Chapter 5 Squares and Square Roots

The unit digit of the square of a number is determined by the unit digit of the number itself. We just need to find the unit digit of the square of the unit digit of the given number.

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(i) 81

The unit digit of 81 is 1.

The square of the unit digit is $1^2 = 1$.

The unit digit of the square of 81 is 1.

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(ii) 272

The unit digit of 272 is 2.

The square of the unit digit is $2^2 = 4$.

The unit digit of the square of 272 is 4.

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(iii) 799

The unit digit of 799 is 9.

The square of the unit digit is $9^2 = 81$.

The unit digit of 81 is 1.

The unit digit of the square of 799 is 1.

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(iv) 3853

The unit digit of 3853 is 3.

The square of the unit digit is $3^2 = 9$.

The unit digit of the square of 3853 is 9.

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(v) 1234

The unit digit of 1234 is 4.

The square of the unit digit is $4^2 = 16$.

The unit digit of 16 is 6.

The unit digit of the square of 1234 is 6.

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(vi) 26387

The unit digit of 26387 is 7.

The square of the unit digit is $7^2 = 49$.

The unit digit of 49 is 9.

The unit digit of the square of 26387 is 9.

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(vii) 52698

The unit digit of 52698 is 8.

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

The unit digit of 64 is 4.

The unit digit of the square of 52698 is 4.

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(viii) 99880

The unit digit of 99880 is 0.

The square of the unit digit is $0^2 = 0$.

The unit digit of the square of 99880 is 0.

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(ix) 12796

The unit digit of 12796 is 6.

The square of the unit digit is $6^2 = 36$.

The unit digit of 36 is 6.

The unit digit of the square of 12796 is 6.

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(x) 55555

The unit digit of 55555 is 5.

The square of the unit digit is $5^2 = 25$.

The unit digit of 25 is 5.

The unit digit of the square of 55555 is 5.

A number is a perfect square if it is the square of an integer. Perfect squares have certain properties based on their unit digit and the number of trailing zeros.

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The unit digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. A number ending in 2, 3, 7, or 8 cannot be a perfect square.

Also, if a number ends with zeros, the number of zeros must be even for it to be a perfect square.

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Let's examine each number:

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(i) 1057

The unit digit is 7. Since perfect squares cannot end in 7, 1057 is not a perfect square.

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(ii) 23453

The unit digit is 3. Since perfect squares cannot end in 3, 23453 is not a perfect square.

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(iii) 7928

The unit digit is 8. Since perfect squares cannot end in 8, 7928 is not a perfect square.

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(iv) 222222

The unit digit is 2. Since perfect squares cannot end in 2, 222222 is not a perfect square.

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(v) 64000

The number ends with three zeros (an odd number of zeros). For a number ending in zeros to be a perfect square, it must have an even number of trailing zeros. Therefore, 64000 is not a perfect square.

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(vi) 89722

The unit digit is 2. Since perfect squares cannot end in 2, 89722 is not a perfect square.

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(vii) 222000

The number ends with three zeros (an odd number of zeros). For a number ending in zeros to be a perfect square, it must have an even number of trailing zeros. Therefore, 222000 is not a perfect square.

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(viii) 505050

The number ends with one zero (an odd number of zeros). For a number ending in zeros to be a perfect square, it must have an even number of trailing zeros. Therefore, 505050 is not a perfect square.

The square of a number is odd if and only if the number itself is odd.

A number is odd if its unit digit is 1, 3, 5, 7, or 9.

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Let's examine the unit digit of each given number:

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(i) 431

The unit digit of 431 is 1.

Since the unit digit (1) is an odd number, the number 431 is odd.

Therefore, the square of 431 would be an odd number.

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(ii) 2826

The unit digit of 2826 is 6.

Since the unit digit (6) is an even number, the number 2826 is even.

Therefore, the square of 2826 would be an even number.

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(iii) 7779

The unit digit of 7779 is 9.

Since the unit digit (9) is an odd number, the number 7779 is odd.

Therefore, the square of 7779 would be an odd number.

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(iv) 82004

The unit digit of 82004 is 4.

Since the unit digit (4) is an even number, the number 82004 is even.

Therefore, the square of 82004 would be an even number.

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The squares of the following numbers would be odd numbers:

(i) 431

(iii) 7779

Let's observe the given pattern:

$11^2 = 121$ (1 zero between 1s in the base, 0 zeros between digits in the square)

$101^2 = 10201$ (1 zero between 1s in the base, 1 zero between digits in the square)

$1001^2 = 1002001$ (2 zeros between 1s in the base, 2 zeros between digits in the square)

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The pattern appears to be: If the number is of the form $10...01$ with 'n' zeros between the two 1s, its square is of the form $10...020...01$, where there are 'n' zeros between 1 and 2, and 'n' zeros between 2 and 1.

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Using this pattern, let's find the missing digits:

$100001^2 = 1 \ ......... \ 2 \ ......... \ 1$

The number is 100001. There are 4 zeros between the two 1s.

Following the pattern, there should be 4 zeros between 1 and 2, and 4 zeros between 2 and 1 in the square.

$100001^2 = 1 \underline{0000} \ 2 \underline{0000} \ 1$

So, $100001^2 = 10000200001$.

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$10000001^2 = \dots \dots \dots \dots$

The number is 10000001. There are 6 zeros between the two 1s.

Following the pattern, there should be 6 zeros between 1 and 2, and 6 zeros between 2 and 1 in the square.

$10000001^2 = 1 \underline{000000} \ 2 \underline{000000} \ 1$

So, $10000001^2 = 100000020000001$.

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The missing digits are:

$100001^2 = 1 \mathbf{0000} \ 2 \mathbf{0000} \ 1$

$10000001^2 = \mathbf{100000020000001}$

Let's observe the given pattern of the squares of numbers consisting of alternating 1s and 0s, starting and ending with 1:

$11^2 = 121$ (Base has 2 ones, 0 zeros between. Square has digits 1-2-1, 0 zeros between.)

$101^2 = 10201$ (Base has 2 ones, 1 zero between. Square has digits 1-2-1, 1 zero between.)

$10101^2 = 102030201$ (Base has 3 ones, 1 zero between each. Square has digits 1-2-3-2-1, 1 zero between each.)

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The pattern observed is that the square of a number consisting of 'n' ones separated by a constant number of zeros (let's say 'm' zeros) will have digits that count up from 1 to 'n' and then back down to 1, with 'm' zeros placed between each of these digits.

In the examples provided, the number of zeros between the 1s in the base is consistently 1 (except for the first case, where there are 0 zeros).

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Let's apply this pattern to the missing numbers:

$1010101^2 = \dots \dots \dots \dots$

The base number is 1010101. It consists of 4 ones, with 1 zero between each pair of ones.

Following the pattern, the digits in the square (ignoring zeros) will count up to 4 and back down: 1, 2, 3, 4, 3, 2, 1.

There should be 1 zero between each of these digits.

So, the square is formed by arranging these digits with a zero in between each pair:

1 0 2 0 3 0 4 0 3 0 2 0 1

Thus, $1010101^2 = 1020304030201$.

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............² = 10203040504030201

Look at the digits in the square, ignoring the zeros: 1, 2, 3, 4, 5, 4, 3, 2, 1.

The sequence counts up to 5 and then back down to 1. This indicates that the base number must contain 5 ones.

Now, look at the number of zeros between the digits in the square. There is 1 zero between each pair of digits (e.g., between 1 and 2, 2 and 3, etc.).

According to the pattern, this means there must be 1 zero between each pair of ones in the base number.

The base number should consist of 5 ones separated by 1 zero each.

The base number is: 1 0 1 0 1 0 1 0 1.

Thus, $101010101^2 = 10203040504030201$.

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The missing numbers are:

$1010101^2 = \mathbf{1020304030201}$

$\mathbf{101010101}^2 = 10203040504030201$

Let's observe the pattern given:

$1^2 + 2^2 + 2^2 = 3^2$

$2^2 + 3^2 + 6^2 = 7^2$

$3^2 + 4^2 + 12^2 = 13^2$

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Let the numbers in the pattern be $a$, $b$, $c$, and $d$ such that $a^2 + b^2 + c^2 = d^2$.

Observe the relationship between the first two numbers $a$ and $b$ in each row:

Row 1: $a=1, b=2$. Here $b = a+1$.

Row 2: $a=2, b=3$. Here $b = a+1$.

Row 3: $a=3, b=4$. Here $b = a+1$.

It seems the second number is always one more than the first number, i.e., $b = a+1$.

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Observe the relationship between the third number $c$ and the first two numbers $a$ and $b$:

Row 1: $a=1, b=2, c=2$. Here $c = a \times b = 1 \times 2 = 2$.

Row 2: $a=2, b=3, c=6$. Here $c = a \times b = 2 \times 3 = 6$.

Row 3: $a=3, b=4, c=12$. Here $c = a \times b = 3 \times 4 = 12$.

It seems the third number is the product of the first two numbers, i.e., $c = a \times b = a(a+1)$.

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Observe the relationship between the fourth number $d$ and the third number $c$:

Row 1: $c=2, d=3$. Here $d = c+1$.

Row 2: $c=6, d=7$. Here $d = c+1$.

Row 3: $c=12, d=13$. Here $d = c+1$.

It seems the fourth number is always one more than the third number, i.e., $d = c+1 = a(a+1) + 1$.

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So the pattern is of the form $a^2 + (a+1)^2 + (a(a+1))^2 = (a(a+1)+1)^2$.

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Now let's use this pattern to find the missing numbers:

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$4^2 + 5^2 + \_\_^{2} = 21^{2}$

Here, the first number $a=4$. The second number is $a+1 = 4+1 = 5$. This matches.

The third number should be $c = a(a+1) = 4 \times 5 = 20$.

The fourth number should be $d = c+1 = 20+1 = 21$. This matches the right side.

So the missing number is 20.

$4^2 + 5^2 + \mathbf{20}^{2} = 21^{2}$

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$5^2 + \_\_^{2} + 30^{2} = 31^{2}$

Here, the first number $a=5$.

The third number is $c=30$. The fourth number is $d=31$. Note that $d = c+1$, which fits the pattern.

The missing second number should be $b = a+1 = 5+1 = 6$.

Let's check if the third number is the product of the first two: $a \times b = 5 \times 6 = 30$. This matches the given third number.

So the missing number is 6.

$5^2 + \mathbf{6}^{2} + 30^{2} = 31^{2}$

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$6^2 + 7^2 + \_\_^{2} = \_\_^{2}$

Here, the first number $a=6$. The second number is $a+1 = 6+1 = 7$. This matches.

The third number should be $c = a(a+1) = 6 \times 7 = 42$.

The fourth number should be $d = c+1 = 42+1 = 43$.

So the missing numbers are 42 and 43.

$6^2 + 7^2 + \mathbf{42}^{2} = \mathbf{43}^{2}$

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The missing numbers are:

$4^2 + 5^2 + \mathbf{20}^{2} = 21^{2}$

$5^2 + \mathbf{6}^{2} + 30^{2} = 31^{2}$

$6^2 + 7^2 + \mathbf{42}^{2} = \mathbf{43}^{2}$

We know that the sum of the first n odd numbers is equal to $n^2$.

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(i) 1 + 3 + 5 + 7 + 9

This is the sum of the first 5 odd numbers (1, 3, 5, 7, 9).

Here, $n=5$.

The sum is $n^2 = 5^2 = 25$.

Sum = 25.

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(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

This is the sum of the first 10 odd numbers (1, 3, 5, 7, 9, 11, 13, 15, 17, 19).

