Chapter 1 Real Numbers

Given:

Two positive integers, 4052 and 12576.

To Find:

The Highest Common Factor (HCF) of 4052 and 12576 using Euclid's algorithm.

Solution:

Euclid's algorithm is based on Euclid's division lemma, which states that for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$, where $0 \le r < b$.

We will apply this lemma repeatedly until the remainder $r$ becomes zero. The divisor at that stage will be the HCF.

Here, let $a = 12576$ and $b = 4052$, since $12576 > 4052$.

Step 1: Apply the division lemma to 12576 and 4052.

Step 2: Since the remainder is not zero, we apply the division lemma to the divisor 4052 and the remainder 420.

Step 3: Apply the division lemma to the new divisor 420 and remainder 272.

Step 4: Apply the division lemma to the new divisor 272 and remainder 148.

Step 5: Apply the division lemma to the new divisor 148 and remainder 124.

Step 6: Apply the division lemma to the new divisor 124 and remainder 24.

Step 7: Apply the division lemma to the new divisor 24 and remainder 4.

The remainder has now become zero, so our procedure stops.

The divisor at this last stage is 4. Therefore, the HCF of 12576 and 4052 is 4.

Final Answer: The HCF of 4052 and 12576 is 4.

To Prove:

Every positive even integer is of the form $2q$, and every positive odd integer is of the form $2q + 1$, where $q$ is some integer.

Proof:

Let 'a' be any positive integer. We will apply Euclid's division lemma to 'a' with the divisor $b = 2$.

According to Euclid's division lemma, there exist unique integers $q$ (quotient) and $r$ (remainder) such that:

where the remainder $r$ must satisfy the condition $0 \leq r < 2$.

The only possible integer values for the remainder $r$ are 0 and 1.

This gives us two possible cases for any positive integer 'a'.

Case 1: When the remainder $r = 0$.

If $r = 0$, the equation becomes:

$a = 2q + 0$

$a = 2q$

This shows that 'a' is a multiple of 2. By definition, any integer that is a multiple of 2 is an even integer.

Case 2: When the remainder $r = 1$.

If $r = 1$, the equation becomes:

$a = 2q + 1$

This shows that 'a' is not a multiple of 2 (it leaves a remainder of 1 when divided by 2). By definition, any integer that is not a multiple of 2 is an odd integer.

Since any positive integer 'a' must fall into one of these two cases, it must be either even (of the form $2q$) or odd (of the form $2q+1$).

Therefore, every positive even integer is of the form $2q$, and every positive odd integer is of the form $2q + 1$, where $q$ is some integer.

Hence Proved.

To Prove:

Any positive odd integer is of the form $4q + 1$ or $4q + 3$, where $q$ is some integer.

Proof:

Let 'a' be any positive integer. We will apply Euclid's division lemma to 'a' with the divisor $b = 4$.

According to Euclid's division lemma, there exist unique integers $q$ and $r$ such that:

where the remainder $r$ must satisfy the condition $0 \leq r < 4$.

The possible integer values for the remainder $r$ are 0, 1, 2, and 3.

This gives four possible forms for any positive integer 'a'.

Case 1: When $r = 0$.

$a = 4q = 2(2q)$. Since 'a' is a multiple of 2, it is an even integer.

Case 2: When $r = 1$.

$a = 4q + 1 = 2(2q) + 1$. Since 'a' is not a multiple of 2, it is an odd integer.

Case 3: When $r = 2$.

$a = 4q + 2 = 2(2q + 1)$. Since 'a' is a multiple of 2, it is an even integer.

Case 4: When $r = 3$.

$a = 4q + 3 = 4q + 2 + 1 = 2(2q + 1) + 1$. Since 'a' is not a multiple of 2, it is an odd integer.

We are asked to consider only positive odd integers. From the four cases above, the integers are odd only when the remainder is 1 or 3.

Thus, any positive odd integer must be of the form $4q + 1$ or $4q + 3$.

Hence Proved.

Given:

Number of kaju barfis = 420.

Number of badam barfis = 130.

