3^3 = 27$ (Odd)

$5^3 = 125$ (Odd)

$7^3 = 343$ (Odd)

In all cases, the cube of an odd number is an odd number.

The cube of an odd number is always an odd number.

Solution:

The cube root of a number $x$ is the value $y$ such that $y^3 = x$.

The standard mathematical notation for the cube root of $x$ is using the radical symbol ($\sqrt{\phantom{x}}$) with a small 3 (called the index) placed above and to the left of the symbol.

So, the cube root of a number $x$ is denoted by $\sqrt[3]{x}$.

The cube root of a number x is denoted by $\sqrt[3]{x}$.

Solution:

We need to find the least number by which 125 must be multiplied so that the resulting product is a perfect square.

First, find the prime factorization of 125.

$\begin{array}{c|cc} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $125 = 5 \times 5 \times 5 = 5^3$.

For a number to be a perfect square, the exponents of all its prime factors in its prime factorization must be even.

In the prime factorization of 125 ($5^3$), the exponent of the prime factor 5 is 3, which is an odd number.

To make the exponent of 5 even, we need to multiply $5^3$ by $5^1$ (or just 5), because $5^3 \times 5^1 = 5^{3+1} = 5^4$. The exponent 4 is even.

The least number we need to multiply 125 by is the factor(s) required to make all exponents in the prime factorization even. In this case, we need one more factor of 5.

The least number is 5.

Multiplying 125 by 5:

$125 \times 5 = 625$

Now, let's check the prime factorization of 625:

$625 = 5 \times 125 = 5 \times 5 \times 25 = 5 \times 5 \times 5 \times 5 = 5^4$.

The exponent of the prime factor 5 is 4, which is even. So, 625 is a perfect square ($625 = 25^2$).

The least number by which 125 must be multiplied to make it a perfect square is 5.

The least number by which 125 be multiplied to make it a perfect square is 5.

Solution:

We need to find the least number by which 72 must be multiplied so that the resulting product is a perfect cube.

First, find the prime factorization of 72.

$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 72 is $72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$.

For a number to be a perfect cube, the exponents of all its prime factors in its prime factorization must be multiples of 3.

In the prime factorization of 72 ($2^3 \times 3^2$), the exponent of the prime factor 2 is 3, which is a multiple of 3. The exponent of the prime factor 3 is 2, which is not a multiple of 3.

To make the exponent of 3 a multiple of 3, we need to multiply $3^2$ by $3^1$ (or just 3), because $3^2 \times 3^1 = 3^{2+1} = 3^3$.

The least number we need to multiply 72 by is the factor(s) required to make all exponents in the prime factorization multiples of 3. In this case, we need one more factor of 3.

The least number is 3.

Multiplying 72 by 3:

$72 \times 3 = 216$

The prime factorization of 216 is $2^3 \times 3^2 \times 3^1 = 2^3 \times 3^3$. Both exponents are multiples of 3, so 216 is a perfect cube ($216 = 6^3$).

The least number by which 72 must be multiplied to make it a perfect cube is 3.

The least number by which 72 be multiplied to make it a perfect cube is 3.

Solution:

We need to find the least number by which 72 must be divided so that the resulting quotient is a perfect cube.

First, find the prime factorization of 72.

$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization of 72 is $72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$.

For a number to be a perfect cube, the exponents of all its prime factors in its prime factorization must be multiples of 3.

In the prime factorization of 72 ($2^3 \times 3^2$), the exponent of the prime factor 2 is 3, which is a multiple of 3. The exponent of the prime factor 3 is 2, which is not a multiple of 3.

To make the number a perfect cube by division, we need to divide by the prime factors that do not form a complete triplet. The prime factor 3 appears twice ($3^2$). To make its exponent a multiple of 3 (specifically, 0 in this case), we must divide by $3^2$.

The least number to divide by is the product of the prime factors with exponents that are not multiples of 3, raised to those powers. In this case, it is $3^2$.

The least number is $3^2 = 3 \times 3 = 9$.

Let's verify by dividing 72 by 9:

$72 \div 9 = 8$

The prime factorization of 8 is $8 = 2 \times 2 \times 2 = 2^3$. The exponent of 2 is 3, which is a multiple of 3. Thus, 8 is a perfect cube ($8 = 2^3$).

The least number by which 72 must be divided to make it a perfect cube is 9.

The least number by which 72 be divided to make it a perfect cube is 9.

Solution:

To find the units digit of the cube of a number, we only need to find the units digit of the cube of the units digit of the original number.

The units digit of the original number is 7.

We need to find the units digit of $7^3$.

$7^3 = 7 \times 7 \times 7$

First, calculate $7 \times 7 = 49$. The units digit is 9.

Next, multiply the units digit (9) by 7: $9 \times 7 = 63$. The units digit is 3.

Therefore, the cube of a number ending in 7 will end in the digit 3.

Let's verify with an example:

Consider the number 17, which ends in 7.

$17^3 = 17 \times 17 \times 17 = 289 \times 17$.

The units digit of $289 \times 17$ is the units digit of $9 \times 7 = 63$, which is 3.

Indeed, $17^3 = 4913$, which ends in 3.

Cube of a number ending in 7 will end in the digit 3.

To find the units digit of the square of a number, we only need to consider the units digit of the original number.

The units digit of the number 86 is 6.

We find the square of the units digit: $6^2 = 36$.

The units digit of 36 is 6.

Therefore, the units digit of the square of 86 will be 6.

The statement is True.

Let's consider an example.

Take two perfect squares, $1^2 = 1$ and $2^2 = 4$.

Their sum is $1 + 4 = 5$.

The number 5 is not a perfect square because there is no integer whose square is 5.

Another example: Take $2^2 = 4$ and $3^2 = 9$.

Their sum is $4 + 9 = 13$.

The number 13 is not a perfect square.

While there are instances where the sum of two perfect squares is a perfect square (e.g., $3^2 + 4^2 = 9 + 16 = 25 = 5^2$), the statement claims it is true for *any* two perfect squares, which is not correct.

The statement is False.

Let the two perfect squares be $m^2$ and $n^2$, where $m$ and $n$ are integers.

The product of these two perfect squares is $m^2 \times n^2$.

Using the properties of exponents, we know that $m^2 \times n^2 = (m \times n)^2$.

Since $m$ and $n$ are integers, their product $m \times n$ is also an integer.

Therefore, $(m \times n)^2$ is the square of an integer, which means it is a perfect square.

Let's take an example:

Consider the perfect squares $9$ ($3^2$) and $16$ ($4^2$).

Their product is $9 \times 16 = 144$.

The number 144 is a perfect square because $12^2 = 144$. Note that $12 = 3 \times 4$, which aligns with $(m \times n)^2$.

The statement is True.

We need to check the perfect squares of integers around this range.

Let's list some perfect squares:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

$6^2 = 36$

$7^2 = 49$

$8^2 = 64$

The perfect square just before 50 is $7^2 = 49$.

The perfect square just after 60 is $8^2 = 64$.

There are no perfect squares between 49 and 64. Since the range is 50 and 60 (exclusive of 50 and 60), there are no perfect squares in this interval.

The statement is True.

To check if the square root of 1521 is 31, we need to calculate the square of 31.

The square of 31 is $31 \times 31$.

$31^2 = 961$.

Since $31^2 = 961$ and not 1521, the statement is incorrect.

(For completeness, the square root of 1521 is 39, as $39^2 = 1521$).

The statement is False.

Let a number be $N$. Let its prime factorization be $N = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$, where $p_1, p_2, \dots, p_k$ are distinct prime numbers and $a_1, a_2, \dots, a_k$ are positive integers representing the number of times each prime factor appears in $N$.

The cube of the number $N$ is $N^3$.

The prime factorization of $N^3$ is obtained by cubing each factor in the prime factorization of $N$:

$N^3 = (p_1^{a_1} p_2^{a_2} \dots p_k^{a_k})^3$

Using the property of exponents $(x^a)^b = x^{ab}$, we raise the power of each prime factor by the exponent 3:

$N^3 = p_1^{a_1 \times 3} p_2^{a_2 \times 3} \dots p_k^{a_k \times 3} = p_1^{3a_1} p_2^{3a_2} \dots p_k^{3a_k}$.

This means that if a prime factor $p_i$ appears $a_i$ times in the prime factorization of $N$, it appears $3a_i$ times in the prime factorization of $N^3$.

For the statement "Each prime factor appears 3 times in its cube" to be true, the exponent $3a_i$ would have to be equal to 3 for every prime factor $p_i$. This would imply that $a_i = 1$ for every prime factor in the original number $N$.

However, the original number $N$ can have prime factors appearing more than once (i.e., $a_i > 1$).

For example, consider the number $N = 12$.

The prime factorization of 12 is $12 = 2^2 \times 3^1$. Here, the prime factor 2 appears $a_1=2$ times, and the prime factor 3 appears $a_2=1$ time.

The cube of 12 is $12^3 = 1728$.

The prime factorization of 1728 is $1728 = (2^2 \times 3^1)^3 = (2^2)^3 \times (3^1)^3 = 2^{2 \times 3} \times 3^{1 \times 3} = 2^6 \times 3^3$.

In the prime factorization of 1728, the prime factor 2 appears 6 times (which is $3 \times 2$), and the prime factor 3 appears 3 times (which is $3 \times 1$).

Since the prime factor 2 appears 6 times (not 3 times) in the cube of 12, the statement is not universally true for all prime factors of any number's cube.

The statement is False.

To check the statement, we need to calculate the square of 2.8.

The square of 2.8 is $(2.8)^2$.

$(2.8)^2 = 2.8 \times 2.8$.

Multiplying 28 by 28 gives 784. Since there is one decimal place in 2.8, there will be $1 + 1 = 2$ decimal places in the product.

So, $2.8 \times 2.8 = 7.84$.

The calculated value of the square of 2.8 is $7.84$.

The statement claims the square of 2.8 is $78.4$.

Since $7.84 \neq 78.4$, the statement is incorrect.

The statement is False.

To verify the statement, we need to calculate the cube of 0.4.

The cube of 0.4 is $(0.4)^3$.

$(0.4)^3 = 0.4 \times 0.4 \times 0.4$.

First, multiply the numbers without considering the decimal point: $4 \times 4 \times 4 = 64$.

Now, count the total number of decimal places in the factors. Each 0.4 has one decimal place. Since we are multiplying three times, the total number of decimal places in the product will be $1 + 1 + 1 = 3$.

Place the decimal point 3 places from the right in the number 64. This gives 0.064.

So, $(0.4)^3 = 0.064$.

The calculated value matches the value given in the statement.

The statement is True.

To check if the square root of 0.9 is 0.3, we need to calculate the square of 0.3.

The square of 0.3 is $(0.3)^2$.

$(0.3)^2 = 0.3 \times 0.3$.

Multiplying the numbers without considering the decimal point: $3 \times 3 = 9$.

Count the total number of decimal places in the factors. Each 0.3 has one decimal place. So, the product will have $1 + 1 = 2$ decimal places.

Place the decimal point 2 places from the right in the number 9. This gives 0.09.

So, $(0.3)^2 = 0.09$.

The statement claims that the square root of 0.9 is 0.3, which means $(0.3)^2$ should be equal to 0.9.

Since $0.09 \neq 0.9$, the statement is incorrect.

The statement is False.

Let $n$ be a natural number. We need to check if $n^2 > n$ for all natural numbers $n$.

Natural numbers are usually considered to be $1, 2, 3, \dots$.

Let's test the statement with the first few natural numbers:

For $n=1$, the square is $1^2 = 1$. Is $1 > 1$? No, $1$ is equal to $1$.

For $n=2$, the square is $2^2 = 4$. Is $4 > 2$? Yes.

For $n=3$, the square is $3^2 = 9$. Is $9 > 3$? Yes.

For any natural number $n > 1$, multiplying $n$ by itself (which is greater than 1) will result in a number larger than $n$. So, $n^2 > n$ for $n > 1$.

However, for $n=1$, we have $1^2 = 1$, which is not strictly greater than 1.

Since the statement claims the square is *always* greater than the number itself for *every* natural number, the case $n=1$ serves as a counterexample.

The statement is False.

To check if the cube root of 8000 is 200, we need to calculate the cube of 200.

The cube of 200 is $(200)^3$.

$(200)^3 = 200 \times 200 \times 200$.

Multiplying the numerical parts: $2 \times 2 \times 2 = 8$.

Count the total number of zeros: 2 zeros in the first 200, 2 zeros in the second 200, and 2 zeros in the third 200. Total zeros = $2 + 2 + 2 = 6$.

So, $(200)^3 = 8,000,000$.

The statement claims that the cube root of 8000 is 200, which means $(200)^3$ should be equal to 8000.

Since $8,000,000 \neq 8000$, the statement is incorrect.

(For completeness, the cube root of 8000 is 20, as $20^3 = 20 \times 20 \times 20 = 8000$).

The statement is False.

We need to find the perfect cubes of natural numbers and check which ones fall between 1 and 100.

Let's list the first few perfect cubes:

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$

$4^3 = 64$

$5^3 = 125$

We are looking for perfect cubes that are strictly greater than 1 and strictly less than 100.

From the list above:

$1^3 = 1$ is not between 1 and 100.

$2^3 = 8$ is between 1 and 100.

$3^3 = 27$ is between 1 and 100.

$4^3 = 64$ is between 1 and 100.

$5^3 = 125$ is not between 1 and 100.

The perfect cubes between 1 and 100 are 8, 27, and 64.

There are 3 perfect cubes between 1 and 100.

The statement claims there are five perfect cubes between 1 and 100.

The statement is False.

We need to find the number of natural numbers strictly between $100^2$ and $101^2$.

First, let's calculate the values of $100^2$ and $101^2$.

$100^2 = 100 \times 100 = 10000$.

$101^2 = 101 \times 101$.

We can calculate $101^2$ as $(100+1)^2 = 100^2 + 2 \times 100 \times 1 + 1^2 = 10000 + 200 + 1 = 10201$.

So, we are looking for the number of natural numbers between 10000 and 10201.

The natural numbers between 10000 and 10201 are the integers $10001, 10002, \dots, 10200$.

To find the count of these numbers, we can subtract the smaller number from the larger number and then subtract 1 (since we are counting numbers *between* them, excluding the endpoints).

Number of natural numbers = $10201 - 10000 - 1 = 201 - 1 = 200$.

Alternatively, we know that there are $2n$ non-perfect square natural numbers between the squares of two consecutive natural numbers $n^2$ and $(n+1)^2$.

Here, $n = 100$.

The number of natural numbers between $100^2$ and $101^2$ is $2 \times 100 = 200$.

This matches the number given in the statement.

The statement is True.

We need to check if the sum of the first $n$ odd natural numbers is equal to $n^2$.

Let's test the statement for small values of $n$.

For $n=1$, the first odd natural number is 1. The sum is 1.

According to the statement, the sum should be $1^2 = 1$.

So, for $n=1$, the statement holds true.

For $n=2$, the first two odd natural numbers are 1 and 3. The sum is $1 + 3 = 4$.

According to the statement, the sum should be $2^2 = 4$.

So, for $n=2$, the statement holds true.

For $n=3$, the first three odd natural numbers are 1, 3, and 5. The sum is $1 + 3 + 5 = 9$.

According to the statement, the sum should be $3^2 = 9$.

So, for $n=3$, the statement holds true.

This pattern suggests that the sum of the first $n$ odd natural numbers is indeed $n^2$.

The sequence of odd natural numbers is an arithmetic progression with first term $a_1 = 1$ and common difference $d = 2$.

The $n$-th term is $a_n = a_1 + (n-1)d = 1 + (n-1)2 = 1 + 2n - 2 = 2n - 1$.