Here, $n=10$.

The sum is $n^2 = 10^2 = 100$.

Sum = 100.

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(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

This is the sum of the first 12 odd numbers (1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23).

Here, $n=12$.

The sum is $n^2 = 12^2 = 144$.

Sum = 144.

We know that the sum of the first $n$ odd natural numbers is $n^2$. This property can be used to express a perfect square as the sum of consecutive odd numbers starting from 1.

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(i) Express 49 as the sum of 7 odd numbers.

The number is 49. We recognise that $49 = 7^2$.

According to the property, $49$ is the sum of the first 7 odd numbers.

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

So, we express 49 as their sum:

$49 = 1 + 3 + 5 + 7 + 9 + 11 + 13$

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(ii) Express 121 as the sum of 11 odd numbers.

The number is 121. We recognise that $121 = 11^2$.

According to the property, $121$ is the sum of the first 11 odd numbers.

The first 11 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, and 21.

So, we express 121 as their sum:

$121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21$

To find the number of non-perfect square numbers between the squares of two consecutive numbers, $n$ and $(n+1)$, we can observe the difference between their squares.

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The square of the first number is $n^2$.

The square of the second number is $(n+1)^2 = n^2 + 2n + 1$.

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The numbers between $n^2$ and $(n+1)^2$ are the integers starting from $n^2 + 1$ up to $(n+1)^2 - 1$.

The list of integers is $n^2+1, n^2+2, \dots, n^2+2n$.

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The number of integers between $n^2$ and $(n+1)^2$ is the total count of numbers from $(n^2+1)$ to $(n^2+2n)$.

Count = (Last number) - (First number) + 1

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

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

Count = $2n$

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So, there are $2n$ non-perfect square numbers between the squares of $n$ and $n+1$. We can use this formula directly.

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(i) 12 and 13

Here, the first number is $n = 12$. The next consecutive number is $n+1 = 13$.

The number of non-perfect square numbers between the square of 12 ($12^2$) and the square of 13 ($13^2$) is $2n$.

Number of numbers = $2 \times 12 = 24$.

There are 24 numbers between the squares of 12 and 13.

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(ii) 25 and 26

Here, the first number is $n = 25$. The next consecutive number is $n+1 = 26$.

The number of non-perfect square numbers between the square of 25 ($25^2$) and the square of 26 ($26^2$) is $2n$.

Number of numbers = $2 \times 25 = 50$.

There are 50 numbers between the squares of 25 and 26.

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(iii) 99 and 100

Here, the first number is $n = 99$. The next consecutive number is $n+1 = 100$.

The number of non-perfect square numbers between the square of 99 ($99^2$) and the square of 100 ($100^2$) is $2n$.

Number of numbers = $2 \times 99 = 198$.

There are 198 numbers between the squares of 99 and 100.

(i) 39

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We can write 39 as the difference of two numbers, for example, $40 - 1$.

Then, the square of 39 is $39^2 = (40 - 1)^2$.

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Using the algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$, where $a=40$ and $b=1$:

$(40 - 1)^2 = 40^2 - 2 \times 40 \times 1 + 1^2$

$= 1600 - 80 + 1$

$= 1520 + 1$

$= 1521$

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So, the square of 39 is 1521.

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(ii) 42

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We can write 42 as the sum of two numbers, for example, $40 + 2$.

Then, the square of 42 is $42^2 = (40 + 2)^2$.

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

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

$= 1600 + 160 + 4$

$= 1760 + 4$

$= 1764$

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So, the square of 42 is 1764.

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

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One way to generate Pythagorean triplets is using the formula $(2m, m^2 - 1, m^2 + 1)$ for any integer $m > 1$. The numbers in the triplet are $2m$, $m^2 - 1$, and $m^2 + 1$.

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We are given that the smallest member of the triplet is 8.

In the formula $(2m, m^2 - 1, m^2 + 1)$, for $m > 1$, $m^2 - 1$ and $m^2 + 1$ are the other two members. Since $m > 1$, $m^2 - 1 \geq 2^2 - 1 = 3$.

If $m=2$, the triplet is $(4, 3, 5)$, where 3 is the smallest.

If $m=3$, the triplet is $(6, 8, 10)$, where 6 is the smallest.

If $m=4$, the triplet is $(8, 15, 17)$. Let's check the members: $2m = 8$, $m^2 - 1 = 15$, $m^2 + 1 = 17$. The smallest member is 8.

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So, we can assume the smallest member, which is 8, corresponds to $2m$ in the formula.

Divide both sides by 2:

$m = \frac{8}{2}$

$m = 4$

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Since $m=4 > 1$, this value is valid for the formula.

Now substitute $m=4$ into the other parts of the triplet formula:

Second member = $m^2 - 1 = 4^2 - 1 = 16 - 1 = 15$.

Third member = $m^2 + 1 = 4^2 + 1 = 16 + 1 = 17$.

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The Pythagorean triplet is (8, 15, 17).

Let's verify if this is a valid Pythagorean triplet:

$8^2 + 15^2 = 64 + 225 = 289$

$17^2 = 289$

Since $8^2 + 15^2 = 17^2$, the triplet (8, 15, 17) is indeed a Pythagorean triplet.

The smallest member in this triplet is 8.

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Therefore, a Pythagorean triplet whose smallest member is 8 is (8, 15, 17).

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

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One way to generate Pythagorean triplets is using the formula $(2m, m^2 - 1, m^2 + 1)$, where $m$ is an integer greater than 1.

The three members of the triplet generated by this formula are $2m$, $m^2 - 1$, and $m^2 + 1$.

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We are given that one member of the triplet is 12.

Let's see if 12 can be obtained by setting one of the formula members equal to 12 and solving for $m$.

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Case 1: Assume $2m = 12$

Divide both sides by 2:

$m = \frac{12}{2}$

$m = 6$

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Since $m=6$ is an integer and $6 > 1$, this is a valid value for $m$.

Now, substitute $m=6$ into the other parts of the triplet formula:

Second member = $m^2 - 1 = 6^2 - 1 = 36 - 1 = 35$.

Third member = $m^2 + 1 = 6^2 + 1 = 36 + 1 = 37$.

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This gives the triplet (12, 35, 37).

Let's verify if this is a Pythagorean triplet:

$12^2 + 35^2 = 144 + 1225 = 1369$

$37^2 = 1369$

Since $12^2 + 35^2 = 37^2$, the triplet (12, 35, 37) is a valid Pythagorean triplet, and one of its members is 12.

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Case 2: Assume $m^2 - 1 = 12$

$m^2 - 1 = 12$

$m^2 = 12 + 1 = 13$

$m = \sqrt{13}$

Since $\sqrt{13}$ is not an integer, this case does not produce a triplet using the formula with integer $m$.

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Case 3: Assume $m^2 + 1 = 12$

$m^2 + 1 = 12$

$m^2 = 12 - 1 = 11$

$m = \sqrt{11}$

Since $\sqrt{11}$ is not an integer, this case also does not produce a triplet using the formula with integer $m$.

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Using the formula $(2m, m^2-1, m^2+1)$ with integer $m>1$, we found one Pythagorean triplet where 12 is a member.

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Therefore, a Pythagorean triplet in which one member is 12 is (12, 35, 37).

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Alternate Solutions:

It's worth noting that there are other Pythagorean triplets that include 12, which may not be generated directly by the $(2m, m^2 - 1, m^2 + 1)$ formula if 12 is not the $2m$ term or if they are multiples of a primitive triplet.

For example, (5, 12, 13) is a Pythagorean triplet since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. Here, 12 is a member.

Also, (9, 12, 15) is a Pythagorean triplet since $9^2 + 12^2 = 81 + 144 = 225 = 15^2$. Here, 12 is a member. This is a multiple of the primitive triplet (3, 4, 5) obtained by multiplying each member by 3.

The question asks for "a" triplet, so any of these is a valid answer. However, based on the preceding example, the method using the $2m$ formula is likely intended.

To find the square of a number, we multiply the number by itself. We can also use algebraic identities to make the calculation easier without direct multiplication of the full numbers.

We will primarily use the identity $(a+b)^2 = a^2 + 2ab + b^2$.

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(i) Find the square of 32.

We can write 32 as a sum of two numbers, for example, $32 = 30 + 2$.

The square of 32 is $32^2 = (30 + 2)^2$.

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$ with $a=30$ and $b=2$:

$(30 + 2)^2 = (30)^2 + 2 \times (30) \times (2) + (2)^2$

$= 900 + 120 + 4$

$= 1024$

So, the square of 32 is 1024.

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(ii) Find the square of 35.

We can write 35 as $30 + 5$.

The square of 35 is $35^2 = (30 + 5)^2$.

Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a=30$ and $b=5$:

$(30 + 5)^2 = (30)^2 + 2 \times (30) \times (5) + (5)^2$

$= 900 + 300 + 25$

$= 1225$

So, the square of 35 is 1225.

Alternate Method for numbers ending in 5:

For a number ending in 5, say $N5$, its square ends in 25. The digits before 25 are obtained by multiplying the number formed by the digits before 5 (which is N) by $(N+1)$.

For 35, N=3. Multiply N by (N+1): $3 \times (3+1) = 3 \times 4 = 12$.

Append 25 to 12. The result is 1225.

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(iii) Find the square of 86.

We can write 86 as $80 + 6$.

The square of 86 is $86^2 = (80 + 6)^2$.

Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a=80$ and $b=6$:

$(80 + 6)^2 = (80)^2 + 2 \times (80) \times (6) + (6)^2$

$= 6400 + 960 + 36$

$= 7360 + 36$

$= 7396$

So, the square of 86 is 7396.

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(iv) Find the square of 93.

We can write 93 as $90 + 3$.

The square of 93 is $93^2 = (90 + 3)^2$.

Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a=90$ and $b=3$:

$(90 + 3)^2 = (90)^2 + 2 \times (90) \times (3) + (3)^2$

$= 8100 + 540 + 9$

$= 8640 + 9$

$= 8649$

So, the square of 93 is 8649.

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(v) Find the square of 71.

We can write 71 as $70 + 1$.

The square of 71 is $71^2 = (70 + 1)^2$.

Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a=70$ and $b=1$:

$(70 + 1)^2 = (70)^2 + 2 \times (70) \times (1) + (1)^2$

$= 4900 + 140 + 1$

$= 5040 + 1$

$= 5041$

So, the square of 71 is 5041.

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(vi) Find the square of 46.

We can write 46 as $40 + 6$.

The square of 46 is $46^2 = (40 + 6)^2$.

Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a=40$ and $b=6$:

$(40 + 6)^2 = (40)^2 + 2 \times (40) \times (6) + (6)^2$

$= 1600 + 480 + 36$

$= 2080 + 36$

$= 2116$

So, the square of 46 is 2116.

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

One way to generate Pythagorean triplets is using the formula $(2m, m^2 - 1, m^2 + 1)$, where $m$ is any integer greater than 1. The three members of the triplet are $2m$, $m^2 - 1$, and $m^2 + 1$.

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(i) One member is 6

Let's use the formula $(2m, m^2 - 1, m^2 + 1)$. We can assume that the given member, 6, is equal to $2m$.

Divide by 2:

$m = \frac{6}{2} = 3$

Since $m=3$ is an integer and $m > 1$, we can use this value to find the other two members of the triplet.

The other two members are:

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

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

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

Check: $6^2 + 8^2 = 36 + 64 = 100$, and $10^2 = 100$. So $6^2 + 8^2 = 10^2$.

A Pythagorean triplet whose one member is 6 is (6, 8, 10).