The condition is that each stack must have the same number of barfis, and this number should be maximum to take up the least area.

To Find:

The maximum number of barfis that can be placed in each stack.

Solution:

To find the maximum number of barfis that can be in each stack, we need to find the largest number that divides both 420 and 130. This is the Highest Common Factor (HCF) of 420 and 130.

We will use Euclid's division algorithm to find the HCF.

Let $a = 420$ and $b = 130$.

Step 1: Apply the division lemma to 420 and 130.

Step 2: Apply the division lemma to the divisor 130 and the remainder 30.

Step 3: Apply the division lemma to the new divisor 30 and remainder 10.

The remainder is now zero, so the algorithm stops. The divisor at this stage is 10.

Therefore, HCF(420, 130) = 10.

Final Answer: The number of barfis that can be placed in each stack is 10.

Concept Used:

We will use Euclid's division algorithm to find the HCF (Highest Common Factor) of the given pairs of numbers. The algorithm is based on Euclid's division lemma, which states that for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that:

$a = bq + r$, where $0 \le r < b$.

The HCF is the last non-zero remainder, or the divisor when the remainder becomes zero.

(i) HCF of 135 and 225

Let $a = 225$ and $b = 135$. Since $225 > 135$, we apply Euclid's division lemma.

Step 1:

Step 2: The remainder is 90. We apply the lemma to 135 and 90.

Step 3: The remainder is 45. We apply the lemma to 90 and 45.

The remainder has become zero. The divisor at this stage is 45.

Therefore, the HCF of 135 and 225 is 45.

(ii) HCF of 196 and 38220

Let $a = 38220$ and $b = 196$. Since $38220 > 196$, we apply Euclid's division lemma.

Step 1:

The remainder is zero in the first step itself. The divisor at this stage is 196.

Therefore, the HCF of 196 and 38220 is 196.

(iii) HCF of 867 and 255

Let $a = 867$ and $b = 255$. Since $867 > 255$, we apply Euclid's division lemma.

Step 1:

Step 2: The remainder is 102. We apply the lemma to 255 and 102.

Step 3: The remainder is 51. We apply the lemma to 102 and 51.

The remainder has become zero. The divisor at this stage is 51.

Therefore, the HCF of 867 and 255 is 51.

To Prove:

Any positive odd integer is of the form $6q + 1$, $6q + 3$, or $6q + 5$, where $q$ is some integer.

Proof:

Let 'a' be any positive integer. We will use Euclid's division lemma with the divisor $b = 6$.

According to the lemma, for any positive integer 'a', there exist unique integers $q$ and $r$ such that:

where the remainder $r$ must satisfy the condition $0 \leq r < 6$.

This means the possible integer values for the remainder $r$ are 0, 1, 2, 3, 4, and 5.

Let's examine each possible case for 'a'.

Case 1: If $r = 0$, then $a = 6q = 2(3q)$. This is a multiple of 2, so it is an even integer.

Case 2: If $r = 1$, then $a = 6q + 1 = 2(3q) + 1$. This is not a multiple of 2, so it is an odd integer.

Case 3: If $r = 2$, then $a = 6q + 2 = 2(3q + 1)$. This is a multiple of 2, so it is an even integer.

Case 4: If $r = 3$, then $a = 6q + 3 = 6q + 2 + 1 = 2(3q + 1) + 1$. This is not a multiple of 2, so it is an odd integer.

Case 5: If $r = 4$, then $a = 6q + 4 = 2(3q + 2)$. This is a multiple of 2, so it is an even integer.

Case 6: If $r = 5$, then $a = 6q + 5 = 6q + 4 + 1 = 2(3q + 2) + 1$. This is not a multiple of 2, so it is an odd integer.

We are concerned with positive odd integers. From the cases above, a positive integer can be odd only if it is of the form $6q + 1$, $6q + 3$, or $6q + 5$.

Conclusion: Thus, any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.

Given:

Number of members in the army contingent = 616.

Number of members in the army band = 32.

The two groups must march in the same number of columns.

To Find:

The maximum number of columns in which they can march.