The sum of the first $n$ terms of an arithmetic progression is given by the formula $S_n = \frac{n}{2}(a_1 + a_n)$.

Substituting $a_1 = 1$ and $a_n = 2n - 1$:

$S_n = \frac{n}{2}(1 + (2n - 1))$

$S_n = \frac{n}{2}(1 + 2n - 1)$

$S_n = \frac{n}{2}(2n)$

$S_n = n \times \frac{2n}{2}$

$S_n = n \times n$

$S_n = n^2$

Thus, the sum of the first $n$ odd natural numbers is $n^2$.

The statement is True.

A perfect square is an integer that is the square of an integer.

To determine if 1000 is a perfect square, we can find its prime factorization.

$1000 = 10 \times 100 = 10 \times 10 \times 10 = (2 \times 5) \times (2 \times 5) \times (2 \times 5)$

$1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$

The prime factorization of 1000 is $2^3 \times 5^3$.

For a number to be a perfect square, the exponents of all prime factors in its prime factorization must be even.

In the prime factorization of 1000, the exponent of 2 is 3 and the exponent of 5 is 3. Both 3 are odd numbers.

Since the exponents are not even, 1000 is not a perfect square.

Alternatively, we can consider the integers whose squares are close to 1000.

$31^2 = 31 \times 31 = 961$.

$32^2 = 32 \times 32 = 1024$.

Since 1000 lies between 961 and 1024, its square root is between 31 and 32. The square root of 1000 is not an integer.

Therefore, 1000 is not a perfect square.

The statement is False.

The units digit of a perfect square is determined by the units digit of the number being squared.

Let's examine the units digits of the squares of the digits from 0 to 9:

$0^2 = 0$ (Units digit is 0)

$1^2 = 1$ (Units digit is 1)

$2^2 = 4$ (Units digit is 4)

$3^2 = 9$ (Units digit is 9)

$4^2 = 16$ (Units digit is 6)

$5^2 = 25$ (Units digit is 5)

$6^2 = 36$ (Units digit is 6)

$7^2 = 49$ (Units digit is 9)

$8^2 = 64$ (Units digit is 4)

$9^2 = 81$ (Units digit is 1)

The possible units digits of a perfect square are 0, 1, 4, 5, 6, and 9.

The digit 8 is not among these possible units digits.

Therefore, a number ending in 8 cannot be a perfect square.

The statement is False.

A Pythagorean triplet consists of three positive integers $a, b,$ and $c$, such that $a^2 + b^2 = c^2$.

The given triplet is $(a, b, c) = (2m - 1, 2m^2 - 2m, 2m^2 - 2m + 1)$.

Let's check if $a^2 + b^2 = c^2$.

$a^2 = (2m - 1)^2 = (2m)^2 - 2(2m)(1) + 1^2 = 4m^2 - 4m + 1$.

$b^2 = (2m^2 - 2m)^2$. We can write $2m^2 - 2m = 2m(m-1)$. So, $b^2 = (2m(m-1))^2 = 4m^2(m-1)^2 = 4m^2(m^2 - 2m + 1) = 4m^4 - 8m^3 + 4m^2$.

$a^2 + b^2 = (4m^2 - 4m + 1) + (4m^4 - 8m^3 + 4m^2) = 4m^4 - 8m^3 + 8m^2 - 4m + 1$.

$c^2 = (2m^2 - 2m + 1)^2$. Let $X = 2m^2 - 2m$. Then $c = X + 1$, and $c^2 = (X+1)^2 = X^2 + 2X + 1$.

$X^2 = (2m^2 - 2m)^2 = 4m^4 - 8m^3 + 4m^2$.

$2X = 2(2m^2 - 2m) = 4m^2 - 4m$.

$c^2 = (4m^4 - 8m^3 + 4m^2) + (4m^2 - 4m) + 1 = 4m^4 - 8m^3 + 8m^2 - 4m + 1$.

We see that $a^2 + b^2 = c^2$ holds true for any value of $m$.

Now we need to check if the terms are positive integers for every natural number $m$. Natural numbers are $1, 2, 3, \dots$.

For $m = 1$:

$a = 2(1) - 1 = 1$. (Positive integer)

$b = 2(1)^2 - 2(1) = 2 - 2 = 0$. (Not a positive integer)

$c = 2(1)^2 - 2(1) + 1 = 2 - 2 + 1 = 1$. (Positive integer)

The triplet for $m=1$ is $(1, 0, 1)$. While this satisfies $1^2 + 0^2 = 1^2$, a standard Pythagorean triplet requires all three numbers to be positive integers.

For $m \ge 2$:

$a = 2m - 1$. If $m \ge 2$, $2m \ge 4$, so $2m-1 \ge 3$. This is a positive integer.

$b = 2m^2 - 2m = 2m(m-1)$. If $m \ge 2$, $m$ and $m-1$ are positive integers, so $2m(m-1)$ is a positive integer.

$c = 2m^2 - 2m + 1 = 2m(m-1) + 1$. Since $2m(m-1)$ is a positive integer for $m \ge 2$, $c$ is a positive integer greater than 1.

For $m \ge 2$, the formula generates a Pythagorean triplet of positive integers. However, the statement claims it is a Pythagorean triplet for every natural number $m$. Since the triplet generated for $m=1$ includes 0, it does not fit the standard definition of a Pythagorean triplet requiring positive integers.

The statement is False.

A Pythagorean triplet consists of three positive integers $a, b,$ and $c$, such that $a^2 + b^2 = c^2$.

Let's analyze the parity (whether a number is odd or even) of the numbers in the equation $a^2 + b^2 = c^2$.

We know the following rules for the square of an integer's parity:

If an integer is Odd, its square is Odd ($Odd^2 = Odd \times Odd = Odd$).

If an integer is Even, its square is Even ($Even^2 = Even \times Even = Even$).

Now consider the sum $a^2 + b^2$:

If both $a$ and $b$ are Odd, then $a^2$ is Odd and $b^2$ is Odd. Their sum is $a^2 + b^2 = Odd + Odd$. The sum of two odd numbers is always Even.

So, if $a$ and $b$ are odd, $a^2 + b^2$ is Even.

If the third number $c$ were also Odd, then $c^2$ would be Odd.

The equation $a^2 + b^2 = c^2$ would then become Even = Odd. This is a contradiction, as an even number cannot equal an odd number.

Therefore, it is impossible for all three numbers in a Pythagorean triplet to be odd. At least one of the numbers $a$ or $b$ must be even.

Consider the most common example of a Pythagorean triplet: (3, 4, 5).

Here, $a=3$, $b=4$, and $c=5$.

Checking the equation: $3^2 + 4^2 = 9 + 16 = 25$.

And $5^2 = 25$.

Since $9 + 16 = 25$, the triplet (3, 4, 5) is a Pythagorean triplet.

However, the numbers in this triplet are 3 (odd), 4 (even), and 5 (odd). Not all numbers are odd.

The statement is False.

We need to check if the statement $a^3 > a^2$ is true for every integer $a$.

Let's consider different cases for the integer $a$.

Case 1: $a$ is a positive integer.

If $a = 1$, $a^2 = 1^2 = 1$ and $a^3 = 1^3 = 1$. Here, $a^3$ is equal to $a^2$, not greater than $a^2$.

If $a = 2$, $a^2 = 2^2 = 4$ and $a^3 = 2^3 = 8$. Here, $a^3 > a^2$ (since $8 > 4$).

If $a > 1$, then $a$ is a positive number greater than 1. Multiplying $a^2$ by $a$ (a number greater than 1) results in a larger number, so $a^3 > a^2$ for $a > 1$.

Case 2: $a$ is zero.

If $a = 0$, $a^2 = 0^2 = 0$ and $a^3 = 0^3 = 0$. Here, $a^3$ is equal to $a^2$, not greater than $a^2$.

Case 3: $a$ is a negative integer.

If $a = -1$, $a^2 = (-1)^2 = 1$ and $a^3 = (-1)^3 = -1$. Here, $a^3 < a^2$ (since $-1 < 1$).

If $a = -2$, $a^2 = (-2)^2 = 4$ and $a^3 = (-2)^3 = -8$. Here, $a^3 < a^2$ (since $-8 < 4$).

If $a$ is any negative integer ($a < 0$), $a^2$ will be positive, and $a^3$ will be negative. A negative number is always less than a positive number, so $a^3 < a^2$ for $a < 0$.

Since the statement "$a^3$ is always greater than $a^2$" is not true for $a=1$, $a=0$, or any negative integer, the statement is false.

The statement is False.

The given statement is: If $x, y$ are integers such that $x^2 > y^2$, then $x^3 > y^3$.

The condition $x^2 > y^2$ implies that $\sqrt{x^2} > \sqrt{y^2}$, which means $|x| > |y|$. This tells us that the absolute value of $x$ is greater than the absolute value of $y$.

We need to check if the implication $x^3 > y^3$ always holds true under this condition for any integers $x$ and $y$.

Let's test the statement with some integer values for $x$ and $y$.

Consider $x = -3$ and $y = 2$. Both are integers.

Check the condition $x^2 > y^2$:

$x^2 = (-3)^2 = 9$

$y^2 = (2)^2 = 4$

Is $9 > 4$? Yes, the condition $x^2 > y^2$ is true for $x=-3$ and $y=2$.

Now, let's check if $x^3 > y^3$ holds for these values:

$x^3 = (-3)^3 = -27$

$y^3 = (2)^3 = 8$

Is $-27 > 8$? No, $-27$ is less than $8$.

In this example, the premise ($x^2 > y^2$) is true, but the conclusion ($x^3 > y^3$) is false.

This counterexample shows that the statement is not true for all integers $x$ and $y$ satisfying the condition.

The statement holds if $x$ and $y$ have the same sign or if $y$ is negative and $x$ is positive with $|x|>|y|$. However, it fails when $x$ is negative and $y$ is positive with $|x|>|y|$, because cubing a negative number results in a negative number, while cubing a positive number results in a positive number.

The statement is False.

We are given that $x$ and $y$ are natural numbers and $x$ divides $y$.

The statement "$x$ divides $y$" means that there exists a natural number $k$ such that $y = kx$.

We need to determine if $x^3$ divides $y^3$. This means we need to check if there exists a natural number (or integer) $m$ such that $y^3 = m x^3$.

Let's cube both sides of the equation $y = kx$:

$y^3 = (kx)^3$

Using the property of exponents $(ab)^n = a^n b^n$, we get:

$y^3 = k^3 x^3$

Since $k$ is a natural number, $k^3$ is also a natural number.

Let $m = k^3$. Since $k$ is a natural number, $m$ is also a natural number.

So we have $y^3 = m x^3$, where $m = k^3$ is a natural number.

This shows that $y^3$ is a multiple of $x^3$, which means $x^3$ divides $y^3$.

Example: Let $x=2$ and $y=6$. $x$ and $y$ are natural numbers. $x$ divides $y$ because $6 = 3 \times 2$ (here $k=3$).

$x^3 = 2^3 = 8$.

$y^3 = 6^3 = 216$.

Does $x^3$ divide $y^3$? Does 8 divide 216?

$216 \div 8 = 27$. Since 27 is an integer ($m=27$, and $k^3 = 3^3 = 27$), 8 divides 216.

The statement holds true.

The statement is True.

For an integer $a$, if $a^2$ ends in 5, this means the units digit of $a^2$ is 5. The only way for the square of an integer to end in 5 is if the integer itself ends in 5.

So, if $a^2$ ends in 5, the units digit of $a$ must be 5.

Let's consider integers $a$ that end in 5 and examine the last two digits of their cubes $a^3$.

Let $a = 5$.

$a^2 = 5^2 = 25$. This ends in 5.

$a^3 = 5^3 = 125$. This ends in 25.

This example supports the statement.

Let $a = 15$.

$a^2 = 15^2 = 225$. This ends in 5.

$a^3 = 15^3 = 15 \times 15 \times 15 = 225 \times 15$.

Calculating the multiplication:

$15^3 = 3375$. This ends in 75, not 25.

Since we found an example ($a=15$) where $a^2$ ends in 5, but $a^3$ does not end in 25, the statement is false.

Note: An integer $a$ ends in 5 if and only if $a$ is of the form $10k + 5$ for some integer $k$.

$a^3 = (10k + 5)^3 = (10k)^3 + 3(10k)^2(5) + 3(10k)(5)^2 + 5^3$

$a^3 = 1000k^3 + 3(100k^2)(5) + 3(10k)(25) + 125$

$a^3 = 1000k^3 + 1500k^2 + 750k + 125$.

The first three terms $1000k^3$, $1500k^2$, and $750k$ end in 0 (or 00 or 000) when considered as integers. The sum of these three terms will end in 0.

So, the last two digits of $a^3$ are determined by the last two digits of $750k + 125$.

$750k + 125 = 100(7k) + 50k + 125 = 100(7k) + 50k + 100 + 25 = 100(7k + 1) + 50k + 25$.

The term $100(7k + 1)$ ends in 00.

The last two digits of $a^3$ are determined by the last two digits of $50k + 25$.

If $k$ is even, let $k=2m$. Then $50k + 25 = 50(2m) + 25 = 100m + 25$, which ends in 25.

If $k$ is odd, let $k=2m+1$. Then $50k + 25 = 50(2m+1) + 25 = 100m + 50 + 25 = 100m + 75$, which ends in 75.

Since $a$ ends in 5, $a = 10k+5$. The value of $k$ depends on $a$.

If $a=5$, $5 = 10(0)+5$, so $k=0$ (even). $a^3=125$, ends in 25.

If $a=15$, $15 = 10(1)+5$, so $k=1$ (odd). $a^3=3375$, ends in 75.

If $a=25$, $25 = 10(2)+5$, so $k=2$ (even). $a^3=15625$, ends in 25.

If $a=35$, $35 = 10(3)+5$, so $k=3$ (odd). $a^3=42875$, ends in 75.

The statement "$a^3$ ends in 25" is only true when $a$ is of the form $10(2m)+5 = 20m+5$ for some integer $m$. It is not true for all $a$ ending in 5.

The statement is False.

If the square of an integer $a$, $a^2$, ends in 9, then the units digit of $a^2$ is 9.

The units digit of a square is determined by the units digit of the original number being squared. Let's look at the squares of single digits:

• $0^2$ ends in 0
• $1^2$ ends in 1
• $2^2$ ends in 4
• $3^2$ ends in 9
• $4^2$ ends in 6
• $5^2$ ends in 5
• $6^2$ ends in 6
• $7^2$ ends in 9
• $8^2$ ends in 4
• $9^2$ ends in 1

From this, we see that if $a^2$ ends in 9, the units digit of $a$ must be either 3 or 7.

Now let's consider the units digit of $a^3$ based on the units digit of $a$.

If the units digit of $a$ is 3, then the units digit of $a^3$ is the units digit of $3^3 = 27$, which is 7. This case aligns with the statement.

If the units digit of $a$ is 7, then the units digit of $a^3$ is the units digit of $7^3 = 343$, which is 3. This case contradicts the statement.

Since there is at least one case where $a^2$ ends in 9 (when $a$ ends in 7), but $a^3$ does not end in 7 (it ends in 3), the statement is false.

Example: Let $a=7$.

$a^2 = 7^2 = 49$. This ends in 9.

$a^3 = 7^3 = 343$. This ends in 3, not 7.

The statement is False.

Let $N$ be a perfect square with $n$ digits, where $n$ is an odd natural number.

The smallest $n$-digit number is $10^{n-1}$. The largest $n$-digit number is $10^n - 1$.

So, $10^{n-1} \le N < 10^n$.

Let $\sqrt{N}$ have $k$ digits.