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(ii) One member is 14

Let's use the formula $(2m, m^2 - 1, m^2 + 1)$. We can assume that the given member, 14, is equal to $2m$.

Divide by 2:

$m = \frac{14}{2} = 7$

Since $m=7$ is an integer and $m > 1$, we can use this value to find the other two members of the triplet.

The other two members are:

$m^2 - 1 = 7^2 - 1 = 49 - 1 = 48$

$m^2 + 1 = 7^2 + 1 = 49 + 1 = 50$

The Pythagorean triplet is (14, 48, 50).

Check: $14^2 + 48^2 = 196 + 2304 = 2500$, and $50^2 = 2500$. So $14^2 + 48^2 = 50^2$.

A Pythagorean triplet whose one member is 14 is (14, 48, 50).

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(iii) One member is 16

Let's use the formula $(2m, m^2 - 1, m^2 + 1)$. We can assume that the given member, 16, is equal to $2m$.

Divide by 2:

$m = \frac{16}{2} = 8$

Since $m=8$ is an integer and $m > 1$, we can use this value to find the other two members of the triplet.

The other two members are:

$m^2 - 1 = 8^2 - 1 = 64 - 1 = 63$

$m^2 + 1 = 8^2 + 1 = 64 + 1 = 65$

The Pythagorean triplet is (16, 63, 65).

Check: $16^2 + 63^2 = 256 + 3969 = 4225$, and $65^2 = 4225$. So $16^2 + 63^2 = 65^2$.

A Pythagorean triplet whose one member is 16 is (16, 63, 65).

---

(iv) One member is 18

Let's use the formula $(2m, m^2 - 1, m^2 + 1)$. We can assume that the given member, 18, is equal to $2m$.

Divide by 2:

$m = \frac{18}{2} = 9$

Since $m=9$ is an integer and $m > 1$, we can use this value to find the other two members of the triplet.

The other two members are:

$m^2 - 1 = 9^2 - 1 = 81 - 1 = 80$

$m^2 + 1 = 9^2 + 1 = 81 + 1 = 82$

The Pythagorean triplet is (18, 80, 82).

Check: $18^2 + 80^2 = 324 + 6400 = 6724$, and $82^2 = 6724$. So $18^2 + 80^2 = 82^2$.

A Pythagorean triplet whose one member is 18 is (18, 80, 82).

We can find the square root of a number using the prime factorization method. This method involves finding the prime factors of the number and then grouping them in pairs. The square root is the product of one factor from each pair.

Step 1: Find the prime factors of 6400.

We divide 6400 by the smallest prime numbers until we are left with 1.

Let's perform the prime factorization:

$$\begin{array}{c|cc} 2 & 6400 \\ \hline 2 & 3200 \\ \hline 2 & 1600 \\ \hline 2 & 800 \\ \hline 2 & 400 \\ \hline 2 & 200 \\ \hline 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

So, the prime factorization of 6400 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5$.

$6400 = 2^8 \times 5^2$

Step 2: Group the prime factors in pairs.

We group identical prime factors in pairs:

$6400 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (5 \times 5)$

Step 3: Take one factor from each pair and multiply them.

To find the square root, we take one factor from each pair and multiply them together.

$\sqrt{6400} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (5 \times 5)}$

$\sqrt{6400} = 2 \times 2 \times 2 \times 2 \times 5$

Now, calculate the product:

$2 \times 2 = 4$

$4 \times 2 = 8$

$8 \times 2 = 16$

$16 \times 5 = 80$

So, $\sqrt{6400} = 80$.

---

Conclusion:

The square root of 6400 is 80.

We can verify this by squaring 80: $80 \times 80 = 6400$.

To check:

Whether 90 is a perfect square.

---

Solution:

A perfect square is a number that is the square of an integer. For example, 9 is a perfect square because $3^2 = 9$, and 16 is a perfect square because $4^2 = 16$.

One way to determine if a number is a perfect square is by using its prime factorization.

Step 1: Find the prime factorization of 90.

We divide 90 by the smallest prime numbers until we reach 1.

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

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

We can write this using exponents: $90 = 2^1 \times 3^2 \times 5^1$.

Step 2: Check the exponents of the prime factors.

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

In the prime factorization of 90 ($2^1 \times 3^2 \times 5^1$), the exponents of the prime factors are:

Exponent of 2 is 1 (which is odd).

Exponent of 3 is 2 (which is even).

Exponent of 5 is 1 (which is odd).

Since the exponents of the prime factors 2 and 5 are odd, 90 is not a perfect square.

Alternate Check: Trailing Zeros

Another simple check for perfect squares is related to trailing zeros.

If a number ends in zeros, it is a perfect square only if the number of trailing zeros is even, and the digits before the zeros form a perfect square.

The number 90 has one trailing zero (it ends in 0, but the number before the zero is 9, which is a perfect square, but the number of zeros must also be even).

The number of trailing zeros in 90 is 1, which is an odd number.

Since the number of trailing zeros is odd, 90 cannot be a perfect square.

---

Conclusion:

Based on the prime factorization (or the rule about trailing zeros), 90 is not a perfect square.

Given:

The number is 2352.

---

To Find:

1. Check if 2352 is a perfect square.

2. If not, find the smallest number to multiply 2352 by to get a perfect square, and find that perfect square number.

3. Find the square root of the perfect square obtained in step 2.

---

Solution:

To check if a number is a perfect square, we find its prime factors. A number is a perfect square if all its prime factors occur in pairs (i.e., have even exponents in the prime factorization).

---

Let's find the prime factorization of 2352 using the division method:

So, the prime factorization of 2352 is $2 \times 2 \times 2 \times 2 \times 3 \times 7 \times 7$.

We can group the factors in pairs: $(2 \times 2) \times (2 \times 2) \times 3 \times (7 \times 7)$.

Writing this in exponential form, we get $2^4 \times 3^1 \times 7^2$.

---

1. Is 2352 a perfect square?

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

In the prime factorization of 2352 ($2^4 \times 3^1 \times 7^2$), the exponent of 2 is 4 (which is even), the exponent of 3 is 1 (which is odd), and the exponent of 7 is 2 (which is even).

Since the prime factor 3 has an odd exponent (1), the number 2352 is not a perfect square.

---

2. Find the smallest multiple of 2352 which is a perfect square.

To make 2352 a perfect square, we need to make the exponent of every prime factor even. The factor with an odd exponent is 3 ($3^1$).

To make the exponent of 3 even, we need to multiply by $3^1$. This will change $3^1$ to $3^1 \times 3^1 = 3^{1+1} = 3^2$, which has an even exponent (2).

The smallest number by which we must multiply 2352 to get a perfect square is 3.

The smallest multiple of 2352 which is a perfect square is $2352 \times 3$.

Calculating the product:

$2352 \times 3 = 7056$.

The prime factorization of the new number, 7056, is $2^4 \times 3^1 \times 7^2 \times 3^1 = 2^4 \times 3^2 \times 7^2$. Now, all exponents are even (4, 2, 2).

---

3. Find the square root of the new number.

The new number is 7056. Its prime factorization is $2^4 \times 3^2 \times 7^2$.

To find the square root of a perfect square from its prime factorization, we take half of the exponent of each prime factor.

$\sqrt{7056} = \sqrt{2^4 \times 3^2 \times 7^2}$

Using the property $\sqrt{a^m \times b^n} = a^{\frac{m}{2}} \times b^{\frac{n}{2}}$:

$\sqrt{7056} = 2^{\frac{4}{2}} \times 3^{\frac{2}{2}} \times 7^{\frac{2}{2}}$

$= 2^2 \times 3^1 \times 7^1$

$= 4 \times 3 \times 7$

$= 12 \times 7$

$= 84$

---

Therefore:

1. 2352 is not a perfect square.

2. The smallest multiple of 2352 which is a perfect square is 7056 (obtained by multiplying 2352 by 3).

3. The square root of 7056 is 84.

Given:

The number 9408.

---

To Find:

1. The smallest number by which 9408 must be divided so that the quotient is a perfect square.

2. The square root of the quotient.

---

Solution:

To find the smallest number by which a given number must be divided to obtain a perfect square, we use the prime factorization method. A number is a perfect square if and only if the exponent of each prime factor in its prime factorization is an even number.

---

Let's find the prime factorization of 9408:

The prime factorization of 9408 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 7 \times 7$, which can be written in exponential form as $2^6 \times 3^1 \times 7^2$.

---

For 9408 to be a perfect square, the exponent of each prime factor must be even.

In the prime factorization $2^6 \times 3^1 \times 7^2$, the exponent of 2 is 6 (even), the exponent of 3 is 1 (odd), and the exponent of 7 is 2 (even).

---

To make the exponent of each prime factor even, we need to divide the number by the prime factors that have odd exponents. The prime factor with an odd exponent is 3 (with exponent 1).

To make the exponent of 3 even, we need to divide by $3^1$.

The smallest number by which we must divide 9408 to get a perfect square is 3.

---

The quotient obtained by dividing 9408 by 3 is:

Quotient = $9408 \div 3 = 3136$.

---

The prime factorization of the quotient 3136 is obtained by removing the factor 3 from the prime factorization of 9408:

$3136 = \frac{2^6 \times 3^1 \times 7^2}{3^1} = 2^6 \times 3^{1-1} \times 7^2 = 2^6 \times 3^0 \times 7^2 = 2^6 \times 7^2$.

In the prime factorization of 3136 ($2^6 \times 7^2$), the exponents of all prime factors (6 and 2) are even. Thus, 3136 is a perfect square.

---

Now, we find the square root of the quotient (3136).

To find the square root from the prime factorization, we take each prime factor and raise its exponent to half of the original exponent.

$\sqrt{3136} = \sqrt{2^6 \times 7^2}$

$= 2^{\frac{6}{2}} \times 7^{\frac{2}{2}}$

$= 2^3 \times 7^1$

$= 8 \times 7$

$= 56$

---

The square root of 3136 is 56.

We can check this: $56^2 = 56 \times 56 = 3136$.

---

Answers:

1. The smallest number by which 9408 must be divided is 3.

2. The square root of the quotient (3136) is 56.

Given:

The numbers 6, 9, and 15.

---

To Find:

The smallest square number which is divisible by each of the numbers 6, 9, and 15.

---

Solution:

The smallest number that is divisible by each of the numbers 6, 9, and 15 is their Least Common Multiple (LCM).

Let's find the prime factorization of each number:

$6 = 2 \times 3 = 2^1 \times 3^1$

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

$15 = 3 \times 5 = 3^1 \times 5^1$

---

The LCM is found by taking the highest power of each prime factor that appears in any of the factorizations:

LCM(6, 9, 15) = $2^{\text{highest power}} \times 3^{\text{highest power}} \times 5^{\text{highest power}}$

LCM(6, 9, 15) = $2^1 \times 3^2 \times 5^1$

LCM(6, 9, 15) = $2 \times 9 \times 5 = 90$

---

So, the smallest number divisible by 6, 9, and 15 is 90.

Now we need to find the smallest perfect square that is a multiple of 90.

---

The prime factorization of 90 is $2^1 \times 3^2 \times 5^1$.

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

In the prime factorization of 90, the exponent of 2 is 1 (odd), the exponent of 3 is 2 (even), and the exponent of 5 is 1 (odd).

---

To make 90 a perfect square, we need to multiply it by the prime factors that have odd exponents, raised to a power that makes the total exponent even.

The prime factors with odd exponents are 2 (exponent 1) and 5 (exponent 1).

We need to multiply by $2^1$ and $5^1$ to make their exponents even.