Solution:

The problem asks for the maximum number of columns. This means we need to find the largest number that can divide both 616 and 32 without leaving a remainder. This largest number is the Highest Common Factor (HCF) of 616 and 32.

We will use Euclid's division algorithm to find the HCF.

Let $a = 616$ and $b = 32$.

Step 1: Apply the division lemma to 616 and 32.

Step 2: Apply the division lemma to the divisor 32 and the remainder 8.

The remainder is now zero. The divisor at this final stage is 8.

Therefore, the HCF of 616 and 32 is 8.

Final Answer: The maximum number of columns in which they can march is 8.

To Prove:

The square of any positive integer is either of the form $3m$ or $3m + 1$ for some integer $m$.

Proof:

Let 'a' be any positive integer. By Euclid's division lemma with divisor $b = 3$, 'a' can be written in one of the following forms: $3q$, $3q + 1$, or $3q + 2$, where $q$ is some integer.

We will now square each of these forms and show that the result can be expressed as $3m$ or $3m + 1$.

Case 1: When $a = 3q$.

Squaring both sides, we get:

$a^2 = (3q)^2 = 9q^2$

This can be written as:

$a^2 = 3(3q^2)$

Let $m = 3q^2$. Since $q$ is an integer, $m$ is also an integer.

Thus, $a^2$ is of the form $3m$.

Case 2: When $a = 3q + 1$.

Squaring both sides, we get:

$a^2 = (3q + 1)^2 = 9q^2 + 6q + 1$

We can factor out 3 from the first two terms:

$a^2 = 3(3q^2 + 2q) + 1$

Let $m = 3q^2 + 2q$. Since $q$ is an integer, $m$ is also an integer.

Thus, $a^2$ is of the form $3m + 1$.

Case 3: When $a = 3q + 2$.

Squaring both sides, we get:

$a^2 = (3q + 2)^2 = 9q^2 + 12q + 4$

We can rewrite the constant term 4 as $3+1$ to facilitate factoring by 3:

$a^2 = 9q^2 + 12q + 3 + 1$

Now, factor out 3 from the first three terms:

$a^2 = 3(3q^2 + 4q + 1) + 1$

Let $m = 3q^2 + 4q + 1$. Since $q$ is an integer, $m$ is also an integer.

Thus, $a^2$ is of the form $3m + 1$.

Conclusion: In all possible cases, the square of any positive integer 'a' can be expressed in the form of either $3m$ or $3m + 1$ for some integer $m$.

Hence Proved.

To Prove:

The cube of any positive integer is of the form $9m$, $9m + 1$, or $9m + 8$, where $m$ is some integer.

Proof:

Let 'a' be any positive integer. By Euclid's division lemma with divisor $b = 3$, 'a' can be written in one of the following forms: $3q$, $3q + 1$, or $3q + 2$, where $q$ is some integer. (We choose $b=3$ because $3^3=27$, which is a multiple of 9, making the algebra simpler).

We will now find the cube of each of these forms.

Case 1: When $a = 3q$.

Cubing both sides:

$a^3 = (3q)^3 = 27q^3$

This can be written as:

$a^3 = 9(3q^3)$

Let $m = 3q^3$. Since $q$ is an integer, $m$ is also an integer.

Thus, $a^3$ is of the form $9m$.

Case 2: When $a = 3q + 1$.

Cubing both sides and using the identity $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$:

$a^3 = (3q + 1)^3 = (3q)^3 + 3(3q)^2(1) + 3(3q)(1)^2 + 1^3$

$a^3 = 27q^3 + 27q^2 + 9q + 1$

Factor out 9 from the first three terms:

$a^3 = 9(3q^3 + 3q^2 + q) + 1$

Let $m = 3q^3 + 3q^2 + q$. Since $q$ is an integer, $m$ is also an integer.

Thus, $a^3$ is of the form $9m + 1$.

Case 3: When $a = 3q + 2$.