The smallest $k$-digit number is $10^{k-1}$. The largest $k$-digit number is $10^k - 1$.

So, $10^{k-1} \le \sqrt{N} < 10^k$.

Squaring all parts of the inequality, we get:

$(10^{k-1})^2 \le (\sqrt{N})^2 < (10^k)^2$

$10^{2(k-1)} \le N < 10^{2k}$

Comparing this with the range for an $n$-digit number, $10^{n-1} \le N < 10^n$, we can relate the exponents:

$2(k-1) \le n-1$

$2k - 2 \le n - 1$

$2k \le n + 1$

$k \le \frac{n+1}{2}$

And

$n \le 2k - 1$ (since $N < 10^{2k}$ and $N \ge 10^{n-1}$, the minimum value $10^{n-1}$ must be less than $10^{2k}$, which is always true if $n-1 < 2k$. Also, the maximum value of an n-digit number $(10^n - 1)$ must be less than $10^{2k}$, which means $10^n < 10^{2k}$, or $n < 2k$)

Consider the inequality $10^{n-1} \le N < 10^n$. Taking the square root: $\sqrt{10^{n-1}} \le \sqrt{N} < \sqrt{10^n}$.

$10^{(n-1)/2} \le \sqrt{N} < 10^{n/2}$.

Let $k$ be the number of digits in $\sqrt{N}$. Then $10^{k-1} \le \sqrt{N} < 10^k$.

Combining the inequalities:

$10^{k-1} < 10^{n/2}$ implies $k-1 < n/2$ implies $k < n/2 + 1$.

$10^{(n-1)/2} < 10^k$ implies $(n-1)/2 < k$.

So we have $\frac{n-1}{2} < k < \frac{n}{2} + 1$.

Since $n$ is odd, let $n = 2m + 1$ for some non-negative integer $m$.

Then the inequality becomes $\frac{(2m+1)-1}{2} < k < \frac{2m+1}{2} + 1$.

$\frac{2m}{2} < k < m + \frac{1}{2} + 1$.

$m < k < m + 1.5$.

Since $k$ must be an integer, the only possible integer value for $k$ in the range $(m, m+1.5)$ is $m+1$.

Now let's express $m+1$ in terms of $n$. Since $n = 2m + 1$, $n+1 = 2m+2$, so $\frac{n+1}{2} = \frac{2m+2}{2} = m+1$.

Thus, the number of digits $k$ is equal to $\frac{n+1}{2}$.

Example: Let $n=3$. A 3-digit perfect square. Smallest is $100 = 10^2$, $\sqrt{100} = 10$ (2 digits). Largest is $961 = 31^2$, $\sqrt{961} = 31$ (2 digits). Using the formula $\frac{n+1}{2} = \frac{3+1}{2} = \frac{4}{2} = 2$. This matches.

Example: Let $n=5$. A 5-digit perfect square. Smallest is $10000 = 100^2$, $\sqrt{10000} = 100$ (3 digits). Largest is $99856 = 316^2$, $\sqrt{99856} = 316$ (3 digits). Using the formula $\frac{n+1}{2} = \frac{5+1}{2} = \frac{6}{2} = 3$. This matches.

The statement is True.

In mathematics, the symbol $\sqrt{\ }$ is used to denote the square root of a number.

For a number $x$, its square root is denoted by $\sqrt{x}$.

Specifically, $\sqrt{x}$ represents the principal (non-negative) square root of $x$ when $x$ is a non-negative real number.

This is the standard and widely accepted notation for the square root.

The statement is True.

To determine the units digit of the square of a number, we only need to consider the units digit of the original number.

The units digit of the number is given as 7.

We need to find the units digit of the square of a number ending in 7. This is the same as finding the units digit of $7^2$.

$7^2 = 49$.

The units digit of 49 is 9.

So, a number having 7 at its ones place will have 9 (not 3) at the units place of its square.

The statement is False.

To determine the units digit of the cube of a number, we only need to consider the units digit of the original number.

The units digit of the number is given as 7.

We need to find the units digit of the cube of a number ending in 7. This is the same as finding the units digit of $7^3$.

$7^3 = 7 \times 7 \times 7 = 49 \times 7$.

To find the units digit of $49 \times 7$, we only need to multiply the units digits: $9 \times 7 = 63$.

The units digit of 63 is 3.

So, the units digit of $7^3$ is 3.

Therefore, a number having 7 at its ones place will have 3 at the ones place of its cube.

The statement is True.

A one-digit number is an integer from 0 to 9. We need to examine the cubes of these numbers and check if any of them result in a two-digit number.

$0^3 = 0$ (One-digit number)

$1^3 = 1$ (One-digit number)

$2^3 = 8$ (One-digit number)

$3^3 = 27$ (Two-digit number)

$4^3 = 64$ (Two-digit number)

$5^3 = 125$ (Three-digit number)

$6^3 = 216$ (Three-digit number)

$7^3 = 343$ (Three-digit number)

$8^3 = 512$ (Three-digit number)

$9^3 = 729$ (Three-digit number)

We can see that the cubes of the one-digit numbers 3 and 4 are 27 and 64, respectively. Both 27 and 64 are two-digit numbers.

Since the cube of a one-digit number can indeed be a two-digit number, the statement is false.

The statement is False.

We need to check the parity (whether a number is odd or even) of the cube of an even number.

An even number can be represented in the form $2k$, where $k$ is an integer.

Let's find the cube of an even number $(2k)$:

$(2k)^3 = (2k) \times (2k) \times (2k)$.

Using the property of multiplication, $(ab)^n = a^n b^n$, we have:

$(2k)^3 = 2^3 \times k^3 = 8 \times k^3$.

Since $8k^3$ is a multiple of 8, it is also a multiple of 2. Any integer that is a multiple of 2 is an even number.

Therefore, the cube of an even number is always an even number.

Example:

Let the even number be 2. Its cube is $2^3 = 8$. 8 is even.

Let the even number be 4. Its cube is $4^3 = 64$. 64 is even.

Let the even number be 6. Its cube is $6^3 = 216$. 216 is even.

The statement claims that the cube of an even number is odd.

The statement is False.

We need to check the parity (whether a number is odd or even) of the cube of an odd number.

An odd number can be represented in the form $2k + 1$, where $k$ is an integer.

Let's find the cube of an odd number $(2k + 1)$:

$(2k + 1)^3 = (2k)^3 + 3(2k)^2(1) + 3(2k)(1)^2 + 1^3$

$(2k + 1)^3 = 8k^3 + 3(4k^2) + 6k + 1$

$(2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1$.

We can factor out 2 from the first three terms:

$8k^3 + 12k^2 + 6k = 2(4k^3 + 6k^2 + 3k)$.

So, $(2k + 1)^3 = 2(4k^3 + 6k^2 + 3k) + 1$.

Let $m = 4k^3 + 6k^2 + 3k$. Since $k$ is an integer, $m$ is also an integer.

The cube is in the form $2m + 1$, which represents an odd number.

Therefore, the cube of an odd number is always an odd number.

Example:

Let the odd number be 3. Its cube is $3^3 = 27$. 27 is odd.

Let the odd number be 5. Its cube is $5^3 = 125$. 125 is odd.

Let the odd number be 7. Its cube is $7^3 = 343$. 343 is odd.

The statement claims that the cube of an odd number is even.

The statement is False.

We need to determine if the cube of an even number is always even.

An even number is an integer that is divisible by 2. It can be written in the form $2k$, where $k$ is an integer.

Let $E$ be an even number. So, $E = 2k$ for some integer $k$.

The cube of $E$ is $E^3 = (2k)^3$.

Using the exponent rule $(ab)^n = a^n b^n$:

$E^3 = 2^3 \times k^3 = 8 \times k^3$.

Since $8k^3$ is a multiple of 8, it is also a multiple of 2 (because $8 = 2 \times 4$).

Any integer that is a multiple of 2 is, by definition, an even number.

Therefore, the cube of an even number is always an even number.

Example:

Let the even number be 4. Its cube is $4^3 = 64$. 64 is an even number.

Let the even number be 10. Its cube is $10^3 = 1000$. 1000 is an even number.

The statement is True.

We need to determine if the cube of an odd number is always odd.

An odd number is an integer that is not divisible by 2. It can be written in the form $2k + 1$, where $k$ is an integer.

Let $O$ be an odd number. So, $O = 2k + 1$ for some integer $k$.

The cube of $O$ is $O^3 = (2k + 1)^3$.

Expanding the cube:

$(2k + 1)^3 = (2k+1)(2k+1)(2k+1)$.

The product of two odd numbers is always an odd number ($Odd \times Odd = Odd$).

So, $(2k+1)(2k+1) = (2k+1)^2$, which is the square of an odd number, and thus is an odd number.

Now we multiply this odd number by the third odd number: $(2k+1)^2 \times (2k+1) = Odd \times Odd$.

The product of two odd numbers is again an odd number.

Therefore, the cube of an odd number is always an odd number.

Example:

Let the odd number be 3. Its cube is $3^3 = 27$. 27 is an odd number.

Let the odd number be 5. Its cube is $5^3 = 125$. 125 is an odd number.

The statement is True.

A perfect cube is an integer that is the cube of an integer. To determine if 999 is a perfect cube, we can find its prime factorization.

We perform the prime factorization of 999:

Divide 999 by the smallest prime factor, which is 3:

The prime factorization of 999 is $3 \times 3 \times 3 \times 37$.

We can write this using exponents: $999 = 3^3 \times 37^1$.

For a number to be a perfect cube, the exponents of all its prime factors must be multiples of 3.

In the prime factorization of 999, the exponent of the prime factor 3 is 3 (which is a multiple of 3), but the exponent of the prime factor 37 is 1 (which is not a multiple of 3).

Since the exponent of at least one prime factor is not a multiple of 3, 999 is not a perfect cube.

The statement is False.

To check if the product $363 \times 81$ is a perfect cube, we need to find the prime factorization of the product. This can be done by finding the prime factorization of each number separately and then combining them.

Find the prime factorization of 363:

So, $363 = 3 \times 11 \times 11 = 3^1 \times 11^2$.

Find the prime factorization of 81:

So, $81 = 3 \times 3 \times 3 \times 3 = 3^4$.

Now, multiply the prime factorizations of 363 and 81:

$363 \times 81 = (3^1 \times 11^2) \times (3^4)$

Using the property of exponents $a^m \times a^n = a^{m+n}$, we combine the powers of the same prime factors:

$363 \times 81 = 3^{1+4} \times 11^2 = 3^5 \times 11^2$.

For a number to be a perfect cube, the exponents of all prime factors in its prime factorization must be multiples of 3.

In the prime factorization of $363 \times 81 = 3^5 \times 11^2$, the exponent of 3 is 5 and the exponent of 11 is 2. Neither 5 nor 2 are multiples of 3.

Since the exponents are not multiples of 3, the number $363 \times 81$ is not a perfect cube.

The statement is False.

We are asked to determine if the cube roots of 8 are +2 and -2.

A cube root of a number $x$ is a number $y$ such that $y^3 = x$.

Let's check if $(+2)^3 = 8$:

$(+2)^3 = 2 \times 2 \times 2 = 8$.

So, +2 is a cube root of 8.

Now let's check if $(-2)^3 = 8$:

$(-2)^3 = (-2) \times (-2) \times (-2)$.

$(-2) \times (-2) = +4$.

$(+4) \times (-2) = -8$.

So, $(-2)^3 = -8$, which is not equal to 8.

Therefore, -2 is not a cube root of 8.

For real numbers, every positive number has exactly one positive real cube root. The cube root of 8 is the unique real number whose cube is 8.

The notation $\sqrt[3]{8}$ denotes the principal (real) cube root of 8, which is +2.

While there are complex cube roots for any non-zero number, when referring to "the cube root" in this context, it typically means the real cube root.

Since -2 is not a cube root of 8, the statement that the cube roots are +2 and -2 is false.

The statement is False.

We need to evaluate both sides of the given equation and check if they are equal.

Consider the left side of the equation: $\sqrt[3]{8\;+\;27}$.

First, perform the addition inside the cube root: $8 + 27 = 35$.

So, the left side is $\sqrt[3]{35}$.

The number 35 is not a perfect cube (since $3^3 = 27$ and $4^3 = 64$). Thus, $\sqrt[3]{35}$ is an irrational number (or not a simple integer).

Now, consider the right side of the equation: $\sqrt[3]{8}$ + $\sqrt[3]{27}$.

First, find the cube root of 8. Since $2^3 = 8$, the cube root of 8 is 2.

$\sqrt[3]{8} = 2$.

Next, find the cube root of 27. Since $3^3 = 27$, the cube root of 27 is 3.

$\sqrt[3]{27} = 3$.

Now, add these cube roots: $\sqrt[3]{8}$ + $\sqrt[3]{27} = 2 + 3 = 5$.

We compare the values of the left side and the right side:

Left side = $\sqrt[3]{35}$.

Right side = 5.

To check if $\sqrt[3]{35} = 5$, we can cube both sides:

$(\sqrt[3]{35})^3 = 35$.

$5^3 = 5 \times 5 \times 5 = 125$.

Since $35 \neq 125$, the statement $\sqrt[3]{35} = 5$ is false.

Therefore, $\sqrt[3]{8\;+\;27} \neq \sqrt[3]{8}$ + $\sqrt[3]{27}$.

In general, for any positive numbers $a$ and $b$, $\sqrt[n]{a+b} \neq \sqrt[n]{a} + \sqrt[n]{b}$ for $n > 1$.

The statement is False.

We need to determine if a negative integer has a cube root. A cube root of a number $x$ is a number $y$ such that $y^3 = x$.

Let's consider a negative integer, for example, -8. We are looking for a number $y$ such that $y^3 = -8$.

Let's consider the cube of a negative number.

Let $y = -2$.

$y^3 = (-2)^3 = (-2) \times (-2) \times (-2) = (4) \times (-2) = -8$.

So, $(-2)^3 = -8$. This means that -2 is a cube root of -8.

In general, for any negative real number $x$, there is exactly one real number $y$ such that $y^3 = x$. This real number $y$ is the cube root of $x$, denoted by $\sqrt[3]{x}$.

If $x$ is a negative integer, say $x = -m$ where $m$ is a positive integer, then $\sqrt[3]{x} = \sqrt[3]{-m}$. If $m$ is a perfect cube (i.e., $m = k^3$ for some integer $k$), then $\sqrt[3]{-m} = \sqrt[3]{-k^3} = \sqrt[3]{(-k)^3} = -k$. Since $k$ is an integer, $-k$ is also an integer.

For example, $\sqrt[3]{-27} = -3$ because $(-3)^3 = -27$.

If $m$ is not a perfect cube, the cube root of $-m$ will be a negative real number, which is not an integer. However, it is still a cube root.

The statement says there is *no* cube root of a negative integer. This is false because every negative integer has exactly one real cube root (which is also negative).

The statement is False.

We are given the statement: If the square of a number $a$ is positive ($a^2 > 0$), then the cube of that number will also be positive ($a^3 > 0$).

The condition $a^2 > 0$ implies that $a^2$ is a positive value.

This occurs when $a$ is any non-zero real number. That is, $a > 0$ or $a < 0$.

Let's consider the two cases for $a$ when $a^2 > 0$:

Case 1: $a$ is a positive number ($a > 0$).

If $a$ is positive, then $a^3 = a \times a \times a$. The product of three positive numbers is positive.

So, if $a > 0$, then $a^3 > 0$. This part of the implication holds true.

Case 2: $a$ is a negative number ($a < 0$).

If $a$ is negative, then $a^3 = a \times a \times a$.

The product of two negative numbers is positive: $a \times a = a^2 > 0$.