The smallest number by which we must multiply 90 to get a perfect square is $2 \times 5 = 10$.

---

The smallest perfect square multiple of 90 is $90 \times 10 = 900$.

---

Let's check the prime factorization of 900:

$900 = 90 \times 10 = (2^1 \times 3^2 \times 5^1) \times (2^1 \times 5^1) = 2^{1+1} \times 3^2 \times 5^{1+1} = 2^2 \times 3^2 \times 5^2$.

Since all the exponents (2, 2, 2) are even, 900 is a perfect square ($30^2 = 900$).

Also, since 900 is a multiple of 90, it is divisible by 6, 9, and 15.

---

The smallest square number which is divisible by each of the numbers 6, 9 and 15 is 900.

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

Let's list the possible unit digits of a number and their squares:

$0^2 = 0$

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16 \implies$ unit digit is 6

$5^2 = 25 \implies$ unit digit is 5

$6^2 = 36 \implies$ unit digit is 6

$7^2 = 49 \implies$ unit digit is 9

$8^2 = 64 \implies$ unit digit is 4

$9^2 = 81 \implies$ unit digit is 1

---

Based on this, we can find the possible unit digit(s) of the square root for each number:

---

(i) 9801

The unit digit of 9801 is 1.

A number squared ends in 1 if the original number ends in 1 or 9.

The possible unit digits of the square root of 9801 are 1 or 9.

---

(ii) 99856

The unit digit of 99856 is 6.

A number squared ends in 6 if the original number ends in 4 or 6.

The possible unit digits of the square root of 99856 are 4 or 6.

---

(iii) 998001

The unit digit of 998001 is 1.

A number squared ends in 1 if the original number ends in 1 or 9.

The possible unit digits of the square root of 998001 are 1 or 9.

---

(iv) 657666025

The unit digit of 657666025 is 5.

A number squared ends in 5 if the original number ends in 5.

The possible unit digit of the square root of 657666025 is 5.

A number is a perfect square only if its unit digit is one of 0, 1, 4, 5, 6, or 9. Numbers ending in 2, 3, 7, or 8 can never be perfect squares.

---

Let's check the unit digit of each given number:

---

(i) 153

The unit digit of 153 is 3. Since numbers ending in 3 are never perfect squares, 153 is surely not a perfect square.

---

(ii) 257

The unit digit of 257 is 7. Since numbers ending in 7 are never perfect squares, 257 is surely not a perfect square.

---

(iii) 408

The unit digit of 408 is 8. Since numbers ending in 8 are never perfect squares, 408 is surely not a perfect square.

---

(iv) 441

Let's look at the unit digit of the number 441.

The unit digit of 441 is 1.

Recall the quick observation rule for perfect squares: Numbers ending in 2, 3, 7, or 8 are surely not perfect squares.

Since the unit digit of 441 is 1, which is not among the digits 2, 3, 7, or 8, we cannot conclude from this quick observation rule alone that 441 is surely not a perfect square.

(In fact, $21 \times 21 = 441$, so 441 is a perfect square).

---

The numbers which are surely not perfect squares (based on the unit digit rule) are:

(i) 153

(ii) 257

(iii) 408

The method of repeated subtraction for finding the square root of a perfect square relies on the fact that the sum of the first $n$ odd natural numbers is $n^2$. We repeatedly subtract consecutive odd numbers (1, 3, 5, 7, ...) from the given number until we reach 0. The number of subtractions performed is the square root of the given number.

---

Finding the square root of 100:

We subtract consecutive odd numbers starting from 1 from 100:

1. $100 - 1 = 99$

2. $99 - 3 = 96$

3. $96 - 5 = 91$

4. $91 - 7 = 84$

5. $84 - 9 = 75$

6. $75 - 11 = 64$

7. $64 - 13 = 51$

8. $51 - 15 = 36$

9. $36 - 17 = 19$

10. $19 - 19 = 0$

We reached 0 after 10 steps.

Therefore, the square root of 100 is 10.

---

Finding the square root of 169:

We subtract consecutive odd numbers starting from 1 from 169:

1. $169 - 1 = 168$

2. $168 - 3 = 165$

3. $165 - 5 = 160$

4. $160 - 7 = 153$

5. $153 - 9 = 144$

6. $144 - 11 = 133$

7. $133 - 13 = 120$

8. $120 - 15 = 105$

9. $105 - 17 = 88$

10. $88 - 19 = 69$

11. $69 - 21 = 48$

12. $48 - 23 = 25$

13. $25 - 25 = 0$

We reached 0 after 13 steps.

Therefore, the square root of 169 is 13.

The prime factorization method for finding the square root of a number involves the following steps:

1. Find the prime factorization of the number.

2. Group the identical prime factors in pairs.

3. For each pair of prime factors, take one factor.

4. Multiply these selected factors together to get the square root.

---

(i) 729

Prime factorization of 729:

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$

Group the factors in pairs: $(3 \times 3) \times (3 \times 3) \times (3 \times 3) = 3^2 \times 3^2 \times 3^2$

Take one factor from each pair: $3 \times 3 \times 3$

Multiply the factors: $3 \times 3 \times 3 = 27$

$\sqrt{729} = 27$

---

(ii) 400

Prime factorization of 400:

$400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 = 2^4 \times 5^2$

Group the factors in pairs: $(2 \times 2) \times (2 \times 2) \times (5 \times 5) = 2^2 \times 2^2 \times 5^2$

Take one factor from each pair: $2 \times 2 \times 5$

Multiply the factors: $2 \times 2 \times 5 = 20$

$\sqrt{400} = 20$

---

(iii) 1764

Prime factorization of 1764:

$1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7 = 2^2 \times 3^2 \times 7^2$

Group the factors in pairs: $(2 \times 2) \times (3 \times 3) \times (7 \times 7) = 2^2 \times 3^2 \times 7^2$

Take one factor from each pair: $2 \times 3 \times 7$

Multiply the factors: $2 \times 3 \times 7 = 42$

$\sqrt{1764} = 42$

---

(iv) 4096

Prime factorization of 4096:

$4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{12}$

Group the factors in pairs: $(2^2)^6$

Take one factor from each pair: $2^{12/2} = 2^6$

Multiply the factors: $2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

$\sqrt{4096} = 64$

---

(v) 7744

Prime factorization of 7744:

$7744 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11 = 2^6 \times 11^2$

Group the factors in pairs: $(2^2)^3 \times 11^2$

Take one factor from each pair: $2^{6/2} \times 11^{2/2} = 2^3 \times 11^1$

Multiply the factors: $8 \times 11 = 88$

$\sqrt{7744} = 88$

---

(vi) 9604

Prime factorization of 9604:

$9604 = 2 \times 2 \times 7 \times 7 \times 7 \times 7 = 2^2 \times 7^4$

Group the factors in pairs: $2^2 \times (7^2)^2$

Take one factor from each pair: $2^{2/2} \times 7^{4/2} = 2^1 \times 7^2$

Multiply the factors: $2 \times 49 = 98$

$\sqrt{9604} = 98$

---

(vii) 5929

Prime factorization of 5929:

$5929 = 7 \times 7 \times 11 \times 11 = 7^2 \times 11^2$

Group the factors in pairs: $(7 \times 7) \times (11 \times 11) = 7^2 \times 11^2$

Take one factor from each pair: $7 \times 11$

Multiply the factors: $7 \times 11 = 77$

$\sqrt{5929} = 77$

---

(viii) 9216

Prime factorization of 9216:

$9216 = 2^{10} \times 3^2$

Group the factors in pairs: $(2^2)^5 \times 3^2$

Take one factor from each pair: $2^{10/2} \times 3^{2/2} = 2^5 \times 3^1$

Multiply the factors: $32 \times 3 = 96$

$\sqrt{9216} = 96$

---

(ix) 529

Prime factorization of 529:

$529 = 23 \times 23 = 23^2$

Group the factors in pairs: $23^2$ (already a pair)

Take one factor from the pair: 23

$\sqrt{529} = 23$

---

(x) 8100

Prime factorization of 8100:

$8100 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 = 2^2 \times 3^4 \times 5^2$

Group the factors in pairs: $(2^2) \times (3^2)^2 \times (5^2)$

Take one factor from each pair: $2^{2/2} \times 3^{4/2} \times 5^{2/2} = 2^1 \times 3^2 \times 5^1$

Multiply the factors: $2 \times 9 \times 5 = 18 \times 5 = 90$

$\sqrt{8100} = 90$

To find the smallest whole number by which a given number should be multiplied to get a perfect square, we use the prime factorization method. A number is a perfect square if the exponent of each prime factor in its prime factorization is an even number.

The steps are: 1. Find the prime factorization. 2. Identify factors with odd exponents. 3. Multiply by the factors needed to make all exponents even. 4. Find the square root of the resulting number.

---

(i) 252

Prime factorization of 252:

$252 = 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7^1$.

The prime factor 7 has an odd exponent (1).

To make the exponent of 7 even, we need to multiply by $7^1 = 7$.

The smallest whole number to multiply by is 7.

The new number (perfect square) is $252 \times 7 = 1764$.

The prime factorization of 1764 is $2^2 \times 3^2 \times 7^2$.

The square root of 1764 is $\sqrt{2^2 \times 3^2 \times 7^2} = 2^{2/2} \times 3^{2/2} \times 7^{2/2} = 2^1 \times 3^1 \times 7^1 = 2 \times 3 \times 7 = 42$.

---

(ii) 180

Prime factorization of 180:

$180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5^1$.

The prime factor 5 has an odd exponent (1).

To make the exponent of 5 even, we need to multiply by $5^1 = 5$.

The smallest whole number to multiply by is 5.

The new number (perfect square) is $180 \times 5 = 900$.

The prime factorization of 900 is $2^2 \times 3^2 \times 5^2$.

The square root of 900 is $\sqrt{2^2 \times 3^2 \times 5^2} = 2^1 \times 3^1 \times 5^1 = 2 \times 3 \times 5 = 30$.

---

(iii) 1008

Prime factorization of 1008:

$1008 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 = 2^4 \times 3^2 \times 7^1$.

The prime factor 7 has an odd exponent (1).

To make the exponent of 7 even, we need to multiply by $7^1 = 7$.

The smallest whole number to multiply by is 7.

The new number (perfect square) is $1008 \times 7 = 7056$.

The prime factorization of 7056 is $2^4 \times 3^2 \times 7^2$.

The square root of 7056 is $\sqrt{2^4 \times 3^2 \times 7^2} = 2^{4/2} \times 3^{2/2} \times 7^{2/2} = 2^2 \times 3^1 \times 7^1 = 4 \times 3 \times 7 = 84$.

---

(iv) 2028

Prime factorization of 2028:

$2028 = 2 \times 2 \times 3 \times 13 \times 13 = 2^2 \times 3^1 \times 13^2$.

The prime factor 3 has an odd exponent (1).

To make the exponent of 3 even, we need to multiply by $3^1 = 3$.

The smallest whole number to multiply by is 3.

The new number (perfect square) is $2028 \times 3 = 6084$.

The prime factorization of 6084 is $2^2 \times 3^2 \times 13^2$.

The square root of 6084 is $\sqrt{2^2 \times 3^2 \times 13^2} = 2^1 \times 3^1 \times 13^1 = 2 \times 3 \times 13 = 6 \times 13 = 78$.

---

(v) 1458

Prime factorization of 1458:

$1458 = 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2^1 \times 3^6$.

The prime factor 2 has an odd exponent (1).

To make the exponent of 2 even, we need to multiply by $2^1 = 2$.