Cubing both sides:

$a^3 = (3q + 2)^3 = (3q)^3 + 3(3q)^2(2) + 3(3q)(2)^2 + 2^3$

$a^3 = 27q^3 + 54q^2 + 36q + 8$

Factor out 9 from the first three terms:

$a^3 = 9(3q^3 + 6q^2 + 4q) + 8$

Let $m = 3q^3 + 6q^2 + 4q$. Since $q$ is an integer, $m$ is also an integer.

Thus, $a^3$ is of the form $9m + 8$.

Conclusion: In all possible cases, the cube of any positive integer 'a' can be expressed in the form of $9m$, $9m + 1$, or $9m + 8$ for some integer $m$.

Hence Proved.

To Check:

Whether there is any value of the natural number $n$ for which the number $4^n$ ends with the digit zero.

Solution:

If a number ends with the digit zero, it must be divisible by 10. For a number to be divisible by 10, its prime factorization must contain both 2 and 5 as prime factors.

Let's analyze the prime factorization of the number $4^n$.

The number 4 can be written as $4 = 2 \times 2 = 2^2$.

Therefore, the number $4^n$ can be written as:

The prime factorization of $4^n$ only contains the prime factor 2. It does not contain the prime factor 5.

According to the Fundamental Theorem of Arithmetic, the prime factorization of any natural number is unique. Since the prime factorization of $4^n$ does not contain the prime number 5, it means that for any natural number $n$, $4^n$ will not be divisible by 5.

As $4^n$ is not divisible by 5, it cannot be divisible by $10$ (which is $2 \times 5$).

Hence, $4^n$ can never end with the digit zero for any natural number $n$.

Conclusion:

There is no value of $n$ for which $4^n$ ends with the digit zero.

To Find:

The LCM (Least Common Multiple) and HCF (Highest Common Factor) of 6 and 20 using the prime factorization method.

Solution:

First, we find the prime factorization of each number.

Prime factorization of 6:

$\begin{array}{c|cc} 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $6 = 2^1 \times 3^1$.

Prime factorization of 20:

$\begin{array}{c|cc} 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $20 = 2^2 \times 5^1$.

Finding the HCF:

The HCF is the product of the smallest power of each common prime factor.

The only common prime factor is 2. The smallest power of 2 is $2^1$.

Therefore, HCF(6, 20) = $2^1 = 2$.

Finding the LCM:

The LCM is the product of the greatest power of each prime factor involved in the numbers.

The prime factors involved are 2, 3, and 5.

The greatest power of 2 is $2^2$.

The greatest power of 3 is $3^1$.

The greatest power of 5 is $5^1$.

Therefore, LCM(6, 20) = $2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$.

Final Answer:

HCF(6, 20) = 2

LCM(6, 20) = 60

To Find:

The HCF of 96 and 404 by prime factorization, and then find their LCM.

Solution:

Step 1: Find the prime factorization of 96 and 404.

Prime factorization of 96:

$\begin{array}{c|cc} 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $96 = 2^5 \times 3^1$.

Prime factorization of 404:

$\begin{array}{c|cc} 2 & 404 \\ \hline 2 & 202 \\ \hline 101 & 101 \\ \hline & 1 \end{array}$

So, $404 = 2^2 \times 101^1$.

Step 2: Find the HCF.

The HCF is the product of the smallest power of each common prime factor.

The only common prime factor is 2. The smallest power of 2 is $2^2$.

Therefore, HCF(96, 404) = $2^2 = 4$.

Step 3: Find the LCM using the HCF.

We know that for any two positive integers $a$ and $b$, the product of the numbers is equal to the product of their HCF and LCM.

HCF$(a, b) \times$ LCM$(a, b) = a \times b$

So, LCM$(96, 404) = \frac{96 \times 404}{\text{HCF}(96, 404)}$

LCM$(96, 404) = \frac{96 \times 404}{4}$

LCM$(96, 404) = 24 \times 404$

$\begin{array}{cc}& & 4 & 0 & 4 \\ \times & & & 2 & 4 \\ \hline & 1 & 6 & 1 & 6 \\ & 8 & 0 & 8 & \times \\ \hline & 9 & 6 & 9 & 6 \\ \hline \end{array}$

So, LCM(96, 404) = 9696.