The product of a positive number and a negative number is negative: $a^2 \times a < 0$.

So, if $a < 0$, then $a^3 < 0$.

This contradicts the conclusion that $a^3$ will also be positive.

The statement claims that if $a^2 > 0$, then $a^3$ *will always* be positive. However, if $a$ is a negative number (which satisfies $a^2 > 0$), its cube $a^3$ is negative.

Consider the example $a = -2$.

$a^2 = (-2)^2 = 4$. Here, $a^2 > 0$ is true.

$a^3 = (-2)^3 = -8$. Here, $a^3 > 0$ is false (since $-8$ is not positive).

Since we found a number ($a=-2$) for which the premise ($a^2 > 0$) is true but the conclusion ($a^3 > 0$) is false, the entire statement is false.

The statement is False.

The first five square numbers are the squares of the first five natural numbers.

The first five natural numbers are 1, 2, 3, 4, and 5.

The square of the first natural number is $1^2 = 1 \times 1 = 1$.

The square of the second natural number is $2^2 = 2 \times 2 = 4$.

The square of the third natural number is $3^2 = 3 \times 3 = 9$.

The square of the fourth natural number is $4^2 = 4 \times 4 = 16$.

The square of the fifth natural number is $5^2 = 5 \times 5 = 25$.

The first five square numbers are 1, 4, 9, 16, and 25.

The first three multiples of 3 are obtained by multiplying 3 by the first three natural numbers (1, 2, and 3).

The first multiple of 3 is $3 \times 1 = 3$.

The second multiple of 3 is $3 \times 2 = 6$.

The third multiple of 3 is $3 \times 3 = 9$.

So, the first three multiples of 3 are 3, 6, and 9.

Now, we need to find the cubes of these numbers.

The cube of the first multiple (3) is $3^3 = 3 \times 3 \times 3 = 27$.

The cube of the second multiple (6) is $6^3 = 6 \times 6 \times 6 = 36 \times 6 = 216$.

The cube of the third multiple (9) is $9^3 = 9 \times 9 \times 9 = 81 \times 9 = 729$.

The cubes of the first three multiples of 3 are 27, 216, and 729.

To show that a number is not a perfect square, we can use its prime factorization. A number is a perfect square if and only if all the exponents in its prime factorization are even.

Find the prime factorization of 500:

The prime factorization of 500 is $2 \times 2 \times 5 \times 5 \times 5$.

In exponential form, this is $500 = 2^2 \times 5^3$.

For 500 to be a perfect square, the exponents of its prime factors (2 and 5) must be even.

The exponent of 2 is 2, which is an even number.

The exponent of 5 is 3, which is an odd number.

Since the exponent of the prime factor 5 is odd, 500 is not a perfect square.

Alternatively, we can look at the units digit. A perfect square cannot end in 0 unless it ends in an even number of zeros. 500 ends in two zeros, but we should still verify with prime factorization.

Another method is to check integers whose squares are close to 500.

$22^2 = 22 \times 22 = 484$.

$23^2 = 23 \times 23 = 529$.

Since 500 lies between 484 and 529, its square root is between 22 and 23. Since the square root of 500 is not an integer, 500 is not a perfect square.

We know that the sum of the first $n$ consecutive odd natural numbers is equal to $n^2$.

In this question, the number is 81. We recognize that $81 = 9^2$.

According to the property, $9^2$ is the sum of the first 9 consecutive odd natural numbers.

The first nine consecutive odd natural numbers are:

1, 3, 5, 7, 9, 11, 13, 15, 17.

Let's find the sum of these numbers:

Sum = $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17$.

We can group them for easier addition:

Sum = $(1 + 17) + (3 + 15) + (5 + 13) + (7 + 11) + 9$

Sum = $18 + 18 + 18 + 18 + 9$

Sum = $4 \times 18 + 9$

Sum = $72 + 9$

Sum = $81$.

So, 81 can be expressed as the sum of the first nine consecutive odd numbers as follows:

$81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17$.

A number is a perfect square if and only if in its prime factorization, all the exponents of the prime factors are even.

(a) 484

Find the prime factorization of 484:

The prime factorization of 484 is $2 \times 2 \times 11 \times 11 = 2^2 \times 11^2$.

The exponents of the prime factors are 2 (for 2) and 2 (for 11). Both exponents are even.

Therefore, 484 is a perfect square ($\sqrt{484} = \sqrt{2^2 \times 11^2} = 2^1 \times 11^1 = 22$).

(b) 11250

Find the prime factorization of 11250:

The prime factorization of 11250 is $2 \times 3 \times 3 \times 5 \times 5 \times 5 \times 5$.

In exponential form, this is $11250 = 2^1 \times 3^2 \times 5^4$.

The exponents of the prime factors are 1 (for 2), 2 (for 3), and 4 (for 5).

The exponent of 2 is 1, which is an odd number.

Therefore, 11250 is not a perfect square.

(c) 841

Find the prime factorization of 841. We can try dividing by small prime numbers. The number 841 is not divisible by 2, 3 (sum of digits is 13), 5. Let's try larger primes.

We know that $20^2 = 400$ and $30^2 = 900$. The number ends in 1, so the square root must end in 1 or 9.

Let's try 21. $21^2 = 441$.

Let's try 29. $29 \times 29$:

$29^2 = 841$.

Since 29 is a prime number, the prime factorization of 841 is $29 \times 29 = 29^2$.

The exponent of the prime factor 29 is 2, which is an even number.

Therefore, 841 is a perfect square ($\sqrt{841} = 29$).

(d) 729

Find the prime factorization of 729:

The prime factorization of 729 is $3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$.

The exponent of the prime factor 3 is 6, which is an even number.

Therefore, 729 is a perfect square ($\sqrt{729} = \sqrt{3^6} = 3^{6/2} = 3^3 = 27$).

Based on the prime factorizations, the perfect squares are 484, 841, and 729.

A number is a perfect cube if and only if in its prime factorization, all the exponents of the prime factors are multiples of 3.

(a) 128

Find the prime factorization of 128:

The prime factorization of 128 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7$.

The exponent of the prime factor 2 is 7. Since 7 is not a multiple of 3, 128 is not a perfect cube.

(b) 343

Find the prime factorization of 343:

The prime factorization of 343 is $7 \times 7 \times 7 = 7^3$.

The exponent of the prime factor 7 is 3. Since 3 is a multiple of 3, 343 is a perfect cube ($\sqrt[3]{343} = 7$).

(c) 729

Find the prime factorization of 729:

The prime factorization of 729 is $3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$.

The exponent of the prime factor 3 is 6. Since 6 is a multiple of 3 ($6 = 3 \times 2$), 729 is a perfect cube ($\sqrt[3]{729} = \sqrt[3]{3^6} = 3^{6/3} = 3^2 = 9$).

(d) 1331

Find the prime factorization of 1331:

The prime factorization of 1331 is $11 \times 11 \times 11 = 11^3$.

The exponent of the prime factor 11 is 3. Since 3 is a multiple of 3, 1331 is a perfect cube ($\sqrt[3]{1331} = 11$).

Based on the prime factorizations, the perfect cubes are 343, 729, and 1331.

The distributive law of multiplication over addition states that $a \times (b + c) = a \times b + a \times c$. We can use this property, often in conjunction with the identity $(x+y)^2 = x^2 + 2xy + y^2$ or $(x-y)^2 = x^2 - 2xy + y^2$, which are derived from the distributive property, to find the squares.

(a) 101

We can write 101 as the sum of two numbers, such as $100 + 1$.

Then, the square of 101 is $101^2 = (100 + 1)^2$.

Using the identity $(x+y)^2 = x^2 + 2xy + y^2$ with $x = 100$ and $y = 1$:

$101^2 = (100)^2 + 2(100)(1) + (1)^2$

$101^2 = 10000 + 200 + 1$

$101^2 = 10201$.

(b) 72

We can write 72 as the sum or difference of two numbers, such as $70 + 2$ or $80 - 8$. Let's use $70 + 2$.

The square of 72 is $72^2 = (70 + 2)^2$.

Using the identity $(x+y)^2 = x^2 + 2xy + y^2$ with $x = 70$ and $y = 2$:

$72^2 = (70)^2 + 2(70)(2) + (2)^2$

$72^2 = 4900 + 280 + 4$

$72^2 = 5180 + 4$

$72^2 = 5184$.

Alternatively, using $80 - 8$:

$72^2 = (80 - 8)^2$.

Using the identity $(x-y)^2 = x^2 - 2xy + y^2$ with $x = 80$ and $y = 8$:

$72^2 = (80)^2 - 2(80)(8) + (8)^2$

$72^2 = 6400 - 1280 + 64$

$72^2 = 5120 + 64$

$72^2 = 5184$.

For three side lengths to form a right triangle, they must satisfy the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides (legs).

Let the side lengths be $a, b,$ and $c$, where $c$ is the longest side. The theorem is $a^2 + b^2 = c^2$.

The given side lengths are 6 cm, 10 cm, and 8 cm.

Identify the longest side. The longest side is 10 cm. This side would be the hypotenuse if the triangle is a right triangle. So, $c = 10$.

The other two sides are 6 cm and 8 cm. These would be the legs. So, $a = 6$ and $b = 8$.

Now, check if the Pythagorean theorem holds true for these side lengths:

$a^2 + b^2 = 6^2 + 8^2$

$6^2 = 6 \times 6 = 36$.

$8^2 = 8 \times 8 = 64$.

$a^2 + b^2 = 36 + 64 = 100$.

Now, calculate the square of the longest side, $c^2$:

$c^2 = 10^2 = 10 \times 10 = 100$.

Comparing $a^2 + b^2$ and $c^2$:

Since $a^2 + b^2 = c^2$ ($100 = 100$), the side lengths satisfy the Pythagorean theorem.

Therefore, a right triangle with sides 6 cm, 10 cm, and 8 cm can be formed.

Reason: The given side lengths satisfy the Pythagorean theorem, which is a fundamental property of right triangles. Specifically, the square of the longest side (10 cm) is equal to the sum of the squares of the other two sides (6 cm and 8 cm), i.e., $10^2 = 6^2 + 8^2$.

A Pythagorean triplet consists of three positive integers $a, b, c$ such that $a^2 + b^2 = c^2$.

We are given that one of the numbers in the triplet is 4. Let's consider the cases where 4 is one of the legs or the hypotenuse.

Case 1: 4 is one of the legs (say $a=4$).

We need to find positive integers $b$ and $c$ such that $4^2 + b^2 = c^2$.

$16 + b^2 = c^2$.

Rearranging the equation, $c^2 - b^2 = 16$.

Using the difference of squares formula, $(c - b)(c + b) = 16$.

Since $b$ and $c$ are positive integers, $c+b$ is a positive integer. Also, since $c^2 = 16 + b^2$, $c^2 > b^2$, and since $c$ and $b$ are positive, $c > b$, which means $c-b$ is a positive integer.

We need to find pairs of factors of 16 whose product is 16.

Possible pairs of positive integer factors of 16 are (1, 16), (2, 8), and (4, 4).

Let $c - b = p$ and $c + b = q$, where $pq = 16$ and $q > p$ (since $c+b > c-b$).

Adding the two equations: $(c - b) + (c + b) = p + q \implies 2c = p + q \implies c = \frac{p+q}{2}$.

Subtracting the first equation from the second: $(c + b) - (c - b) = q - p \implies 2b = q - p \implies b = \frac{q-p}{2}$.

For $b$ and $c$ to be integers, $p+q$ and $q-p$ must both be even. This happens when both $p$ and $q$ are either both even or both odd.

Consider the factor pairs of 16:

Pair (1, 16): $p=1, q=16$. $p$ is odd, $q$ is even. Not both even or both odd. (Also $q > p$ is satisfied).

Pair (2, 8): $p=2, q=8$. Both are even. ($q > p$ is satisfied).

$c = \frac{2+8}{2} = \frac{10}{2} = 5$.

$b = \frac{8-2}{2} = \frac{6}{2} = 3$.

This gives the triplet (4, 3, 5). Since $3^2 + 4^2 = 9 + 16 = 25 = 5^2$, this is a Pythagorean triplet. Conventionally, we list them in ascending order: (3, 4, 5).

Pair (4, 4): $p=4, q=4$. Both are even. However, $q > p$ is not satisfied (they are equal). This case would lead to $b=0$, which is not a positive integer.

Case 2: 4 is the hypotenuse ($c=4$).

We need to find positive integers $a$ and $b$ such that $a^2 + b^2 = 4^2$.

$a^2 + b^2 = 16$.

Since $a$ and $b$ are positive integers, $a \ge 1$ and $b \ge 1$.

$a^2 \ge 1$ and $b^2 \ge 1$.

$a^2 + b^2 \ge 1 + 1 = 2$.

We need to find two perfect squares that add up to 16.

Possible perfect squares less than 16 are 1, 4, 9.

Can we find two of these that sum to 16?

• $1 + 4 = 5 \neq 16$
• $1 + 9 = 10 \neq 16$
• $4 + 9 = 13 \neq 16$
• $4 + 4 = 8 \neq 16$
• $9 + 4 = 13 \neq 16$

Also, we must have $a < 4$ and $b < 4$. The possible integer values for $a$ and $b$ are 1, 2, 3. Their squares are $1^2=1, 2^2=4, 3^2=9$.

Sum of two squares from $\{1, 4, 9\}$ that equals 16? There are none.

So, 4 cannot be the hypotenuse of a Pythagorean triplet.

The Pythagorean triplet whose one of the numbers is 4 is (3, 4, 5).

To find the square root of a number using prime factorization, we first find the prime factorization of the number. Then, we group the identical prime factors into pairs. The square root is obtained by taking one factor from each pair and multiplying them.

(a) 11025

Find the prime factorization of 11025:

The prime factorization of 11025 is $3 \times 3 \times 5 \times 5 \times 7 \times 7$.

Group the prime factors into pairs: $(3 \times 3) \times (5 \times 5) \times (7 \times 7) = 3^2 \times 5^2 \times 7^2$.

To find the square root, take one factor from each pair: $3 \times 5 \times 7$.

Multiply these factors: $3 \times 5 \times 7 = 15 \times 7 = 105$.

So, $\sqrt{11025} = 105$.

(b) 4761

Find the prime factorization of 4761.

The sum of the digits is $4 + 7 + 6 + 1 = 18$, which is divisible by 3, so 4761 is divisible by 3.

The prime factorization of 4761 is $3 \times 3 \times 23 \times 23$.

Group the prime factors into pairs: $(3 \times 3) \times (23 \times 23) = 3^2 \times 23^2$.

To find the square root, take one factor from each pair: $3 \times 23$.

Multiply these factors: $3 \times 23 = 69$.

So, $\sqrt{4761} = 69$.

To find the cube root of a number using prime factorization, we first find the prime factorization of the number. Then, we group the identical prime factors into triplets. The cube root is obtained by taking one factor from each triplet and multiplying them.

(a) 512

Find the prime factorization of 512:

The prime factorization of 512 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9$.

Group the prime factors into triplets: $(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 2^3 \times 2^3 \times 2^3$.

To find the cube root, take one factor from each triplet: $2 \times 2 \times 2$.

Multiply these factors: $2 \times 2 \times 2 = 8$.

So, $\sqrt[3]{512} = 8$.

(b) 2197

Find the prime factorization of 2197. We can try dividing by small prime numbers. The number is not divisible by 2, 3 (sum of digits 19), 5.

Let's consider the units digit of the number, which is 7. The units digit of the cube root must be a number whose cube ends in 7. Checking the cubes of single digits, only $3^3 = 27$ ends in 7. So, the units digit of the cube root must be 3.

Let's try prime numbers ending in 3. The prime numbers are 3, 13, 23, etc. We already checked 3 (not divisible). Let's try 13.