The smallest whole number to multiply by is 2.

The new number (perfect square) is $1458 \times 2 = 2916$.

The prime factorization of 2916 is $2^2 \times 3^6$.

The square root of 2916 is $\sqrt{2^2 \times 3^6} = 2^{2/2} \times 3^{6/2} = 2^1 \times 3^3 = 2 \times 27 = 54$.

---

(vi) 768

Prime factorization of 768:

$768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^8 \times 3^1$.

The prime factor 3 has an odd exponent (1).

To make the exponent of 3 even, we need to multiply by $3^1 = 3$.

The smallest whole number to multiply by is 3.

The new number (perfect square) is $768 \times 3 = 2304$.

The prime factorization of 2304 is $2^8 \times 3^2$.

The square root of 2304 is $\sqrt{2^8 \times 3^2} = 2^{8/2} \times 3^{2/2} = 2^4 \times 3^1 = 16 \times 3 = 48$.

To find the smallest whole number by which a given number should be divided to make the quotient a perfect square, we use the prime factorization method. A number is a perfect square if the exponent of each prime factor in its prime factorization is an even number.

The steps are: 1. Find the prime factorization. 2. Identify factors with odd exponents. 3. The number to divide by is the product of these prime factors with odd exponents (each raised to the power of 1). 4. Divide the original number by this number to get the perfect square quotient. 5. Find the square root of the resulting number by halving the exponents in its prime factorization.

---

(i) 252

Prime factorization of 252:

$252 = 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7^1$.

The prime factor 7 has an odd exponent (1). To make the exponent of 7 even, we must divide by $7^1 = 7$.

The smallest whole number to divide by is 7.

The quotient is $252 \div 7 = 36$.

The prime factorization of the quotient 36 is $2^2 \times 3^2$. All exponents are even, so it is a perfect square.

The square root of 36 is $\sqrt{2^2 \times 3^2} = 2^{2/2} \times 3^{2/2} = 2^1 \times 3^1 = 2 \times 3 = 6$.

---

(ii) 2925

Prime factorization of 2925:

$2925 = 3 \times 3 \times 5 \times 5 \times 13 = 3^2 \times 5^2 \times 13^1$.

The prime factor 13 has an odd exponent (1). To make the exponent of 13 even, we must divide by $13^1 = 13$.

The smallest whole number to divide by is 13.

The quotient is $2925 \div 13 = 225$.

The prime factorization of the quotient 225 is $3^2 \times 5^2$. All exponents are even, so it is a perfect square.

The square root of 225 is $\sqrt{3^2 \times 5^2} = 3^1 \times 5^1 = 3 \times 5 = 15$.

---

(iii) 396

Prime factorization of 396:

$396 = 2 \times 2 \times 3 \times 3 \times 11 = 2^2 \times 3^2 \times 11^1$.

The prime factor 11 has an odd exponent (1). To make the exponent of 11 even, we must divide by $11^1 = 11$.

The smallest whole number to divide by is 11.

The quotient is $396 \div 11 = 36$.

The prime factorization of the quotient 36 is $2^2 \times 3^2$. All exponents are even, so it is a perfect square.

The square root of 36 is $\sqrt{2^2 \times 3^2} = 2^1 \times 3^1 = 2 \times 3 = 6$.

---

(iv) 2645

Prime factorization of 2645:

$2645 = 5 \times 23 \times 23 = 5^1 \times 23^2$.

The prime factor 5 has an odd exponent (1). To make the exponent of 5 even, we must divide by $5^1 = 5$.

The smallest whole number to divide by is 5.

The quotient is $2645 \div 5 = 529$.

The prime factorization of the quotient 529 is $23^2$. The exponent is even, so it is a perfect square.

The square root of 529 is $\sqrt{23^2} = 23^1 = 23$.

---

(v) 2800

Prime factorization of 2800:

$2800 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 7 = 2^4 \times 5^2 \times 7^1$.

The prime factor 7 has an odd exponent (1). To make the exponent of 7 even, we must divide by $7^1 = 7$.

The smallest whole number to divide by is 7.

The quotient is $2800 \div 7 = 400$.

The prime factorization of the quotient 400 is $2^4 \times 5^2$. All exponents are even, so it is a perfect square.

The square root of 400 is $\sqrt{2^4 \times 5^2} = 2^{4/2} \times 5^{2/2} = 2^2 \times 5^1 = 4 \times 5 = 20$.

---

(vi) 1620

Prime factorization of 1620:

$1620 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 = 2^2 \times 3^4 \times 5^1$.

The prime factor 5 has an odd exponent (1). To make the exponent of 5 even, we must divide by $5^1 = 5$.

The smallest whole number to divide by is 5.

The quotient is $1620 \div 5 = 324$.

The prime factorization of the quotient 324 is $2^2 \times 3^4$. All exponents are even, so it is a perfect square.

The square root of 324 is $\sqrt{2^2 \times 3^4} = 2^{2/2} \times 3^{4/2} = 2^1 \times 3^2 = 2 \times 9 = 18$.

Given:

Total donation made by students of Class VIII = $\textsf{₹} 2401$.

Each student donated as many rupees as the number of students in the class.

---

To Find:

The number of students in the class.

---

Solution:

Let the number of students in the class be $n$.

According to the problem, each student donated as many rupees as the number of students in the class.

So, the amount donated by each student = $\textsf{₹} n$.

---

The total donation is the number of students multiplied by the amount donated by each student.

Total donation = (Number of students) $\times$ (Amount donated by each student)

Total donation = $n \times n = n^2$.

---

We are given that the total donation is $\textsf{₹} 2401$.

---

To find the number of students, we need to find the square root of 2401.

$n = \sqrt{2401}$

---

We can find the square root of 2401 using the prime factorization method.

Prime factorization of 2401:

$2401 = 7 \times 7 \times 7 \times 7 = 7^4$.

To find the square root, we take half of the exponents:

$\sqrt{2401} = \sqrt{7^4} = 7^{4/2} = 7^2$

$= 7 \times 7 = 49$.

---

So, $n = 49$.

The number of students in the class is 49.

---

We can check this: If there are 49 students and each donated $\textsf{₹} 49$, the total donation is $49 \times 49 = 49^2 = 2401$. This matches the given total donation.

---

The number of students in the class is 49.

Given:

Total number of plants to be planted in a garden = 2025.

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

---

To Find:

The number of rows and the number of plants in each row.

---

Solution:

Let the number of rows in the garden be $r$.

According to the problem, the number of plants in each row is equal to the number of rows.

So, the number of plants in each row = $r$.

---

The total number of plants is the number of rows multiplied by the number of plants in each row.

Total number of plants = (Number of rows) $\times$ (Number of plants in each row)

Total number of plants = $r \times r = r^2$.

---

We are given that the total number of plants is 2025.

---

To find the number of rows (r), we need to find the square root of 2025.

$r = \sqrt{2025}$

---

We can find the square root of 2025 using the prime factorization method.

Prime factorization of 2025:

$2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5 = 3^4 \times 5^2$.

To find the square root, we take half of the exponents:

$\sqrt{2025} = \sqrt{3^4 \times 5^2} = 3^{4/2} \times 5^{2/2} = 3^2 \times 5^1$

$= 9 \times 5 = 45$.

---

So, $r = 45$.

The number of rows is 45.

Since the number of plants in each row is equal to the number of rows, the number of plants in each row is also 45.

---

We can check this: If there are 45 rows with 45 plants in each row, the total number of plants is $45 \times 45 = 45^2 = 2025$. This matches the given total number of plants.

---

The number of rows is 45, and the number of plants in each row is 45.

Given:

The numbers 4, 9, and 10.

---

To Find:

The smallest square number that is divisible by each of the numbers 4, 9, and 10.

---

Solution:

To find the smallest number that is divisible by each of the given numbers, we need to find their Least Common Multiple (LCM).

---

Let's find the prime factorization of each number:

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

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

$10 = 2 \times 5 = 2^1 \times 5^1$

---

The LCM is found by taking the highest power of each prime factor present in the factorizations:

LCM(4, 9, 10) = $2^{\text{highest power}} \times 3^{\text{highest power}} \times 5^{\text{highest power}}$

LCM(4, 9, 10) = $2^2 \times 3^2 \times 5^1$

LCM(4, 9, 10) = $4 \times 9 \times 5 = 36 \times 5 = 180$

---

The smallest number divisible by 4, 9, and 10 is their LCM, which is 180.

Now we need to find the smallest multiple of 180 that is a perfect square.

---

The prime factorization of 180 is $2^2 \times 3^2 \times 5^1$.

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

In the prime factorization of 180, the exponent of 2 is 2 (even), the exponent of 3 is 2 (even), but the exponent of 5 is 1 (odd).

---

To make 180 a perfect square, we need to multiply it by the prime factors that have odd exponents, raised to a power that makes the total exponent even.

The prime factor with an odd exponent is 5 (with exponent 1).

To make the exponent of 5 even, we need to multiply by $5^1 = 5$.

The smallest multiple of 180 that is a perfect square is obtained by multiplying 180 by 5.

Smallest square number = $180 \times 5 = 900$.

---

Let's check the prime factorization of 900:

$900 = 180 \times 5 = (2^2 \times 3^2 \times 5^1) \times 5^1 = 2^2 \times 3^2 \times 5^{1+1} = 2^2 \times 3^2 \times 5^2$.

In the prime factorization of 900, all the exponents (2, 2, and 2) are even. Thus, 900 is a perfect square ($30^2 = 900$).

Since 900 is a multiple of 180, it is divisible by 4, 9, and 10.

---

The smallest square number that is divisible by each of the numbers 4, 9 and 10 is 900.

Given:

The numbers 8, 15, and 20.

---

To Find:

The smallest square number that is divisible by each of the numbers 8, 15, and 20.

---

Solution:

The smallest number that is divisible by each of the given numbers (8, 15, and 20) is their Least Common Multiple (LCM).

---

Let's find the prime factorization of each number:

$8 = 2 \times 2 \times 2 = 2^3$

$15 = 3 \times 5 = 3^1 \times 5^1$

$20 = 2 \times 2 \times 5 = 2^2 \times 5^1$

---

To find the LCM, we take the highest power of each prime factor present in the factorizations:

Prime factors involved are 2, 3, and 5.

Highest power of 2 is $2^3$ (from 8).

Highest power of 3 is $3^1$ (from 15).

Highest power of 5 is $5^1$ (from 15 and 20).

LCM(8, 15, 20) = $2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$.

---

The smallest number divisible by 8, 15, and 20 is 120.

Now, we need to find the smallest multiple of 120 that is 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 exponent of each prime factor in its prime factorization must be an even number.

In the prime factorization of 120 ($2^3 \times 3^1 \times 5^1$), the exponents of the prime factors 2 (exponent 3), 3 (exponent 1), and 5 (exponent 1) are all odd.

---

To make the number a perfect square, we need to multiply it by the prime factors that have odd exponents, each raised to a power that makes the total exponent even. In this case, we need to multiply by $2^1$, $3^1$, and $5^1$ to make the exponents of 2, 3, and 5 even (resulting exponents will be $3+1=4$ for 2, $1+1=2$ for 3, and $1+1=2$ for 5).

The smallest number by which 120 must be multiplied to get a perfect square is $2^1 \times 3^1 \times 5^1 = 2 \times 3 \times 5 = 30$.

---

The smallest square number divisible by 8, 15, and 20 is the product of 120 and 30.

Smallest square number = $120 \times 30 = 3600$.