Final Answer:

HCF(96, 404) = 4

LCM(96, 404) = 9696

To Find:

The HCF and LCM of 6, 72, and 120 using the prime factorization method.

Solution:

Step 1: Find the prime factorization of each number.

Prime factorization of 6: $6 = 2^1 \times 3^1$.

Prime factorization of 72:

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

So, $72 = 2^3 \times 3^2$.

Prime factorization of 120:

$\begin{array}{c|cc} 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $120 = 2^3 \times 3^1 \times 5^1$.

Step 2: Find the HCF.

The HCF is the product of the smallest power of each common prime factor.

The common prime factors are 2 and 3.

Smallest power of 2: $2^1$ (from the factorization of 6).

Smallest power of 3: $3^1$ (from the factorizations of 6 and 120).

HCF(6, 72, 120) = $2^1 \times 3^1 = 6$.

Step 3: Find the LCM.

The LCM is the product of the greatest power of each prime factor involved.

The prime factors involved are 2, 3, and 5.

Greatest power of 2: $2^3$ (from 72 and 120).

Greatest power of 3: $3^2$ (from 72).

Greatest power of 5: $5^1$ (from 120).

LCM(6, 72, 120) = $2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360$.

Final Answer:

HCF(6, 72, 120) = 6

LCM(6, 72, 120) = 360

(i) 140

Prime factorization of 140:

$\begin{array}{c|cc} 2 & 140 \\ \hline 2 & 70 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

Thus, the product of its prime factors is $140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7$.

(ii) 156

Prime factorization of 156:

$\begin{array}{c|cc} 2 & 156 \\ \hline 2 & 78 \\ \hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

Thus, the product of its prime factors is $156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13$.

(iii) 3825

Prime factorization of 3825:

$\begin{array}{c|cc} 3 & 3825 \\ \hline 3 & 1275 \\ \hline 5 & 425 \\ \hline 5 & 85 \\ \hline 17 & 17 \\ \hline & 1 \end{array}$

Thus, the product of its prime factors is $3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17$.

(iv) 5005

Prime factorization of 5005:

$\begin{array}{c|cc} 5 & 5005 \\ \hline 7 & 1001 \\ \hline 11 & 143 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

Thus, the product of its prime factors is $5005 = 5 \times 7 \times 11 \times 13$.

(v) 7429

Prime factorization of 7429:

$\begin{array}{c|cc} 17 & 7429 \\ \hline 19 & 437 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$

Thus, the product of its prime factors is $7429 = 17 \times 19 \times 23$.

(i) 26 and 91

Prime Factorization:

$26 = 2 \times 13$

$91 = 7 \times 13$

HCF and LCM:

HCF(26, 91) = (Product of smallest powers of common prime factors) = $13^1 = 13$.

LCM(26, 91) = (Product of greatest powers of all prime factors) = $2^1 \times 7^1 \times 13^1 = 182$.

Verification:

Product of the two numbers = $26 \times 91 = 2366$.

Product of HCF and LCM = $13 \times 182 = 2366$.

Since Product of numbers = Product of HCF and LCM, the relation is verified.

(ii) 510 and 92

Prime Factorization:

$510 = 2 \times 3 \times 5 \times 17$

$92 = 2 \times 2 \times 23 = 2^2 \times 23$

HCF and LCM:

HCF(510, 92) = $2^1 = 2$.

LCM(510, 92) = $2^2 \times 3 \times 5 \times 17 \times 23 = 4 \times 3 \times 5 \times 17 \times 23 = 23460$.

Verification:

Product of the two numbers = $510 \times 92 = 46920$.

Product of HCF and LCM = $2 \times 23460 = 46920$.

Since Product of numbers = Product of HCF and LCM, the relation is verified.

(iii) 336 and 54

Prime Factorization:

$336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 = 2^4 \times 3 \times 7$

$54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3$

HCF and LCM:

HCF(336, 54) = $2^1 \times 3^1 = 6$.