$13 \times 13 = 169$.

$169 \times 13$:

$13 \times 13 \times 13 = 2197$.

Since 13 is a prime number, the prime factorization of 2197 is $13 \times 13 \times 13 = 13^3$.

Group the prime factors into triplets: $(13 \times 13 \times 13) = 13^3$.

To find the cube root, take one factor from the triplet: 13.

So, $\sqrt[3]{2197} = 13$.

To determine if 176 is a perfect square and, if not, find the smallest multiplier to make it one, we use prime factorization.

Find the prime factorization of 176:

The prime factorization of 176 is $2 \times 2 \times 2 \times 2 \times 11$.

In exponential form, this is $176 = 2^4 \times 11^1$.

For a number to be a perfect square, the exponents of all its prime factors must be even.

In the factorization $2^4 \times 11^1$:

• The exponent of 2 is 4, which is even.
• The exponent of 11 is 1, which is odd.

Since the exponent of the prime factor 11 is odd, 176 is not a perfect square.

To make 176 a perfect square by multiplication, we need to make the exponents of all prime factors even. The exponent of 2 is already even (4). The exponent of 11 is 1 (odd). To make the exponent of 11 even, we need to multiply by another factor of 11, which will make the exponent $1+1=2$.

So, we need to multiply 176 by 11.

The new number will be $176 \times 11 = (2^4 \times 11^1) \times 11^1 = 2^4 \times 11^{1+1} = 2^4 \times 11^2$.

In the factorization $2^4 \times 11^2$, the exponents are 4 (even) and 2 (even). This is a perfect square.

The square root of the new number is $\sqrt{2^4 \times 11^2} = 2^{4/2} \times 11^{2/2} = 2^2 \times 11^1 = 4 \times 11 = 44$.

The smallest number by which 176 should be multiplied to get a perfect square is 11.

The resulting perfect square is $176 \times 11 = 1936$. ($44^2 = 1936$).

Answer: 176 is not a perfect square. The smallest number by which it should be multiplied to get a perfect square is 11.

To determine if 9720 is a perfect cube and, if not, find the smallest number by which it should be divided to make it one, we use prime factorization.

Find the prime factorization of 9720:

The prime factorization of 9720 is $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5$.

In exponential form, this is $9720 = 2^3 \times 3^5 \times 5^1$.

For a number to be a perfect cube, the exponents of all its prime factors must be multiples of 3.

In the factorization $2^3 \times 3^5 \times 5^1$:

• The exponent of 2 is 3, which is a multiple of 3.
• The exponent of 3 is 5, which is not a multiple of 3.
• The exponent of 5 is 1, which is not a multiple of 3.

Since the exponents of the prime factors 3 and 5 are not multiples of 3, 9720 is not a perfect cube.

To make 9720 a perfect cube by division, we need to remove the prime factors that do not have exponents as multiples of 3.

The exponent of 3 is 5. The largest multiple of 3 less than or equal to 5 is 3. So we need to divide by $3^{5-3} = 3^2$.

The exponent of 5 is 1. The largest multiple of 3 less than or equal to 1 is 0. So we need to divide by $5^{1-0} = 5^1$.

The smallest number by which 9720 should be divided is $3^2 \times 5^1 = 9 \times 5 = 45$.

Dividing 9720 by 45:

$\frac{9720}{45} = \frac{2^3 \times 3^5 \times 5^1}{3^2 \times 5^1} = 2^3 \times 3^{5-2} \times 5^{1-1} = 2^3 \times 3^3 \times 5^0 = 2^3 \times 3^3 \times 1 = (2 \times 3)^3 = 6^3$.

The resulting number is $2^3 \times 3^3 = 8 \times 27 = 216$.

The number 216 is a perfect cube ($6^3 = 216$).

Answer: 9720 is not a perfect cube. The smallest number by which it should be divided to get a perfect cube is 45.

A Pythagorean triplet consists of three positive integers $a, b, c$ such that $a^2 + b^2 = c^2$. We are looking for triplets where one of the numbers is 5.

Case 1: 5 is one of the legs (say $a=5$).

We need to find positive integers $b$ and $c$ such that $5^2 + b^2 = c^2$.

$25 + b^2 = c^2$.

Rearranging, $c^2 - b^2 = 25$.

Using the difference of squares formula, $(c - b)(c + b) = 25$.

Since $b$ and $c$ are positive integers, $c+b$ and $c-b$ are positive integers. Also, $c > b$, so $c+b > c-b$.

We need to find pairs of factors of 25 whose product is 25.

Possible pairs of positive integer factors of 25 are (1, 25) and (5, 5).

Let $c - b = p$ and $c + b = q$, where $pq = 25$ and $q > p$.

$c = \frac{p+q}{2}$ and $b = \frac{q-p}{2}$. For $b$ and $c$ to be integers, $p$ and $q$ must have the same parity (both even or both odd).

Consider the factor pairs of 25:

Pair (1, 25): $p=1, q=25$. Both are odd. $q > p$ is satisfied.

$c = \frac{1+25}{2} = \frac{26}{2} = 13$.

$b = \frac{25-1}{2} = \frac{24}{2} = 12$.

This gives the triplet (5, 12, 13). Let's check: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. This is a Pythagorean triplet.

Pair (5, 5): $p=5, q=5$. Both are odd. However, $q > p$ is not satisfied (they are equal). This case would lead to $b = \frac{5-5}{2} = 0$, which is not a positive integer.

Case 2: 5 is the hypotenuse ($c=5$).

We need to find positive integers $a$ and $b$ such that $a^2 + b^2 = 5^2$.

$a^2 + b^2 = 25$.

We need to find two perfect squares that add up to 25.

Possible perfect squares less than 25 are 1, 4, 9, 16.

Let's check sums of pairs from $\{1, 4, 9, 16\}$:

• $1 + 4 = 5 \neq 25$
• $1 + 9 = 10 \neq 25$
• $1 + 16 = 17 \neq 25$
• $4 + 9 = 13 \neq 25$
• $4 + 16 = 20 \neq 25$
• $9 + 16 = 25$. Yes!

So, $a^2$ and $b^2$ could be 9 and 16.

If $a^2 = 9$, then $a = \sqrt{9} = 3$ (since $a$ is positive).

If $b^2 = 16$, then $b = \sqrt{16} = 4$ (since $b$ is positive).

This gives the triplet (3, 4, 5). Let's check: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$. This is a Pythagorean triplet.

We have found two Pythagorean triplets where 5 is one of the numbers: (3, 4, 5) and (5, 12, 13).

The order of the legs does not matter, so (4, 3, 5) is the same triplet as (3, 4, 5).

Two Pythagorean triplets each having one of the numbers as 5 are (3, 4, 5) and (5, 12, 13).

To find the smallest number by which 216 should be divided to make it a perfect square, we need to find the prime factorization of 216.

Prime factorization of 216:

So, the prime factorization of 216 is $2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3$.

For a number to be a perfect square, the exponents of all prime factors in its prime factorization must be even.

In the prime factorization of 216 ($2^3 \times 3^3$), the exponents of both 2 and 3 are 3, which is odd.

To make the exponents even by division, we need to divide by the prime factors that have odd exponents, each raised to the power that makes the resulting exponent even. In this case, we need to divide by $2^1$ and $3^1$ to reduce the exponents of 2 and 3 from 3 to 2.

The smallest number to divide by is the product of these factors:

Thus, 216 should be divided by 6 to get a perfect square.

Now, let's find the quotient when 216 is divided by 6:

The quotient is 36.

Let's check its prime factorization: $36 = 2^2 \times 3^2$. The exponents are even, so 36 is a perfect square.

Finally, we need to find the square root of the quotient (36).

The square root of the quotient is 6.

To find the smallest number by which 3600 should be multiplied to make it a perfect cube, we first need to find the prime factorization of 3600.

Prime factorization of 3600:

$3600 = 36 \times 100 = (6 \times 6) \times (10 \times 10) = (2 \times 3 \times 2 \times 3) \times (2 \times 5 \times 2 \times 5)$

Rearranging the factors:

For a number to be a perfect cube, the exponents of all prime factors in its prime factorization must be a multiple of 3.

In the prime factorization of 3600 ($2^4 \times 3^2 \times 5^2$), the exponents are 4, 2, and 2.

To make the exponents multiples of 3 by multiplication, we need to add factors to make each exponent the next multiple of 3. The next multiple of 3 after 4 is 6, after 2 is 3.

• For $2^4$, we need $2^{6-4} = 2^2$ more factors of 2.
• For $3^2$, we need $3^{3-2} = 3^1$ more factors of 3.
• For $5^2$, we need $5^{3-2} = 5^1$ more factors of 5.

The smallest number to multiply by is the product of these needed factors:

Thus, 3600 should be multiplied by 60 to get a perfect cube.

Now, let's find the quotient (product in this case) when 3600 is multiplied by 60:

The product is $3600 \times 60 = 216000$.

Finally, we need to find the cube root of the product (216000).

The cube root of $2^6 \times 3^3 \times 5^3$ is found by dividing each exponent by 3:

The cube root of the quotient (product) is 60.

(a) Finding the square root of 1369 by long division:

First, we pair the digits of 1369 from right to left. We get the pairs 13 and 69.

Now, we perform the long division:

Steps:

1. Pair the digits from the right: 13 69.

2. Find the largest number whose square is less than or equal to the first pair (13). $3^2 = 9$. Write 3 as the divisor and quotient. Subtract 9 from 13 to get 4.

3. Bring down the next pair (69). The new dividend is 469.

4. Double the current quotient (3 $\times$ 2 = 6). Write 6 followed by a blank space as the next divisor (6_).

5. Find a digit to fill the blank space such that (6_ $\times$ _) is less than or equal to 469. We try 7: $67 \times 7 = 469$.

6. Write 7 as the next digit in the quotient and in the divisor. Subtract 469 from 469 to get 0.

7. The remainder is 0. The quotient is 37.

Thus, the square root of 1369 is 37.

(b) Finding the square root of 5625 by long division:

First, we pair the digits of 5625 from right to left. We get the pairs 56 and 25.

Now, we perform the long division:

Steps:

1. Pair the digits from the right: 56 25.

2. Find the largest number whose square is less than or equal to the first pair (56). $7^2 = 49$. Write 7 as the divisor and quotient. Subtract 49 from 56 to get 7.

3. Bring down the next pair (25). The new dividend is 725.

4. Double the current quotient (7 $\times$ 2 = 14). Write 14 followed by a blank space as the next divisor (14_).

5. Find a digit to fill the blank space such that (14_ $\times$ _) is less than or equal to 725. We try 5: $145 \times 5 = 725$.

6. Write 5 as the next digit in the quotient and in the divisor. Subtract 725 from 725 to get 0.

7. The remainder is 0. The quotient is 75.

Thus, the square root of 5625 is 75.

(a) Finding the square root of 27.04 by long division:

First, we pair the digits of 27.04 starting from the decimal point. For the integer part, we pair from right to left (27). For the decimal part, we pair from left to right (04).

Now, we perform the long division:

Steps:

1. Pair the digits: 27 . 04.

2. Find the largest number whose square is less than or equal to the first pair (27). $5^2 = 25$. Write 5 as the divisor and quotient. Subtract 25 from 27 to get 2.

3. Bring down the next pair (04). Since we are bringing down the decimal part, place a decimal point in the quotient. The new dividend is 204.

4. Double the current quotient (5 $\times$ 2 = 10). Write 10 followed by a blank space as the next divisor (10_).

5. Find a digit to fill the blank space such that (10_ $\times$ _) is less than or equal to 204. We try 2: $102 \times 2 = 204$.

6. Write 2 as the next digit in the quotient and in the divisor. Subtract 204 from 204 to get 0.

7. The remainder is 0. The quotient is 5.2.

Thus, the square root of 27.04 is 5.2.

(b) Finding the square root of 1.44 by long division:

First, we pair the digits of 1.44 starting from the decimal point. For the integer part, we pair from right to left (1). For the decimal part, we pair from left to right (44).

Now, we perform the long division:

Steps:

1. Pair the digits: 1 . 44.

2. Find the largest number whose square is less than or equal to the first pair (1). $1^2 = 1$. Write 1 as the divisor and quotient. Subtract 1 from 1 to get 0.

3. Bring down the next pair (44). Since we are bringing down the decimal part, place a decimal point in the quotient. The new dividend is 44.

4. Double the current quotient (1 $\times$ 2 = 2). Write 2 followed by a blank space as the next divisor (2_).

5. Find a digit to fill the blank space such that (2_ $\times$ _) is less than or equal to 44. We try 2: $22 \times 2 = 44$.

6. Write 2 as the next digit in the quotient and in the divisor. Subtract 44 from 44 to get 0.

7. The remainder is 0. The quotient is 1.2.

Thus, the square root of 1.44 is 1.2.

To find the least number that should be subtracted from 1385 to get a perfect square, we will use the long division method to find the square root of 1385.

First, we pair the digits of 1385 from right to left. We get the pairs 13 and 85.

Now, we perform the long division:

Steps:

1. Pair the digits from the right: 13 85.

2. Find the largest number whose square is less than or equal to the first pair (13). $3^2 = 9$. Write 3 as the divisor and quotient. Subtract 9 from 13 to get 4.

3. Bring down the next pair (85). The new dividend is 485.

4. Double the current quotient (3 $\times$ 2 = 6). Write 6 followed by a blank space as the next divisor (6_).

5. Find a digit to fill the blank space such that (6_ $\times$ _) is less than or equal to 485. We find that $67 \times 7 = 469$, which is less than 485. $68 \times 8 = 544$, which is greater than 485.

6. Write 7 as the next digit in the quotient and in the divisor. Subtract 469 from 485 to get 16.

7. The remainder is 16. Since there are no more pairs, the process stops.

The remainder obtained in the long division is 16. This means that 1385 is 16 more than a perfect square.

Therefore, the least number that should be subtracted from 1385 to get a perfect square is the remainder, which is 16.

The perfect square obtained after subtracting 16 is:

The square root of the perfect square (1369) is the quotient obtained in the long division method, which is 37.

The square root of the perfect square is 37.

To find the least number that should be added to 6200 to make it a perfect square, we will use the long division method to find the square root of 6200.

First, we pair the digits of 6200 from right to left. We get the pairs 62 and 00.

Now, we perform the long division:

Steps:

1. Pair the digits from the right: 62 00.

2. Find the largest number whose square is less than or equal to the first pair (62). $7^2 = 49$. Write 7 as the divisor and quotient. Subtract 49 from 62 to get 13.

3. Bring down the next pair (00). The new dividend is 1300.

4. Double the current quotient (7 $\times$ 2 = 14). Write 14 followed by a blank space as the next divisor (14_).

5. Find a digit to fill the blank space such that (14_ $\times$ _) is less than or equal to 1300. We find that $148 \times 8 = 1184$, which is less than 1300. $149 \times 9 = 1341$, which is greater than 1300.

6. Write 8 as the next digit in the quotient and in the divisor. Subtract 1184 from 1300 to get 116.

7. The remainder is 116. The quotient is 78.

From the long division, we find that $78^2 < 6200$. Specifically, $78^2 = 6084$.

The remainder 116 shows that 6200 is not a perfect square.

To get a perfect square, we need to find the square of the next consecutive integer after the quotient 78, which is 79.

The square of 79 is $79^2$.

Let's calculate $79 \times 79$:

The smallest perfect square greater than 6200 is 6241.

The least number that should be added to 6200 to make it a perfect square is the difference between this perfect square and 6200.

The least number that should be added to 6200 is 41.

The least number of four digits is 1000.