---

Let's check the prime factorization of 3600:

$3600 = 120 \times 30 = (2^3 \times 3^1 \times 5^1) \times (2^1 \times 3^1 \times 5^1) = 2^{3+1} \times 3^{1+1} \times 5^{1+1} = 2^4 \times 3^2 \times 5^2$.

Since all the exponents (4, 2, and 2) are even, 3600 is a perfect square.

Since 3600 is a multiple of 120, it is divisible by 8, 15, and 20.

---

The smallest square number that is divisible by each of the numbers 8, 15 and 20 is 3600.

Solution:

We will find the square root of the given numbers using the long division method.

---

(i) Find the square root of 729.

We use the long division method:

Steps:

1. Pair the digits from the right: $\overline{7} \; \overline{29}$.

2. Consider the first pair (or single digit) from the left (7). Find the largest number whose square is less than or equal to 7. This is 2 ($2^2 = 4$). Write 2 as the divisor and the first digit of the quotient. Subtract 4 from 7 to get 3.

3. Bring down the next pair (29). The new dividend is 329.

4. Double the current quotient (2), which is 4. Write 4 followed by a blank digit (_). This forms the trial divisor 4_.

5. Find the largest digit to place in the blank such that when 4_ is multiplied by that digit, the product is less than or equal to 329. Since the unit digit of 329 is 9, the unit digit of the square root is either 3 ($3^2=9$) or 7 ($7^2=49$, unit digit 9). Let's try 7: $47 \times 7 = 329$.

6. Write 7 in the blank of the divisor (making it 47) and as the next digit of the quotient (making the quotient 27). Multiply $47 \times 7 = 329$. Subtract 329 from 329, leaving a remainder of 0.

Since the remainder is 0, 729 is a perfect square, and its square root is the quotient.

---

The square root of 729 is 27.

---

(ii) Find the square root of 1296.

We use the long division method:

Steps:

1. Pair the digits from the right: $\overline{12} \; \overline{96}$.

2. Consider the first pair from the left (12). Find the largest number whose square is less than or equal to 12. $3^2 = 9$, $4^2 = 16$. So, the number is 3. Write 3 as the divisor and the first digit of the quotient. Subtract 9 from 12 to get 3.

3. Bring down the next pair (96). The new dividend is 396.

4. Double the current quotient (3), which is 6. Write 6 followed by a blank digit (_). This forms the trial divisor 6_.

5. Find the largest digit to place in the blank such that when 6_ is multiplied by that digit, the product is less than or equal to 396. Since the unit digit of 396 is 6, the unit digit of the square root is either 4 ($4^2=16$, unit digit 6) or 6 ($6^2=36$, unit digit 6). Let's try 6: $66 \times 6 = 396$.

6. Write 6 in the blank of the divisor (making it 66) and as the next digit of the quotient (making the quotient 36). Multiply $66 \times 6 = 396$. Subtract 396 from 396, leaving a remainder of 0.

Since the remainder is 0, 1296 is a perfect square, and its square root is the quotient.

---

The square root of 1296 is 36.

Given:

The number 5607.

---

To Find:

1. The least number that must be subtracted from 5607 to get a perfect square.

2. The square root of the perfect square obtained.

---

Solution:

To find the greatest perfect square less than a given number, we use the long division method for finding the square root. The remainder obtained in this method is the smallest number that must be subtracted from the original number to get a perfect square.

---

Let's find the square root of 5607 using the long division method:

Steps:

1. Pair the digits from the right: $\overline{56} \; \overline{07}$.

2. Find the largest number whose square is less than or equal to the first pair (56). That number is 7 ($7^2 = 49$). Write 7 as the divisor and first digit of the quotient. Subtract 49 from 56 to get 7.

3. Bring down the next pair (07). The new dividend is 707.

4. Double the current quotient (7), which is 14. Write 14 followed by a blank digit (_). Find the largest digit to place in the blank such that when 14_ is multiplied by the digit, the result is less than or equal to 707. $144 \times 4 = 576$ and $145 \times 5 = 725$ (too large). So, the digit is 4.

5. Write 4 in the blank of the divisor (making it 144) and as the next digit of the quotient (making the quotient 74). Multiply $144 \times 4 = 576$. Subtract 576 from 707: $707 - 576 = 131$.

---

The quotient is 74 and the remainder is 131.

This means that $74^2 = 5607 - 131$.

The remainder, 131, is the excess amount in 5607 over the largest perfect square less than 5607.

So, to get a perfect square, we must subtract this remainder from 5607.

The least number that must be subtracted from 5607 is the remainder, which is 131.

---

The perfect square number obtained after subtracting 131 is:

$5607 - 131 = 5476$.

---

The square root of this perfect square (5476) is the quotient we obtained in the long division, which is 74.

$\sqrt{5476} = 74$.

We can verify this: $74 \times 74 = 5476$.

---

Answers:

1. The least number that must be subtracted from 5607 is 131.

2. The perfect square obtained is 5476, and its square root is 74.

To Find:

The greatest 4-digit number which is a perfect square.

---

Solution:

The greatest 4-digit number is 9999.

To find the greatest perfect square less than or equal to 9999, we find the square root of 9999 using the long division method.

---

Let's find the square root of 9999:

Steps:

1. Pair the digits from the right: $\overline{99} \; \overline{99}$.

2. Find the largest number whose square is less than or equal to the first pair (99). That number is 9 ($9^2 = 81$). Write 9 as the divisor and first digit of the quotient. Subtract 81 from 99 to get 18.

3. Bring down the next pair (99). The new dividend is 1899.

4. Double the current quotient (9), which is 18. Write 18 followed by a blank digit (_). This forms the trial divisor 18_.

5. Find the largest digit to place in the blank such that when 18_ is multiplied by the digit, the result is less than or equal to 1899. We find that $189 \times 9 = 1701$. $190 \times 10$ is too large. So, the digit is 9.

6. Write 9 in the blank of the divisor (making it 189) and as the next digit of the quotient (making the quotient 99). Multiply $189 \times 9 = 1701$. Subtract 1701 from 1899: $1899 - 1701 = 198$.

---

The quotient is 99 and the remainder is 198.

This indicates that $99^2 = 9999 - 198$. So, $99^2 = 9801$.

The remainder 198 is the amount by which 9999 exceeds the greatest perfect square less than it.

To get the greatest perfect square which is a 4-digit number, we subtract this remainder from 9999.

Greatest 4-digit perfect square = $9999 - 198 = 9801$.

---

The number 9801 is a 4-digit number. Its square root is the quotient obtained, which is 99 ($99^2 = 9801$).

The next perfect square after $99^2$ is $100^2 = 10000$, which is a 5-digit number.

Therefore, 9801 is indeed the greatest 4-digit perfect square.

---

The greatest 4-digit number which is a perfect square is 9801.

Given:

The number 1300.

---

To Find:

1. The least number that must be added to 1300 to get a perfect square.

2. The square root of the perfect square obtained.

---

Solution:

To find the least number that must be added to a given number to make it a perfect square, we find the square root of the number using the long division method. This helps us determine between which two consecutive perfect squares the given number lies. The smallest perfect square greater than the given number is the one we seek.

---

Let's find the square root of 1300 using the long division method:

Steps:

1. Pair the digits from the right: $\overline{13} \; \overline{00}$.

2. Find the largest number whose square is less than or equal to the first pair (13). $3^2 = 9$, $4^2 = 16$. So, the number is 3. Write 3 as the divisor and the first digit of the quotient. Subtract $3^2 = 9$ from 13, leaving a remainder of 4.

3. Bring down the next pair (00). The new dividend is 400.

4. Double the current quotient (3), which is 6. Write 6 followed by a blank digit (_). This forms the trial divisor 6_.

5. Find the largest digit to put in the blank such that when 6_ is multiplied by that digit, the product is less than or equal to 400. $66 \times 6 = 396$, which is less than 400. $67 \times 7 = 469$, which is greater than 400. So, the digit is 6.

6. Write 6 in the blank of the divisor (making it 66) and as the next digit of the quotient (making the quotient 36). Multiply $66 \times 6 = 396$. Subtract 396 from 400, leaving a remainder of 4.

---

The long division shows that $36^2 < 1300$. The remainder is 4. This means $36^2 = 1300 - 4 = 1296$.

The perfect square just greater than 1300 will be the square of the next integer after the quotient 36, which is 37.

The next perfect square is $37^2$.

---

Let's calculate $37^2$:

$37^2 = 37 \times 37 = 1369$.

---

The smallest perfect square greater than 1300 is 1369.

To find the least number that must be added to 1300 to get 1369, we find the difference:

Number to be added = $1369 - 1300 = 69$.

---

The perfect square number obtained is $1300 + 69 = 1369$.

The square root of this perfect square (1369) is 37.

$\sqrt{1369} = 37$.

---

Answers:

1. The least number that must be added to 1300 is 69.

2. The perfect square obtained is 1369, and its square root is 37.

To Find:

The square root of 12.25.

---

Solution:

We can find the square root of a decimal number using the long division method. We pair digits from the right for the whole number part and from the left for the decimal part. The decimal point is placed in the quotient when we bring down the first pair of digits from the decimal part of the number.

---

Let's find the square root of 12.25 using the long division method:

Steps:

1. Pair the digits. Pair the whole number part from the right ($\overline{12}$). Pair the decimal part from the left ($\overline{25}$).

2. Consider the first pair (12). Find the largest number whose square is less than or equal to 12. $3^2 = 9$. Write 3 as the first digit of the quotient and the divisor. Subtract 9 from 12, leaving 3.

3. Bring down the next pair (25). Since this is the first pair after the decimal point, place a decimal point in the quotient after 3.

4. The new dividend is 325.

5. Double the current quotient (3), which is 6. Write 6 followed by a blank digit (_). This forms the trial divisor 6_.

6. Find the largest digit to put in the blank such that when 6_ is multiplied by that digit, the product is less than or equal to 325. The unit digit of the dividend is 5, so the unit digit of the root is 5 ($5^2=25$). Let's try 5: $65 \times 5 = 325$.

7. Write 5 in the blank of the divisor (making it 65) and as the next digit of the quotient (making the quotient 3.5). Multiply $65 \times 5 = 325$. Subtract 325 from 325, leaving a remainder of 0.

---

Since the remainder is 0, 12.25 is a perfect square, and its square root is the quotient obtained.

$\sqrt{12.25} = 3.5$

We can verify this: $3.5 \times 3.5 = 12.25$.

---

The square root of 12.25 is 3.5.

---

Alternate Solution (using fractions):

We can express the decimal number as a fraction:

$12.25 = \frac{1225}{100}$

---

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

$\sqrt{12.25} = \sqrt{\frac{1225}{100}} = \frac{\sqrt{1225}}{\sqrt{100}}$

---

Find the square root of the numerator (1225). Using prime factorization:

$1225 = 5 \times 5 \times 7 \times 7 = 5^2 \times 7^2$.

$\sqrt{1225} = \sqrt{5^2 \times 7^2} = 5^{\frac{2}{2}} \times 7^{\frac{2}{2}} = 5^1 \times 7^1 = 5 \times 7 = 35$.

---

Find the square root of the denominator (100).

$\sqrt{100} = 10$ (since $10 \times 10 = 100$).

---

Substitute the square roots back into the fraction:

$\sqrt{12.25} = \frac{35}{10} = 3.5$

---

Both methods give the same result.

Given:

Area of the square plot = $2304 \text{ m}^2$.

---

To Find:

The side of the square plot.

---

Solution:

Let the side of the square plot be $s$ meters.

The area of a square is given by the formula:

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

Given Area = $2304 \text{ m}^2$.