To find the least perfect square of four digits, we need to find the smallest integer whose square is a four-digit number. This integer will be greater than the square root of the smallest four-digit number (1000).

Let's find the square root of 1000 using the long division method.

Pair the digits of 1000 from right to left: 10 00.

Now, perform the long division:

Steps:

1. Pair the digits from the right: 10 00.

2. Find the largest number whose square is less than or equal to the first pair (10). $3^2 = 9$. Write 3 as the divisor and quotient. Subtract 9 from 10 to get 1.

3. Bring down the next pair (00). The new dividend is 100.

4. Double the current quotient (3 $\times$ 2 = 6). Write 6 followed by a blank space as the next divisor (6_).

5. Find a digit to fill the blank space such that (6_ $\times$ _) is less than or equal to 100. We find that $61 \times 1 = 61$, which is less than 100. $62 \times 2 = 124$, which is greater than 100.

6. Write 1 as the next digit in the quotient and in the divisor. Subtract 61 from 100 to get 39.

7. The remainder is 39. The quotient is 31.

The long division shows that $\sqrt{1000}$ is between 31 and 32.

This means that $31^2 < 1000 < 32^2$.

Let's calculate $31^2$ and $32^2$:

The number $31^2 = 961$ is a 3-digit number.

The number $32^2 = 1024$ is a 4-digit number.

Since 32 is the smallest integer whose square is 1000 or greater, $32^2$ is the smallest perfect square that is a four-digit number.

The least number of four digits that is a perfect square is 1024.

The greatest number of three digits is 999.

To find the greatest perfect square of three digits, we need to find the largest perfect square that is less than or equal to 999. We can do this by finding the square root of 999 using the long division method.

Pair the digits of 999 from right to left. We get the pairs 9 and 99.

Now, we perform the long division:

Steps:

1. Pair the digits from the right: 9 99.

2. Find the largest number whose square is less than or equal to the first pair (9). $3^2 = 9$. Write 3 as the divisor and quotient. Subtract 9 from 9 to get 0.

3. Bring down the next pair (99). The new dividend is 99.

4. Double the current quotient (3 $\times$ 2 = 6). Write 6 followed by a blank space as the next divisor (6_).

5. Find a digit to fill the blank space such that (6_ $\times$ _) is less than or equal to 99. We find that $61 \times 1 = 61$, which is less than 99. $62 \times 2 = 124$, which is greater than 99.

6. Write 1 as the next digit in the quotient and in the divisor. Subtract 61 from 99 to get 38.

7. The remainder is 38. The quotient is 31.

The remainder obtained in the long division is 38. This means that $31^2 \le 999$, and the next perfect square is $32^2$.

The largest integer whose square is less than or equal to 999 is 31.

So, the greatest perfect square less than or equal to 999 is $31^2$.

Let's calculate $31 \times 31$:

The number $961$ is a three-digit number and is a perfect square.

The next perfect square is $32^2 = 1024$, which is a four-digit number.

Therefore, the greatest number of three digits that is a perfect square is 961.

To find the least square number which is exactly divisible by 3, 4, 5, 6 and 8, we first need to find the least common multiple (LCM) of these numbers.

The LCM of 3, 4, 5, 6, and 8 is the smallest number that is divisible by each of these numbers.

Let's find the prime factorization of each number:

• $3 = 3^1$
• $4 = 2 \times 2 = 2^2$
• $5 = 5^1$
• $6 = 2 \times 3 = 2^1 \times 3^1$
• $8 = 2 \times 2 \times 2 = 2^3$

To find the LCM, we take the highest power of each prime factor that appears in any of the factorizations:

• Highest power of 2 is $2^3$ (from 8).
• Highest power of 3 is $3^1$ (from 3 and 6).
• Highest power of 5 is $5^1$ (from 5).

The least number divisible by 3, 4, 5, 6, and 8 is 120.

Now, we need to find the least square number that is divisible by 120. This means the required number must be a multiple of 120, and it must be a perfect square.

The prime factorization of 120 is $2^3 \times 3^1 \times 5^1$.

For a number to be a perfect square, the exponents of all prime factors in its prime factorization must be even.

In the factorization of 120, the exponents are 3, 1, and 1, which are all odd.

To make it a perfect square by multiplying by the smallest possible number, we need to multiply by the factors that will make each exponent even. We need to make the exponent of 2 even (from 3 to 4), the exponent of 3 even (from 1 to 2), and the exponent of 5 even (from 1 to 2).

The missing factors required are $2^1$, $3^1$, and $5^1$.

The smallest number to multiply 120 by to make it a perfect square is the product of these missing factors:

The least square number is the LCM multiplied by this multiplier:

Let's check the prime factorization of 3600: $3600 = 120 \times 30 = (2^3 \times 3^1 \times 5^1) \times (2^1 \times 3^1 \times 5^1) = 2^{3+1} \times 3^{1+1} \times 5^{1+1} = 2^4 \times 3^2 \times 5^2$.

The exponents (4, 2, 2) are all even, so 3600 is a perfect square ($3600 = 60^2$).

Since 3600 is a multiple of 120, it is divisible by 3, 4, 5, 6, and 8.

It is the least such number because we used the LCM and the smallest necessary factors to make the exponents even.

The least square number which is exactly divisible by 3, 4, 5, 6 and 8 is 3600.

Given that the length of the diagonal of the square is 10 cm.

Let the side length of the square be $s$ cm.

In a square, the diagonal divides the square into two right-angled triangles. The sides of the square are the legs of the right triangle, and the diagonal is the hypotenuse.

By the Pythagorean theorem, the relationship between the side ($s$) and the diagonal ($d$) of a square is:

Substitute the given diagonal length $d = 10$ cm into the equation:

Simplify the equation:

Divide by 2:

To find the side length $s$, take the square root of both sides:

Simplify the square root:

Alternatively, using the direct formula $s = \frac{d}{\sqrt{2}}$:

Rationalize the denominator:

The length of the side of the square is $5\sqrt{2}$ cm.

Let the decimal number be $x$.

According to the question, the number multiplied by itself is 51.84.

This can be written as:

To find the number $x$, we need to find the square root of 51.84.

We will use the long division method to find the square root of 51.84.

First, we pair the digits from the decimal point. The integer part (51) is paired from right to left. The decimal part (84) is paired from left to right.

The pairs are 51 and 84.

Now, we perform the long division:

Steps:

1. Pair the digits: 51 . 84.

2. Find the largest number whose square is less than or equal to the first pair (51). $7^2 = 49$. Write 7 as the divisor and quotient. Subtract 49 from 51 to get 2.

3. Bring down the decimal point and the next pair (84). Place a decimal point in the quotient after 7. The new dividend is 284.

4. Double the current quotient (7 $\times$ 2 = 14). Write 14 followed by a blank space as the next divisor (14_).

5. Find a digit to fill the blank space such that (14_ $\times$ _) is less than or equal to 284. We find that $142 \times 2 = 284$.

6. Write 2 as the next digit in the quotient and in the divisor. Subtract 284 from 284 to get 0.

7. The remainder is 0. The quotient is 7.2.

The square root of 51.84 is 7.2.

We can verify this by multiplying 7.2 by itself:

The decimal number is 7.2.

Let the decimal fraction be $y$.

According to the question, the decimal fraction multiplied by itself is 84.64.

This can be written as:

To find the decimal fraction $y$, we need to find the square root of 84.64.

We will use the long division method to find the square root of 84.64.

First, we pair the digits from the decimal point. The integer part (84) is paired from right to left. The decimal part (64) is paired from left to right.

The pairs are 84 and 64.

Now, we perform the long division:

Steps:

1. Pair the digits: 84 . 64.

2. Find the largest number whose square is less than or equal to the first pair (84). $9^2 = 81$. Write 9 as the divisor and quotient. Subtract 81 from 84 to get 3.

3. Bring down the decimal point and the next pair (64). Place a decimal point in the quotient after 9. The new dividend is 364.

4. Double the current quotient (9 $\times$ 2 = 18). Write 18 followed by a blank space as the next divisor (18_).

5. Find a digit to fill the blank space such that (18_ $\times$ _) is less than or equal to 364. We find that $182 \times 2 = 364$.

6. Write 2 as the next digit in the quotient and in the divisor. Subtract 364 from 364 to get 0.

7. The remainder is 0. The quotient is 9.2.

The square root of 84.64 is 9.2.

We can verify this by multiplying 9.2 by itself:

The decimal fraction is 9.2.

Given that the field is a square.

The length of the side of the square field is 150 m.

To find the area the farmer will have to plough, we need to calculate the area of the square field.

The formula for the area of a square with side length $s$ is:

Substitute the given side length, $s = 150$ m, into the formula:

Calculate the product:

The farmer will have to plough an area of 22500 $\text{m}^2$.

Let the number of unit squares on each side of the square graph paper be $n$.

Since the graph paper is square, the total number of unit squares is the area of the square, which is given by the square of the number of unit squares on one side.

We are given that the total number of unit squares is 256.

So, we have the equation:

To find the number of unit squares on each side ($n$), we need to find the square root of 256.

We will use the long division method to find the square root of 256.

First, we pair the digits of 256 from right to left. We get the pairs 2 and 56.

Now, we perform the long division:

Steps:

1. Pair the digits from the right: 2 56.

2. Find the largest number whose square is less than or equal to the first pair (2). $1^2 = 1$. Write 1 as the divisor and quotient. Subtract 1 from 2 to get 1.

3. Bring down the next pair (56). The new dividend is 156.

4. Double the current quotient (1 $\times$ 2 = 2). Write 2 followed by a blank space as the next divisor (2_).

5. Find a digit to fill the blank space such that (2_ $\times$ _) is less than or equal to 156. We find that $26 \times 6 = 156$.

6. Write 6 as the next digit in the quotient and in the divisor. Subtract 156 from 156 to get 0.

7. The remainder is 0. The quotient is 16.

The square root of 256 is 16.

The number of unit squares on each side of the square graph paper will be 16.

Given:

The length of one side of a cube is 15 m.

To Find:

The volume of the cube.

Solution:

The formula for the volume of a cube with side length $s$ is given by:

Substitute the given side length, $s = 15$ m, into the formula:

Calculate the value:

The volume of the cube is 3375 $\text{m}^3$.

Given:

Length of the rectangular field ($l$) = 80 m

Width of the rectangular field ($w$) = 18 m

To Find:

The length of the diagonal of the rectangular field.

Solution:

Let the length of the diagonal be $d$.

In a rectangle, the diagonal divides the rectangle into two right-angled triangles. The sides of the rectangle are the legs of the right triangle, and the diagonal is the hypotenuse.

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Applying the Pythagorean theorem to the rectangular field:

Substitute the given values of length ($l = 80$ m) and width ($w = 18$ m) into the equation:

Add the squares of the dimensions:

To find the length of the diagonal $d$, take the square root of both sides of the equation:

We can find the square root of 6724 using the long division method.

Pair the digits from the right: 67 24.

Steps:

1. Pair digits: 67 24.

2. Largest square $\le 67$ is $8^2 = 64$. Quotient is 8. Remainder is $67 - 64 = 3$.

3. Bring down 24. New dividend is 324.

4. Double quotient (8 $\times$ 2 = 16). New divisor form is 16_.

5. $162 \times 2 = 324$. Use digit 2. Quotient is 82. Remainder is $324 - 324 = 0$.

So, $\sqrt{6724} = 82$.

The length of the diagonal of the rectangular field is 82 m.

Given:

The perimeter of the square field is 96 m.

To Find:

The area of the square field.

Solution:

Let the length of each side of the square field be $s$ meters.

The formula for the perimeter of a square is:

We are given that the perimeter is 96 m.

So, we can set up the equation:

To find the side length $s$, divide both sides by 4:

Now that we have the side length, we can find the area of the square field.

The formula for the area of a square is:

Substitute the value of $s = 24$ m into the formula:

Calculate the value:

The area of the square field is 576 $\text{m}^2$.

Given:

The volume of the cube is 512 cm$^3$.

To Find:

The length of each side of the cube.

Solution:

Let the length of each side of the cube be $s$ cm.

The formula for the volume of a cube with side length $s$ is given by:

We are given that the volume $V = 512$ cm$^3$.

Substitute the given volume into the formula:

To find the side length $s$, we need to find the cube root of 512.

We need to find a number which, when multiplied by itself three times, equals 512.

We know that $8 \times 8 \times 8 = 64 \times 8 = 512$.

Therefore, the cube root of 512 is 8.

The length of each side of the cube is 8 cm.

Given:

The ratio of three numbers is 1 : 2 : 3.

The sum of their cubes is 4500.

To Find:

The three numbers.

Solution:

Let the three numbers be $x$, $2x$, and $3x$, where $x$ is a common factor.

According to the problem, the sum of the cubes of these numbers is 4500.

Cube each term:

Combine the terms with $x^3$:

Solve for $x^3$ by dividing both sides by 36:

Simplify the fraction:

Take the cube root of both sides to find $x$:

Now, find the three numbers using the value of $x=5$:

• First number = $x = 5$
• Second number = $2x = 2 \times 5 = 10$
• Third number = $3x = 3 \times 5 = 15$

The three numbers are 5, 10, and 15.

Given:

The room is square-shaped.

The length of the side of the square room is 6.5 m.

To Find:

The area of carpet required to cover the room.

Solution:

The area of carpet required is equal to the area of the square room.

Let the side length of the square room be $s$.

The formula for the area of a square is:

Substitute the given side length, $s = 6.5$ m, into the formula:

Calculate the product:

The number of square metres of carpet required for the room is 42.25 $\text{m}^2$.

Given:

The area of a square is equal to the area of a rectangle.

Dimensions of the rectangle: length ($l$) = 6.4 m, width ($w$) = 2.5 m.

To Find:

The length of the side of the square.

Solution:

First, find the area of the rectangle.

The formula for the area of a rectangle is:

Substitute the given dimensions:

Calculate the product:

Let the side of the square be $s$ meters.

The area of the square is given by the formula:

According to the problem, the area of the square is equal to the area of the rectangle:

To find the side length $s$, take the square root of both sides:

The length of the side of the square is 4 m.

Given:

The difference between two perfect cubes is 189.

The cube root of the smaller of the two numbers is 3.

To Find:

The cube root of the larger number.

Solution:

Let the two perfect cubes be $a^3$ and $b^3$, where $a$ and $b$ are integers.

Let the smaller number be $S$ and the larger number be $L$. So, $S = a^3$ and $L = b^3$.

The difference between the two perfect cubes is 189.

We are given that the cube root of the smaller number is 3.

So, the smaller perfect cube is $S = a^3 = 3^3 = 3 \times 3 \times 3 = 27$.

Now substitute the value of the smaller cube into the difference equation (i):

Solve for the larger number $L$:

The larger perfect cube is 216.

We need to find the cube root of the larger number ($L$).

We know that $6 \times 6 \times 6 = 36 \times 6 = 216$.

Therefore, the cube root of 216 is 6.

The cube root of the larger number is 6.

Given:

Total number of plants = 1024.

The number of plants in a row is the same as the number of rows.

To Find:

The number of plants in each row.

Solution:

Let the number of rows be $n$.

Let the number of plants in each row be $m$.

According to the problem, the number of plants in a row is the same as the number of rows. So, $m = n$.

The total number of plants in the arrangement is the product of the number of rows and the number of plants in each row.

Since $m=n$:

We are given that the total number of plants is 1024.

So, we have the equation:

To find the number of rows ($n$), we need to find the square root of 1024.

We can use the long division method to find the square root of 1024.

Pair the digits of 1024 from right to left. We get the pairs 10 and 24.

Steps:

1. Pair the digits: 10 24.

2. Find the largest number whose square is less than or equal to the first pair (10). $3^2 = 9$. Write 3 as the divisor and quotient. Subtract 9 from 10 to get 1.