So, we have the equation:

To find the side $s$, we need to calculate the square root of 2304.

$s = \sqrt{2304}$

---

We can find the square root of 2304 using the long division method:

Steps:

1. Pair the digits from the right: $\overline{23} \; \overline{04}$.

2. Find the largest number whose square is less than or equal to 23. This is 4 ($4^2 = 16$). Write 4 as the divisor and first digit of the quotient. Subtract 16 from 23 to get 7.

3. Bring down the next pair 04. The new dividend is 704.

4. Double the current quotient (4), which is 8. Write 8 followed by a blank digit (_). This forms the trial divisor 8_.

5. Find the largest digit such that when 8_ is multiplied by that digit, the product is less than or equal to 704. The unit digit of 704 is 4, so the unit digit of the square root is either 2 ($2^2=4$) or 8 ($8^2=64$, unit digit 4). Let's try 8: $88 \times 8 = 704$.

6. Write 8 in the blank of the divisor (making it 88) and as the next digit of the quotient (making the quotient 48). Multiply $88 \times 8 = 704$. Subtract 704 from 704, leaving a remainder of 0.

---

The square root of 2304 is 48.

So, $s = 48$.

The side of the square plot is 48 meters.

Check: $48 \times 48 = 2304$.

---

The side of the square plot is 48 m.

Given:

Total number of students in the school = 2401.

Arrangement: Number of rows = Number of columns.

---

To Find:

The number of rows.

---

Solution:

Let the number of rows be $n$.

Since the number of rows is equal to the number of columns, the number of columns is also $n$.

---

The total number of students is found by multiplying the number of rows by the number of columns.

Total number of students = Number of rows $\times$ Number of columns

Total number of students $= n \times n = n^2$.

---

We are given that the total number of students is 2401.

To find the number of rows ($n$), we need to find the square root of 2401.

$n = \sqrt{2401}$

---

We can find the square root of 2401 using the long division method:

Steps:

1. Pair the digits from the right: $\overline{24} \; \overline{01}$.

2. Find the largest number whose square is less than or equal to 24. This is 4 ($4^2 = 16$). Write 4 as the divisor and first digit of the quotient. Subtract 16 from 24 to get 8.

3. Bring down the next pair 01. The new dividend is 801.

4. Double the current quotient (4), which is 8. Write 8 followed by a blank digit (_). This forms the trial divisor 8_.

5. Find the largest digit such that when 8_ is multiplied by that digit, the product is less than or equal to 801. The unit digit of 801 is 1, so the unit digit of the square root could be 1 ($1^2=1$) or 9 ($9^2=81$, unit digit 1). Let's try 9: $89 \times 9 = 801$.

6. Write 9 in the blank of the divisor (making it 89) and as the next digit of the quotient (making the quotient 49). Multiply $89 \times 9 = 801$. Subtract 801 from 801, leaving a remainder of 0.

---

The square root of 2401 is 49.

So, $n = 49$.

---

The number of rows is 49. The number of columns is also 49.

Check: $49 \times 49 = 2401$.

---

The number of rows is 49.

We will find the square root of each number using the long division method.

---

(i) 2304

Steps:

1. Pair the digits from the right: $\overline{23} \; \overline{04}$.

2. The largest square less than or equal to 23 is $4^2=16$. Write 4 as the first digit of the quotient. $23-16 = 7$.

3. Bring down 04. The dividend is 704. Double the quotient: $4 \times 2 = 8$. Write 8_.

4. Find digit x such that $8x \times x \leq 704$. $88 \times 8 = 704$. Write 8 as the next digit of the quotient. $704 - 704 = 0$.

The remainder is 0. The square root of 2304 is 48.

---

(ii) 4489

Steps:

1. Pair the digits from the right: $\overline{44} \; \overline{89}$.

2. The largest square less than or equal to 44 is $6^2=36$. Write 6 as the first digit of the quotient. $44-36 = 8$.

3. Bring down 89. The dividend is 889. Double the quotient: $6 \times 2 = 12$. Write 12_.

4. Find digit x such that $12x \times x \leq 889$. $127 \times 7 = 889$. Write 7 as the next digit of the quotient. $889 - 889 = 0$.

The remainder is 0. The square root of 4489 is 67.

---

(iii) 3481

Steps:

1. Pair the digits from the right: $\overline{34} \; \overline{81}$.

2. The largest square less than or equal to 34 is $5^2=25$. Write 5 as the first digit of the quotient. $34-25 = 9$.

3. Bring down 81. The dividend is 981. Double the quotient: $5 \times 2 = 10$. Write 10_.

4. Find digit x such that $10x \times x \leq 981$. $109 \times 9 = 981$. Write 9 as the next digit of the quotient. $981 - 981 = 0$.

The remainder is 0. The square root of 3481 is 59.

---

(iv) 529

Steps:

1. Pair the digits from the right: $\overline{5} \; \overline{29}$.

2. The largest square less than or equal to 5 is $2^2=4$. Write 2 as the first digit of the quotient. $5-4 = 1$.

3. Bring down 29. The dividend is 129. Double the quotient: $2 \times 2 = 4$. Write 4_.

4. Find digit x such that $4x \times x \leq 129$. $43 \times 3 = 129$. Write 3 as the next digit of the quotient. $129 - 129 = 0$.

The remainder is 0. The square root of 529 is 23.

---

(v) 3249

Steps:

1. Pair the digits from the right: $\overline{32} \; \overline{49}$.

2. The largest square less than or equal to 32 is $5^2=25$. Write 5 as the first digit of the quotient. $32-25 = 7$.

3. Bring down 49. The dividend is 749. Double the quotient: $5 \times 2 = 10$. Write 10_.

4. Find digit x such that $10x \times x \leq 749$. $107 \times 7 = 749$. Write 7 as the next digit of the quotient. $749 - 749 = 0$.

The remainder is 0. The square root of 3249 is 57.

---

(vi) 1369

Steps:

1. Pair the digits from the right: $\overline{13} \; \overline{69}$.

2. The largest square less than or equal to 13 is $3^2=9$. Write 3 as the first digit of the quotient. $13-9 = 4$.

3. Bring down 69. The dividend is 469. Double the quotient: $3 \times 2 = 6$. Write 6_.

4. Find digit x such that $6x \times x \leq 469$. $67 \times 7 = 469$. Write 7 as the next digit of the quotient. $469 - 469 = 0$.

The remainder is 0. The square root of 1369 is 37.

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(vii) 5776

Steps:

1. Pair the digits from the right: $\overline{57} \; \overline{76}$.

2. The largest square less than or equal to 57 is $7^2=49$. Write 7 as the first digit of the quotient. $57-49 = 8$.

3. Bring down 76. The dividend is 876. Double the quotient: $7 \times 2 = 14$. Write 14_.

4. Find digit x such that $14x \times x \leq 876$. $146 \times 6 = 876$. Write 6 as the next digit of the quotient. $876 - 876 = 0$.

The remainder is 0. The square root of 5776 is 76.

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(viii) 7921

Steps:

1. Pair the digits from the right: $\overline{79} \; \overline{21}$.

2. The largest square less than or equal to 79 is $8^2=64$. Write 8 as the first digit of the quotient. $79-64 = 15$.

3. Bring down 21. The dividend is 1521. Double the quotient: $8 \times 2 = 16$. Write 16_.

4. Find digit x such that $16x \times x \leq 1521$. $169 \times 9 = 1521$. Write 9 as the next digit of the quotient. $1521 - 1521 = 0$.

The remainder is 0. The square root of 7921 is 89.

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(ix) 576

Steps:

1. Pair the digits from the right: $\overline{5} \; \overline{76}$.

2. The largest square less than or equal to 5 is $2^2=4$. Write 2 as the first digit of the quotient. $5-4 = 1$.

3. Bring down 76. The dividend is 176. Double the quotient: $2 \times 2 = 4$. Write 4_.

4. Find digit x such that $4x \times x \leq 176$. $44 \times 4 = 176$. Write 4 as the next digit of the quotient. $176 - 176 = 0$.

The remainder is 0. The square root of 576 is 24.

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(x) 1024

Steps:

1. Pair the digits from the right: $\overline{10} \; \overline{24}$.

2. The largest square less than or equal to 10 is $3^2=9$. Write 3 as the first digit of the quotient. $10-9 = 1$.

3. Bring down 24. The dividend is 124. Double the quotient: $3 \times 2 = 6$. Write 6_.

4. Find digit x such that $6x \times x \leq 124$. $62 \times 2 = 124$. Write 2 as the next digit of the quotient. $124 - 124 = 0$.

The remainder is 0. The square root of 1024 is 32.

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(xi) 3136

Steps:

1. Pair the digits from the right: $\overline{31} \; \overline{36}$.

2. The largest square less than or equal to 31 is $5^2=25$. Write 5 as the first digit of the quotient. $31-25 = 6$.

3. Bring down 36. The dividend is 636. Double the quotient: $5 \times 2 = 10$. Write 10_.

4. Find digit x such that $10x \times x \leq 636$. $106 \times 6 = 636$. Write 6 as the next digit of the quotient. $636 - 636 = 0$.

The remainder is 0. The square root of 3136 is 56.

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(xii) 900

Steps:

1. Pair the digits from the right: $\overline{9} \; \overline{00}$.

2. The largest square less than or equal to 9 is $3^2=9$. Write 3 as the first digit of the quotient. $9-9 = 0$.

3. Bring down 00. The dividend is 000. Double the quotient: $3 \times 2 = 6$. Write 6_.

4. Find digit x such that $6x \times x \leq 0$. $60 \times 0 = 0$. Write 0 as the next digit of the quotient. $0 - 0 = 0$.

The remainder is 0. The square root of 900 is 30.

To find the number of digits in the square root of a perfect square number without calculation, we can count the number of digits in the given number and use the following rule:

If the number of digits in the perfect square is $n$, then the number of digits in its square root is:

• $\frac{n}{2}$ if $n$ is an even number.
• $\frac{n+1}{2}$ if $n$ is an odd number.

---

(i) 64

The number of digits in 64 is $n=2$.

Since $n$ is even, the number of digits in the square root is $\frac{n}{2} = \frac{2}{2} = 1$.

The number of digits in the square root of 64 is 1. ($\sqrt{64} = 8$).

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(ii) 144

The number of digits in 144 is $n=3$.

Since $n$ is odd, the number of digits in the square root is $\frac{n+1}{2} = \frac{3+1}{2} = \frac{4}{2} = 2$.

The number of digits in the square root of 144 is 2. ($\sqrt{144} = 12$).

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(iii) 4489

The number of digits in 4489 is $n=4$.

Since $n$ is even, the number of digits in the square root is $\frac{n}{2} = \frac{4}{2} = 2$.

The number of digits in the square root of 4489 is 2. ($\sqrt{4489} = 67$).

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(iv) 27225

The number of digits in 27225 is $n=5$.

Since $n$ is odd, the number of digits in the square root is $\frac{n+1}{2} = \frac{5+1}{2} = \frac{6}{2} = 3$.

The number of digits in the square root of 27225 is 3. ($\sqrt{27225} = 165$).

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(v) 390625

The number of digits in 390625 is $n=6$.

Since $n$ is even, the number of digits in the square root is $\frac{n}{2} = \frac{6}{2} = 3$.

The number of digits in the square root of 390625 is 3. ($\sqrt{390625} = 625$).

We will find the square root of each decimal number using the long division method. In this method, we pair the digits from the right for the whole number part and from the left for the decimal part. We place the decimal point in the quotient when we bring down the first pair of digits from the decimal part.