3. Bring down the next pair (24). The new dividend is 124.

4. Double the current quotient (3 $\times$ 2 = 6). Write 6 followed by a blank space as the next divisor (6_).

5. Find a digit to fill the blank space such that (6_ $\times$ _) is less than or equal to 124. We find that $62 \times 2 = 124$.

6. Write 2 as the next digit in the quotient and in the divisor. Subtract 124 from 124 to get 0.

7. The remainder is 0. The quotient is 32.

The square root of 1024 is 32.

Since the number of plants in a row is the same as the number of rows, the number of plants in each row is also 32.

The number of plants in each row is 32.

Given:

Total capacity of the hall (total number of seats) = 2704.

The number of rows is equal to the number of seats in each row.

To Find:

The number of seats in each row.

Solution:

Let the number of rows be $x$.

According to the problem, the number of seats in each row is also $x$.

The total number of seats in the hall is the product of the number of rows and the number of seats in each row.

We are given that the total number of seats is 2704.

So, we have the equation:

To find the number of seats in each row ($x$), we need to find the square root of 2704.

We can use the long division method to find the square root of 2704.

Pair the digits of 2704 from right to left. We get the pairs 27 and 04.

Steps:

1. Pair the digits from the right: 27 04.

2. Find the largest number whose square is less than or equal to the first pair (27). $5^2 = 25$. Write 5 as the divisor and quotient. Subtract 25 from 27 to get 2.

3. Bring down the next pair (04). The new dividend is 204.

4. Double the current quotient (5 $\times$ 2 = 10). Write 10 followed by a blank space as the next divisor (10_).

5. Find a digit to fill the blank space such that (10_ $\times$ _) is less than or equal to 204. We find that $102 \times 2 = 204$.

6. Write 2 as the next digit in the quotient and in the divisor. Subtract 204 from 204 to get 0.

7. The remainder is 0. The quotient is 52.

The square root of 2704 is 52.

Since the number of seats in each row is $x$, the number of seats in each row is 52.

The number of seats in each row is 52.

Given:

Total number of soldiers = 7500.

The General wants to arrange the soldiers in the form of a square.

To Find:

The number of soldiers left out after the arrangement.

Solution:

Arranging soldiers in the form of a square means that the number of rows is equal to the number of soldiers in each row. Let this number be $n$. The total number of soldiers in a perfect square formation would be $n^2$.

Since some soldiers were left out from 7500, this means 7500 is not a perfect square. We need to find the largest perfect square number less than 7500.

To find the largest perfect square less than or equal to 7500, we can find the square root of 7500 using the long division method.

First, we pair the digits of 7500 from right to left. We get the pairs 75 and 00.

Now, we perform the long division:

Steps:

1. Pair the digits from the right: 75 00.

2. Find the largest number whose square is less than or equal to the first pair (75). $8^2 = 64$. Write 8 as the divisor and quotient. Subtract 64 from 75 to get 11.

3. Bring down the next pair (00). The new dividend is 1100.

4. Double the current quotient (8 $\times$ 2 = 16). Write 16 followed by a blank space as the next divisor (16_).

5. Find a digit to fill the blank space such that (16_ $\times$ _) is less than or equal to 1100. We find that $166 \times 6 = 996$, which is less than 1100. $167 \times 7 = 1169$, which is greater than 1100.

6. Write 6 as the next digit in the quotient and in the divisor. Subtract 996 from 1100 to get 104.

7. The remainder is 104. The quotient is 86.

From the long division, we see that the quotient is 86 and the remainder is 104.

This means that $86^2 < 7500$. The largest perfect square less than 7500 is $86^2$.

Calculate $86^2$:

Let's calculate $86 \times 86$:

So, the largest number of soldiers that can be arranged in a perfect square is 7396.

The number of soldiers left out is the difference between the total number of soldiers and the largest perfect square number of soldiers that can be arranged.

Calculate the difference:

The remainder from the long division (104) also directly gives the number of soldiers left out.

The number of soldiers left out was 104.

Given:

Total number of students in the lecture room = 8649.

The arrangement is such that the number of rows is equal to the number of students in each row.

To Find:

The number of students in each row.

Solution:

Let the number of rows in the lecture room be $n$.

According to the problem, the number of students in each row is also $n$.

The total number of students is found by multiplying the number of rows by the number of students in each row.

We are given that the total number of students is 8649.

So, we can write the equation:

To find the number of students in each row ($n$), we need to find the square root of 8649.

We will use the long division method to find the square root of 8649.

First, we pair the digits of 8649 from right to left. We get the pairs 86 and 49.

Now, we perform the long division:

Steps:

1. Pair the digits from the right: 86 49.

2. Find the largest number whose square is less than or equal to the first pair (86). $9^2 = 81$. Write 9 as the divisor and quotient. Subtract 81 from 86 to get 5.

3. Bring down the next pair (49). The new dividend is 549.

4. Double the current quotient (9 $\times$ 2 = 18). Write 18 followed by a blank space as the next divisor (18_).

5. Find a digit to fill the blank space such that (18_ $\times$ _) is less than or equal to 549. We find that $183 \times 3 = 549$.

6. Write 3 as the next digit in the quotient and in the divisor. Subtract 549 from 549 to get 0.

7. The remainder is 0. The quotient is 93.

The square root of 8649 is 93.

Since the number of students in each row is $n$, the number of students in each row is 93.

There were 93 students in each row of the lecture room.

Given:

Distance walked North = 12 m

Distance walked West = 35 m

The return path is diagonal from the friend's house to Rahul's house.

To Find:

The distance Rahul walked while returning.

Solution:

Let Rahul's house be at point A. He walks 12 m North to point B, and then turns West to walk 35 m to reach his friend's house at point C.

Since North and West directions are perpendicular to each other, the path forms a right-angled triangle ABC, where AB is the distance walked North, BC is the distance walked West, and $\angle\text{ABC} = 90^\circ$.

The return path is the diagonal distance directly from point C (friend's house) to point A (Rahul's house). This is the hypotenuse of the right-angled triangle ABC.

Let the distance Rahul walked while returning (the length of the hypotenuse AC) be $d$ meters.

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Substitute the given distances: AB = 12 m and BC = 35 m.

Calculate the squares:

Substitute these values back into the equation:

Add the numbers:

To find the distance $d$, take the square root of both sides:

We need to find the square root of 1369. Using the long division method (as shown in Question 103(a)):

So, $\sqrt{1369} = 37$.

The distance Rahul walked while returning is 37 m.

Given:

Length of the ladder = 5.5 m

Height the ladder reaches on the wall = 4.4 m

To Find:

The distance between the wall and the foot of the ladder.

Solution:

Let the wall be vertical and the ground be horizontal. A ladder leaned against the wall forms a right-angled triangle with the wall and the ground. The ladder is the hypotenuse, the height on the wall is one leg, and the distance from the wall to the foot of the ladder is the other leg.

Let:

• $c$ be the length of the ladder ($c = 5.5$ m).
• $a$ be the height the ladder reaches on the wall ($a = 4.4$ m).
• $b$ be the distance between the wall and the foot of the ladder (what we need to find).

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:

Substitute the given values into the equation:

Calculate the squares:

Substitute these values back into the equation:

Solve for $b^2$ by subtracting 19.36 from both sides:

Calculate the difference:

To find the distance $b$, take the square root of both sides:

We can find the square root of 10.89 using the long division method.

Pair the digits from the decimal point: 10 . 89.

So, $\sqrt{10.89} = 3.3$.

The distance between the wall and the foot of the ladder is 3.3 m.

Given:

Number of gold coins received each day follows a pattern: 1, 3, 5, ...

The reward is for a month, which is stated as 30 days.

To Find:

Total number of coins the advisor will get in that month (30 days), without making calculations.

Solution:

The pattern of coins received each day is 1, 3, 5, ...

This sequence represents the consecutive odd numbers.

• Day 1: 1 coin (1st odd number)
• Day 2: 3 coins (2nd odd number)
• Day 3: 5 coins (3rd odd number)
• ...

The total number of coins received over a certain number of days is the sum of the odd numbers for that many days.

There is a known mathematical property about the sum of the first $n$ consecutive odd numbers.

The advisor receives coins for 30 days, following this pattern. This means we need to find the sum of the first 30 odd numbers.

Using the property mentioned above, the total number of coins for 30 days ($n=30$) is $30^2$.

Now, we calculate $30^2$:

So, the total number of coins the advisor will get is 900.

The phrase "Without making calculations" refers to the direct application of the property $\sum_{i=1}^{n} (2i-1) = n^2$, rather than summing 1 + 3 + 5 + ... up to the 30th term manually.

The advisor will get a total of 900 gold coins in that month.

Given:

Three numbers are in the ratio 2 : 3 : 5.

The sum of the squares of these numbers is 608.

To Find:

The three numbers.

Solution:

Let the three numbers be represented by $2k$, $3k$, and $5k$, where $k$ is a non-zero common factor.

According to the problem, the sum of the squares of these numbers is 608.

Square each term:

Combine the terms with $k^2$:

Solve for $k^2$ by dividing both sides by 38:

Perform the division:

To find the value of $k$, take the square root of both sides:

Now, find the three numbers using the value of $k=4$:

• First number = $2k = 2 \times 4 = 8$
• Second number = $3k = 3 \times 4 = 12$
• Third number = $5k = 5 \times 4 = 20$

The three numbers are 8, 12, and 20.

Verification: Sum of squares $= 8^2 + 12^2 + 20^2 = 64 + 144 + 400 = 208 + 400 = 608$. This matches the given condition.

To find the smallest square number divisible by each of the numbers 8, 9, and 10, we first need to find the least common multiple (LCM) of these numbers.

The LCM of 8, 9, and 10 is the smallest number that is divisible by each of these numbers.

Let's find the prime factorization of each number:

• $8 = 2 \times 2 \times 2 = 2^3$
• $9 = 3 \times 3 = 3^2$
• $10 = 2 \times 5 = 2^1 \times 5^1$

To find the LCM, we take the highest power of each prime factor that appears in any of the factorizations:

• Highest power of 2 is $2^3$ (from 8).
• Highest power of 3 is $3^2$ (from 9).
• Highest power of 5 is $5^1$ (from 10).

The least number divisible by 8, 9, and 10 is 360.

Now, we need to find the smallest perfect square number that is divisible by 360. This means the required number must be a multiple of 360, and it must be a perfect square.

The prime factorization of 360 is $2^3 \times 3^2 \times 5^1$.

For a number to be a perfect square, the exponents of all prime factors in its prime factorization must be even.

In the factorization of 360, the exponents are 3 (for 2), 2 (for 3), and 1 (for 5).

To make it a perfect square by multiplying by the smallest possible number, we need to multiply by the factors that will make each exponent even. We need to make the exponent of 2 even (from 3 to 4) and the exponent of 5 even (from 1 to 2). The exponent of 3 (which is 2) is already even.

The missing factors required are $2^{4-3} = 2^1$ and $5^{2-1} = 5^1$.

The smallest number to multiply 360 by to make it a perfect square is the product of these missing factors:

The smallest square number is the LCM multiplied by this multiplier:

Let's check the prime factorization of 3600: $3600 = 360 \times 10 = (2^3 \times 3^2 \times 5^1) \times (2^1 \times 5^1) = 2^{3+1} \times 3^2 \times 5^{1+1} = 2^4 \times 3^2 \times 5^2$.

The exponents (4, 2, 2) are all even, so 3600 is a perfect square ($3600 = 60^2$).

Since 3600 is a multiple of 360, it is divisible by 8, 9, and 10.

It is the least such number because we used the LCM and the smallest necessary factors to make the exponents even.

The smallest square number divisible by each one of the numbers 8, 9 and 10 is 3600.

Given:

The area of a square plot is $101\frac{1}{400}$ m$^2$.

To Find:

The length of one side of the square plot.

Solution:

Let the length of one side of the square plot be $s$ meters.

The formula for the area of a square with side length $s$ is:

We are given the area in mixed fraction form. First, convert the mixed fraction to an improper fraction:

Now, substitute the area into the formula for the area of a square:

To find the side length $s$, take the square root of both sides:

Calculate the square root of the denominator:

Now, calculate the square root of the numerator, 40401, using the long division method.

Pair the digits from the right: 01, 04, 4 (make it 04). Pairs: 04 04 01.

The square root of 40401 is 201.

Substitute the square roots back into the equation for $s$:

Convert the improper fraction to a mixed number or a decimal:

As a decimal: $\frac{1}{20} = 0.05$. So, $s = 10.05$ m.

The length of one side of the plot is $10\frac{1}{20}$ m or 10.05 m.

To find the square root of 324 by the method of repeated subtraction, we repeatedly subtract consecutive odd numbers starting from 1 from 324 until the result is 0. The number of steps required to reach 0 is the square root.

Let's perform the repeated subtraction:

Step 1: $324 - 1 = 323$

Step 2: $323 - 3 = 320$

Step 3: $320 - 5 = 315$

Step 4: $315 - 7 = 308$

Step 5: $308 - 9 = 299$

Step 6: $299 - 11 = 288$

Step 7: $288 - 13 = 275$

Step 8: $275 - 15 = 260$

Step 9: $260 - 17 = 243$

Step 10: $243 - 19 = 224$

Step 11: $224 - 21 = 203$

Step 12: $203 - 23 = 180$

Step 13: $180 - 25 = 155$

Step 14: $155 - 27 = 128$

Step 15: $128 - 29 = 99$

Step 16: $99 - 31 = 68$

Step 17: $68 - 33 = 35$

Step 18: $35 - 35 = 0$

We reached 0 in 18 steps.

Therefore, the square root of 324 is the number of steps taken.

The square root of 324 is 18.

Given:

The ratio of three numbers is 2 : 3 : 4.

The sum of the cubes of these numbers is 0.334125.

To Find:

The three numbers.

Solution:

Let the three numbers be represented by $2k$, $3k$, and $4k$, where $k$ is a non-zero common factor.

According to the problem, the sum of the cubes of these numbers is 0.334125.

Cube each term:

Combine the terms with $k^3$:

Solve for $k^3$ by dividing both sides by 99:

To simplify the division, we can write the decimal as a fraction:

So, the equation becomes:

Divide 334125 by 99:

Substitute this back into the equation for $k^3$:

To find the value of $k$, take the cube root of both sides:

Find the cube roots:

• Prime factorization of 3375: $3375 = 5 \times 675 = 5 \times 5 \times 135 = 5 \times 5 \times 5 \times 3 \times 3 \times 3 = 3^3 \times 5^3 = (3 \times 5)^3 = 15^3$. So, $\sqrt[3]{3375} = 15$.
• Cube root of 1000000: $1000000 = 10^6 = (10^2)^3 = 100^3$. So, $\sqrt[3]{1000000} = 100$.

Substitute the cube roots back into the equation for $k$:

Simplify the fraction or convert to decimal:

Now, find the three numbers using the value of $k=0.15$:

• First number $= 2k = 2 \times 0.15 = 0.3$
• Second number $= 3k = 3 \times 0.15 = 0.45$
• Third number $= 4k = 4 \times 0.15 = 0.6$

The three numbers are 0.3, 0.45, and 0.6.

Verification: Sum of cubes $= (0.3)^3 + (0.45)^3 + (0.6)^3 = 0.027 + 0.091125 + 0.216 = 0.334125$. This matches the given condition.

We need to evaluate the expression $\sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064}$.

First, let's find the cube root of each term separately.

1. Find $\sqrt[3]{27}$:

2. Find $\sqrt[3]{0.008}$:

We can write 0.008 as a fraction:

Now, find the cube root:

We know that $\sqrt[3]{8} = 2$ (since $2^3 = 8$) and $\sqrt[3]{1000} = 10$ (since $10^3 = 1000$).

3. Find $\sqrt[3]{0.064}$:

We can write 0.064 as a fraction:

Now, find the cube root:

We know that $\sqrt[3]{64} = 4$ (since $4^3 = 64$) and $\sqrt[3]{1000} = 10$.

Now, substitute the values back into the original expression and add them:

The evaluated value of the expression is 3.6.

We need to evaluate the expression $\left\{ \left( 5^2 \;+\;(12^2)^{\frac{1}{2}} \right) \right\}^{3}$.

Follow the order of operations (Parentheses/Brackets, Exponents, Multiplication and Division, Addition and Subtraction).

First, evaluate the terms inside the innermost parentheses and exponents.

Calculate $5^2$:

Calculate $(12^2)^{\frac{1}{2}}$. Remember that raising to the power of $\frac{1}{2}$ is the same as taking the square root.

Now, substitute these values back into the expression inside the main parentheses:

Perform the addition:

Finally, substitute this result back into the original expression and evaluate the outermost exponent (power of 3):

Calculate the value:

Let's perform the multiplication:

The evaluated value of the expression is 50653.

We need to evaluate the expression $\left\{ \left( 6^2 \;+\;(8^2)^{\frac{1}{2}} \right) \right\}^{3}$.

We follow the order of operations (BODMAS/PEMDAS).

First, evaluate the terms inside the innermost parentheses and exponents.

Calculate $6^2$:

Calculate $(8^2)^{\frac{1}{2}}$. Raising to the power of $\frac{1}{2}$ is equivalent to taking the square root.

Now, substitute these values back into the expression inside the main parentheses:

Perform the addition:

Finally, substitute this result back into the original expression and evaluate the outermost exponent (power of 3):

Calculate the value:

Let's perform the multiplication:

The evaluated value of the expression is 85184.

Given:

We are looking for a 4-digit number with the following properties:

• It is a perfect square.
• It has exactly four digits.
• None of its digits is zero.
• The digits from left to right are: Even, Even, Odd, Even.

To Find:

The number.

Solution:

Let the 4-digit number be $N$. Since $N$ is a 4-digit number, it must be between 1000 and 9999 (inclusive).

Since $N$ is a perfect square, its value is $s^2$ for some integer $s$.

The square root of the smallest 4-digit number is $\sqrt{1000} \approx 31.6$.

The square root of the largest 4-digit number is $\sqrt{9999} \approx 99.99$.

So, the square root $s$ must be an integer between 32 and 99 (inclusive).

Let the four digits of the number from left to right be $d_1$, $d_2$, $d_3$, and $d_4$. The number is $1000d_1 + 100d_2 + 10d_3 + d_4$.

According to the problem:

• $d_1$ is Even and non-zero: $d_1 \in \{2, 4, 6, 8\}$
• $d_2$ is Even and non-zero: $d_2 \in \{2, 4, 6, 8\}$
• $d_3$ is Odd: $d_3 \in \{1, 3, 5, 7, 9\}$
• $d_4$ is Even and non-zero: $d_4 \in \{2, 4, 6, 8\}$

Consider the last digit ($d_4$) of a perfect square. The possible last digits of a perfect square are 0, 1, 4, 5, 6, 9. Since $d_4$ must be an even and non-zero digit, the possible values for $d_4$ are 4 or 6.

Also, the last two digits of a perfect square formed by an even tens digit and an even units digit ($... \text{even} \text{ even}$) must be 00, 04, 16, 24, 36, 44, 56, 64, 76, 84, 96. The last two digits formed by an odd tens digit ($d_3$) and an even units digit ($d_4$) must follow a pattern related to the last digit of the square root.

• If $d_4 = 4$, the square root must end in 2 or 8. A perfect square ending in ...$d_34$ where $d_3$ is odd is the square of a number ending in 8 (e.g., $18^2=324$, $38^2=1444$).
• If $d_4 = 6$, the square root must end in 4 or 6. A perfect square ending in ...$d_36$ where $d_3$ is odd is the square of a number ending in 4 (e.g., $14^2=196$, $34^2=1156$).

So, the square root $s$ must end in either 8 (if the number ends in odd-4) or 4 (if the number ends in odd-6).

Let's test square roots $s$ between 32 and 99 that end in 4 or 8, and check if their squares meet all the digit criteria (Even, Even, Odd, Even, with no zeros).

Case 1: Square root $s$ ends in 4. Possible values for $s$: 34, 44, 54, 64, 74, 84, 94.

• $34^2 = 1156$ (Digits: 1, 1, 5, 6 - Odd, Odd, Odd, Even. Fails)
• $44^2 = 1936$ (Digits: 1, 9, 3, 6 - Odd, Odd, Odd, Even. Fails)
• $54^2 = 2916$ (Digits: 2, 9, 1, 6 - Even, Odd, Odd, Even. Fails $d_2$)
• $64^2 = 4096$ (Digits: 4, 0, 9, 6 - Even, Zero, Odd, Even. Fails $d_2$)
• $74^2 = 5476$ (Digits: 5, 4, 7, 6 - Odd, Even, Odd, Even. Fails $d_1$)
• $84^2 = 7056$ (Digits: 7, 0, 5, 6 - Odd, Zero, Odd, Even. Fails $d_1, d_2$)
• $94^2 = 8836$ (Digits: 8, 8, 3, 6 - Even, Even, Odd, Even. **Matches!**)

Case 2: Square root $s$ ends in 8. Possible values for $s$: 38, 48, 58, 68, 78, 88, 98.

• $38^2 = 1444$ (Digits: 1, 4, 4, 4 - Odd, Even, Even, Even. Fails $d_1, d_3$)
• $48^2 = 2304$ (Digits: 2, 3, 0, 4 - Even, Odd, Zero, Even. Fails $d_2, d_3$)
• $58^2 = 3364$ (Digits: 3, 3, 6, 4 - Odd, Odd, Even, Even. Fails $d_1, d_2, d_3$)
• $68^2 = 4624$ (Digits: 4, 6, 2, 4 - Even, Even, Even, Even. Fails $d_3$)
• $78^2 = 6084$ (Digits: 6, 0, 8, 4 - Even, Zero, Even, Even. Fails $d_2, d_3$)
• $88^2 = 7744$ (Digits: 7, 7, 4, 4 - Odd, Odd, Even, Even. Fails $d_1, d_2, d_3$)
• $98^2 = 9604$ (Digits: 9, 6, 0, 4 - Odd, Even, Zero, Even. Fails $d_1, d_3$)

The only perfect square number that satisfies all the given conditions is 8836.

Let's verify the digits of 8836:

• First digit (8): Even, non-zero (Yes).
• Second digit (8): Even, non-zero (Yes).
• Third digit (3): Odd (Yes).
• Fourth digit (6): Even, non-zero (Yes).
• None of the digits are zero (Yes).
• It is a 4-digit number (Yes).
• It is a perfect square ($94^2 = 8836$) (Yes).

The number is 8836.

Solution

Let the three numbers to be placed in the circles be $n_1$, $n_2$, and $n_3$. These numbers must be different from each other ($n_1 \neq n_2$, $n_1 \neq n_3$, $n_2 \neq n_3$).

The diagram shows three lines, each connecting a circle to a square containing a number. Let's assume the connections are fixed as implied by the layout:

• Circle 1 is connected to the square with the number 1.
• Circle 2 is connected to the square with the number 8.
• Circle 3 is connected to the square with the number 15.

Let the number placed in Circle 1 be $a$, the number in Circle 2 be $b$, and the number in Circle 3 be $c$. The numbers $a$, $b$, and $c$ must be distinct.

According to the problem, when we add the number in a circle to the number in the connected square, the result is a perfect square. So, we have the following conditions:

We need to find three distinct numbers $a$, $b$, and $c$ that satisfy these equations. We are looking for positive integer solutions for $a, b, c$ (as is typical for such puzzles).

From the equations, we can express $a$, $b$, and $c$ in terms of perfect squares:

Let's list some possible values for $a$, $b$, and $c$ by choosing integer values for $k_1$, $k_2$, and $k_3$ such that $a, b, c$ are positive integers:

• For $a = k_1^2 - 1$, we need $k_1^2 > 1$, so $k_1 \ge 2$. Possible values for $a$: $2^2-1=3$, $3^2-1=8$, $4^2-1=15$, $5^2-1=24$, ...
• For $b = k_2^2 - 8$, we need $k_2^2 > 8$, so $k_2 \ge 3$. Possible values for $b$: $3^2-8=1$, $4^2-8=8$, $5^2-8=17$, $6^2-8=28$, ...
• For $c = k_3^2 - 15$, we need $k_3^2 > 15$, so $k_3 \ge 4$. Possible values for $c$: $4^2-15=1$, $5^2-15=10$, $6^2-15=21$, $7^2-15=34$, ...

We need to find a combination of one value from the list of possible $a$'s, one from the list of possible $b$'s, and one from the list of possible $c$'s, such that the three chosen values are distinct.

Let's try to pick the smallest possible distinct values:

• Choose $a=3$ (from $a+1=4=2^2$).
• Choose $b=1$ (from $b+8=9=3^2$). $b=1$ is distinct from $a=3$.
• Now we need to choose $c$ from its list $\{1, 10, 21, 34, ...\}$ such that $c$ is distinct from $a=3$ and $b=1$. The smallest value in the list is 1, which is not distinct from $b$. The next value is 10. Let's choose $c=10$ (from $c+15=25=5^2$). $c=10$ is distinct from $a=3$ and $b=1$.

So, the three numbers we found are $a=3$, $b=1$, and $c=10$. These are three different numbers.

Let's verify the sums with the connected squares:

• Circle connected to 1 contains 3. Sum = $3 + 1 = 4$, which is $2^2$ (perfect square).
• Circle connected to 8 contains 1. Sum = $1 + 8 = 9$, which is $3^2$ (perfect square).
• Circle connected to 15 contains 10. Sum = $10 + 15 = 25$, which is $5^2$ (perfect square).

The conditions are satisfied, and the numbers 1, 3, and 10 are all different.

The three different numbers to be put in the circles are 1, 3, and 10.

To match the diagram's connections: put 3 in the circle connected to 1, put 1 in the circle connected to 8, and put 10 in the circle connected to 15.

Given:

Perimeter of the first square ($P_1$) = 40 m.

Perimeter of the second square ($P_2$) = 96 m.

To Find:

The perimeter of a third square whose area is equal to the sum of the areas of the first two squares.

Solution:

Let $s_1$ be the side length and $A_1$ be the area of the first square.

The formula for the perimeter of a square with side length $s$ is:

Given $P_1 = 40$ m, we can find $s_1$ using formula (i):

The formula for the area of a square with side length $s$ is:

Using $s_1 = 10$ m, the area of the first square is, using formula (ii):

Let $s_2$ be the side length and $A_2$ be the area of the second square.

Given $P_2 = 96$ m, using formula (i):

Using $s_2 = 24$ m, the area of the second square is, using formula (ii):

Let $s_3$ be the side length and $A_3$ be the area of the third square.

The area of the third square is equal to the sum of the areas of the first two squares:

Using the area formula (ii) for the third square, we have:

To find the side length $s_3$, take the square root of $A_3$:

We find that $26 \times 26 = 676$.

Finally, find the perimeter of the third square using formula (i) with side length $s_3 = 26$ m:

The perimeter of the third square is 104 metres.

We are looking for a 3-digit perfect square composed only of digits that are still recognisable when viewed upside down. These digits are 0, 1, 6, 8, and 9 (where 6 and 9 swap places, 0, 1, 8 remain the same).

Let's list the 3-digit perfect squares ($10^2=100$ to $31^2=961$) and check their digits:

• $10^2 = 100$. Digits are 1, 0, 0. All are interpretable upside down. Upside down, 100 remains 100. Is 100 a perfect square? Yes ($10^2$). So, 100 is a solution.
• $11^2 = 121$. Contains digit 2 (not interpretable).
• $13^2 = 169$. Digits are 1, 6, 9. All are interpretable upside down. Upside down, 169 becomes 196 (1->1, 6->9, 9->6). Is 196 a perfect square? Yes ($14^2$). So, 169 is a solution.
• $14^2 = 196$. Digits are 1, 9, 6. All are interpretable upside down. Upside down, 196 becomes 169 (1->1, 9->6, 6->9). Is 169 a perfect square? Yes ($13^2$). So, 196 is a solution.
• Other 3-digit perfect squares ($12^2, 15^2, \dots, 30^2, 31^2$) contain digits other than 0, 1, 6, 8, 9 (like 2, 3, 4, 5, 7) or result in non-perfect squares when flipped (e.g., $30^2=900$ flips to 600, which is not a perfect square; $31^2=961$ flips to 169, which is a perfect square, so 961 is also a potential answer).

The 3-digit perfect squares that fit the criteria are 100, 169, 196, and 961. The hint about 6 and 9 swapping strongly suggests that 169 or 196 is the intended answer, as they flip into each other.

One such number is 169.

Another such number is 196.

We are looking for pairs of numbers $(a, b)$ such that $b$ is the number $a$ with its digits reversed, and the square of $b$ ($b^2$) is the number $a^2$ with its digits reversed.

The given example is the pair (13, 31).

• The number $a = 13$. Its reverse is $b = 31$.
• The square of $a$ is $13^2 = 169$.
• The square of $b$ is $31^2 = 961$.
• The reverse of $a^2 = 169$ is 961.

Since $b^2 = 961$ and the reverse of $a^2$ is 961, the condition is satisfied for the pair (13, 31).

Let's search for other such pairs.

We can test pairs of numbers whose digits are reversed, starting with small numbers.

Consider pairs of two-digit numbers $(ab, ba)$ where $a, b \in \{1, 2, \dots, 9\}$.

Let's test the pair (12, 21).

• $a = 12$. Its reverse is $b = 21$.
• $a^2 = 12^2 = 144$.
• $b^2 = 21^2 = 441$.
• The reverse of $a^2 = 144$ is 441.

Since $b^2 = 441$ and the reverse of $a^2$ is 441, the condition is satisfied. The pair (12, 21) is one such pair.

Let's test the pair (14, 41).

• $a = 14$. Its reverse is $b = 41$.
• $a^2 = 14^2 = 196$.
• $b^2 = 41^2 = 1681$.
• The reverse of $a^2 = 196$ is 691.

Since $b^2 = 1681$ and the reverse of $a^2$ is 691 ($1681 \neq 691$), the condition is not satisfied for (14, 41).

Let's consider pairs involving digits that can result in particular end digits for squares upon reversal. The pairs found so far (12, 21) and (13, 31) lead to 3-digit squares. Let's try 3-digit numbers whose squares are 5-digit numbers, etc.

Consider the pair (102, 201).

• $a = 102$. Its reverse is $b = 201$.
• $a^2 = 102^2 = 10404$.
• $b^2 = 201^2 = 40401$.
• The reverse of $a^2 = 10404$ is 40401.

Since $b^2 = 40401$ and the reverse of $a^2$ is 40401, the condition is satisfied. The pair (102, 201) is another such pair.

We have found two pairs different from (13, 31): (12, 21) and (102, 201).

Let's verify the pairs:

Pair 1: (12, 21)

$(21)^2 = \text{Reverse}((12)^2)$. This pair works.

Pair 2: (102, 201)

$(201)^2 = \text{Reverse}((102)^2)$. This pair works.

Other such pairs also exist, for example, (103, 301): $103^2=10609$, $301^2=90601$, and $\text{Reverse}(10609)=90601$.

Two other such pairs are (12, 21) and (102, 201).