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(i) 2.56

Steps:

1. Pair the digits: $\overline{2} \; .\overline{56}$.

2. Find the largest number whose square is $\leq 2$: $1^2 = 1$. Write 1 as the first digit of the quotient. $2 - 1 = 1$.

3. Bring down the next pair $\overline{56}$. Since this is the first pair after the decimal, place a decimal point in the quotient. The new dividend is 156.

4. Double the quotient (1): $1 \times 2 = 2$. Write 2_.

5. Find digit x such that $2x \times x \leq 156$. $26 \times 6 = 156$. Write 6 as the next digit of the quotient. $156 - 156 = 0$.

The remainder is 0. The square root of 2.56 is 1.6.

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(ii) 7.29

The square root of 7.29 is 2.7.

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(iii) 51.84

The square root of 51.84 is 7.2.

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(iv) 42.25

The square root of 42.25 is 6.5.

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(v) 31.36

The square root of 31.36 is 5.6.

To find the least number that must be subtracted from a number to get a perfect square, we find the square root of the number using the long division method. The remainder obtained is the least number to be subtracted.

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(i) 402

The remainder is 2. The least number to subtract is 2.

The perfect square is $402 - 2 = 400$.

The square root of 400 is the quotient, which is 20. ($\sqrt{400} = 20$).

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(ii) 1989

The remainder is 53. The least number to subtract is 53.

The perfect square is $1989 - 53 = 1936$.

The square root of 1936 is the quotient, which is 44. ($\sqrt{1936} = 44$).

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(iii) 3250

The remainder is 1. The least number to subtract is 1.

The perfect square is $3250 - 1 = 3249$.

The square root of 3249 is the quotient, which is 57. ($\sqrt{3249} = 57$).

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(iv) 825

The remainder is 41. The least number to subtract is 41.

The perfect square is $825 - 41 = 784$.

The square root of 784 is the quotient, which is 28. ($\sqrt{784} = 28$).

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(v) 4000

The remainder is 31. The least number to subtract is 31.

The perfect square is $4000 - 31 = 3969$.

The square root of 3969 is the quotient, which is 63. ($\sqrt{3969} = 63$).

To find the least number that must be added to a number to make it a perfect square, we find the square root of the number using the long division method. We continue the division until we have the largest perfect square less than or equal to the given number. The difference between the next perfect square (the square of the next integer after the quotient) and the original number is the least number to be added.

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(i) 525

The quotient is 22 and the remainder is 41.

This means $22^2 < 525$. The largest perfect square less than 525 is $22^2 = 484$.

The next integer after the quotient 22 is 23.

The next perfect square is $23^2 = 529$.

The least number to be added is the difference between the next perfect square and the original number:

$529 - 525 = 4$.

The least number to add is 4.

The perfect square obtained is $525 + 4 = 529$.

The square root of the perfect square 529 is 23.

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(ii) 1750

The quotient is 41 and the remainder is 69.

This means $41^2 < 1750$. The largest perfect square less than 1750 is $41^2 = 1681$.

The next integer after the quotient 41 is 42.

The next perfect square is $42^2 = 1764$.

The least number to be added is the difference between the next perfect square and the original number:

$1764 - 1750 = 14$.

The least number to add is 14.

The perfect square obtained is $1750 + 14 = 1764$.

The square root of the perfect square 1764 is 42.

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(iii) 252

The quotient is 15 and the remainder is 27.

This means $15^2 < 252$. The largest perfect square less than 252 is $15^2 = 225$.

The next integer after the quotient 15 is 16.

The next perfect square is $16^2 = 256$.

The least number to be added is the difference between the next perfect square and the original number:

$256 - 252 = 4$.

The least number to add is 4.

The perfect square obtained is $252 + 4 = 256$.

The square root of the perfect square 256 is 16.

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(iv) 1825

The quotient is 42 and the remainder is 61.

This means $42^2 < 1825$. The largest perfect square less than 1825 is $42^2 = 1764$.

The next integer after the quotient 42 is 43.

The next perfect square is $43^2 = 1849$.

The least number to be added is the difference between the next perfect square and the original number:

$1849 - 1825 = 24$.

The least number to add is 24.

The perfect square obtained is $1825 + 24 = 1849$.

The square root of the perfect square 1849 is 43.

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(v) 6412

The quotient is 80 and the remainder is 12.

This means $80^2 < 6412$. The largest perfect square less than 6412 is $80^2 = 6400$.

The next integer after the quotient 80 is 81.

The next perfect square is $81^2 = 6561$.

The least number to be added is the difference between the next perfect square and the original number:

$6561 - 6412 = 149$.

The least number to add is 149.

The perfect square obtained is $6412 + 149 = 6561$.

The square root of the perfect square 6561 is 81.

Given:

Area of the square = $441 \text{ m}^2$.

---

To Find:

The length of the side of the square.

---

Solution:

Let the length of the side of the square be $s$ meters.

The area of a square is given by the formula:

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

Given Area = $441 \text{ m}^2$.

So, we have the equation:

$s^2 = 441$

To find the side $s$, we need to calculate the square root of 441:

$s = \sqrt{441}$

We can find the square root of 441 using the long division method:

Steps:

1. Pair the digits from the right: $\overline{4} \; \overline{41}$.

2. Find the largest number whose square is less than or equal to the first pair (4). This is 2 ($2^2 = 4$). Write 2 as the divisor and the first digit of the quotient. Subtract 4 from 4 to get 0.

3. Bring down the next pair 41. The new dividend is 41.

4. Double the current quotient (2), which is 4. Write 4 followed by a blank digit (_). This forms the trial divisor 4_.

5. Find the largest digit such that when 4_ is multiplied by that digit, the product is less than or equal to 41. $41 \times 1 = 41$.

6. Write 1 in the blank of the divisor (making it 41) and as the next digit of the quotient (making the quotient 21). Multiply $41 \times 1 = 41$. Subtract 41 from 41, leaving a remainder of 0.

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The square root of 441 is 21.

Thus, the side of the square is 21 meters.

Check: $21 \times 21 = 441$.

---

The length of the side of the square is 21 m.

Given:

In a right triangle ABC, $\angle$B = $90^\circ$. This means AC 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.

Hypotenuse$^2$ = Base$^2$ + Perpendicular$^2$

In $\triangle$ABC with $\angle$B = $90^\circ$, the hypotenuse is AC, and the other two sides are AB and BC.

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(a) If AB = 6 cm, BC = 8 cm, find AC

Given AB = 6 cm and BC = 8 cm.

Substitute these values into the Pythagorean theorem:

AC$^2$ = $(6 \text{ cm})^2$ + $(8 \text{ cm})^2$

AC$^2$ = $36 \text{ cm}^2 + 64 \text{ cm}^2$

AC$^2$ = $100 \text{ cm}^2$

To find AC, take the square root of both sides:

AC = $\sqrt{100 \text{ cm}^2}$

AC = 10 cm

The length of AC is 10 cm.

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(b) If AC = 13 cm, BC = 5 cm, find AB

Given AC = 13 cm and BC = 5 cm.

Substitute these values into the Pythagorean theorem:

AC$^2$ = AB$^2$ + BC$^2$

$(13 \text{ cm})^2$ = AB$^2$ + $(5 \text{ cm})^2$

$169 \text{ cm}^2$ = AB$^2$ + $25 \text{ cm}^2$

To find AB$^2$, subtract $25 \text{ cm}^2$ from both sides:

AB$^2$ = $169 \text{ cm}^2 - 25 \text{ cm}^2$

AB$^2$ = $144 \text{ cm}^2$

To find AB, take the square root of both sides:

AB = $\sqrt{144 \text{ cm}^2}$

AB = 12 cm

The length of AB is 12 cm.

Given:

Total number of plants the gardener has = 1000.

The gardener wants to plant the plants in rows and columns such that the number of rows is equal to the number of columns.

---

To Find:

The minimum number of plants the gardener needs more to arrange them in this way.

---

Solution:

If the number of rows is equal to the number of columns, say $n$, then the total number of plants must be $n \times n = n^2$. This means the total number of plants must be a perfect square.

The gardener has 1000 plants. He needs to add a minimum number of plants to make the total number a perfect square. This means we need to find the smallest perfect square number that is greater than 1000.

---

We can find the largest perfect square less than or equal to 1000 by finding the square root of 1000 using the long division method.

The steps are:

1. Pair the digits from the right: $\overline{10} \; \overline{00}$.

2. The largest square less than or equal to 10 is $3^2 = 9$. Write 3 as the first digit of the quotient. $10 - 9 = 1$.

3. Bring down the next pair 00. The new dividend is 100.

4. Double the current quotient (3), which is 6. Write 6 followed by a blank digit (_).

5. Find the largest digit x such that $6x \times x \leq 100$. $61 \times 1 = 61$. $62 \times 2 = 124 > 100$. So the digit is 1.

6. Write 1 in the blank of the divisor (making it 61) and as the next digit of the quotient (making the quotient 31). $100 - 61 = 39$.

---

The quotient is 31, and the remainder is 39.

This means that $31^2 = 1000 - 39 = 961$.

The largest perfect square less than 1000 is 961 ($31^2$).

---

To form a perfect square arrangement with more plants, the total number of plants must be the next perfect square after 961. The square root of this next perfect square will be the next integer after 31, which is 32.

The smallest perfect square greater than 1000 is $32^2$.

$32^2 = 32 \times 32 = 1024$.

---

The total number of plants needed for this arrangement is 1024.

The gardener currently has 1000 plants.

The minimum number of plants he needs more is the difference between the required total and the current number of plants.

Number of plants needed more = $1024 - 1000 = 24$.

---

With 1024 plants, the gardener can plant them in $32$ rows and $32$ columns.

---

The minimum number of plants he needs more is 24.

Given:

Total number of children in the school = 500.

The children have to stand in rows and columns such that the number of rows is equal to the number of columns.

---

To Find:

How many children would be left out in this arrangement.

---

Solution:

If the number of rows is equal to the number of columns, say $n$, then the children standing in the arrangement will form a perfect square, with a total of $n \times n = n^2$ children.

The total number of children is 500. We need to find the largest number of children less than or equal to 500 that can be arranged in a perfect square (i.e., find the largest perfect square less than or equal to 500).

---

We can find the largest perfect square less than or equal to 500 by finding the square root of 500 using the long division method.

The steps are:

1. Pair the digits from the right: $\overline{5} \; \overline{00}$.

2. The largest square less than or equal to 5 is $2^2 = 4$. Write 2 as the first digit of the quotient. $5 - 4 = 1$.

3. Bring down the next pair 00. The new dividend is 100.

4. Double the current quotient (2), which is 4. Write 4 followed by a blank digit (_).

5. Find the largest digit x such that $4x \times x \leq 100$. $42 \times 2 = 84$. $43 \times 3 = 129 > 100$. So the digit is 2.

6. Write 2 in the blank of the divisor (making it 42) and as the next digit of the quotient (making the quotient 22). $100 - 84 = 16$.

---

The quotient is 22, and the remainder is 16.

This means that $22^2 = 500 - 16 = 484$.

The largest perfect square less than or equal to 500 is 484 ($22^2$).

---

This means that 484 children can be arranged in $22$ rows and $22$ columns.

The total number of children is 500.

The number of children who can be arranged is 484.

The number of children left out is the total number of children minus the number of children arranged.

Number of children left out = $500 - 484 = 16$.

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The remainder (16) directly represents the number of children left out after forming the largest possible square arrangement.

---

16 children would be left out in this arrangement.