LCM(336, 54) = $2^4 \times 3^3 \times 7^1 = 16 \times 27 \times 7 = 3024$.

Verification:

Product of the two numbers = $336 \times 54 = 18144$.

Product of HCF and LCM = $6 \times 3024 = 18144$.

Since Product of numbers = Product of HCF and LCM, the relation is verified.

(i) 12, 15 and 21

Prime Factorization:

$12 = 2 \times 2 \times 3 = 2^2 \times 3$

$15 = 3 \times 5$

$21 = 3 \times 7$

HCF and LCM:

HCF(12, 15, 21) = (Smallest power of common prime factors) = $3^1 = 3$.

LCM(12, 15, 21) = (Greatest power of all prime factors) = $2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 420$.

HCF = 3, LCM = 420

(ii) 17, 23 and 29

Prime Factorization:

17, 23, and 29 are all prime numbers.

$17 = 17$

$23 = 23$

$29 = 29$

HCF and LCM:

Since there are no common prime factors, the HCF is 1.

HCF(17, 23, 29) = 1.

The LCM is the product of the numbers themselves.

LCM(17, 23, 29) = $17 \times 23 \times 29 = 11339$.

HCF = 1, LCM = 11339

(iii) 8, 9 and 25

Prime Factorization:

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

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

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

HCF and LCM:

These numbers are co-prime (no common prime factors). Therefore, their HCF is 1.

HCF(8, 9, 25) = 1.

The LCM is the product of the highest powers of all prime factors.

LCM(8, 9, 25) = $2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$.

HCF = 1, LCM = 1800

Given:

The two numbers are 306 and 657.

HCF(306, 657) = 9.

To Find:

LCM(306, 657).

Solution:

We know the fundamental relationship between the HCF and LCM of two numbers:

Product of two numbers = HCF $\times$ LCM

So, $306 \times 657 = \text{HCF}(306, 657) \times \text{LCM}(306, 657)$.

Substituting the given HCF value:

$306 \times 657 = 9 \times \text{LCM}(306, 657)$

Rearranging the formula to find the LCM:

$\text{LCM}(306, 657) = \frac{306 \times 657}{9}$

We can divide 306 by 9:

$\frac{\cancel{306}^{34}}{\cancel{9}_{1}} = 34$.

Now, calculate the product:

$\text{LCM}(306, 657) = 34 \times 657$

$\begin{array}{cc}& & 6 & 5 & 7 \\ \times & & & 3 & 4 \\ \hline & 2 & 6 & 2 & 8 \\ 1 & 9 & 7 & 1 & \times \\ \hline 2 & 2 & 3 & 3 & 8 \\ \hline \end{array}$

LCM(306, 657) = 22338.

Final Answer: The LCM of 306 and 657 is 22338.

To Check:

Whether $6^n$ can end with the digit 0 for any natural number $n$.

Explanation:

For a number to end with the digit 0, it must be divisible by 10. This means its prime factorization must include both 2 and 5 as factors.

Let's find the prime factorization of $6^n$.

The prime factors of 6 are 2 and 3.

$6 = 2 \times 3$

Therefore, for any natural number $n$, the prime factorization of $6^n$ is:

$6^n = (2 \times 3)^n = 2^n \times 3^n$

The prime factors of $6^n$ are only 2 and 3. According to the Fundamental Theorem of Arithmetic, this factorization is unique.

For $6^n$ to end in 0, the prime 5 must be one of its factors. However, the prime factorization of $6^n$ does not contain the prime 5.

Since 5 is not a prime factor of $6^n$, it cannot be divisible by 5, and therefore cannot be divisible by 10.

Conclusion: $6^n$ cannot end with the digit 0 for any natural number $n$.

A composite number is a positive integer that has at least one divisor other than 1 and itself. We need to show that both given expressions result in composite numbers.

Expression 1: $7 \times 11 \times 13 + 13$

We can factor out the common term, 13, from the expression:

$7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 \times 1 + 1)$

$= 13 \times (77 + 1)$

$= 13 \times 78$

The expression is a product of two integers, 13 and 78. Since the number has factors other than 1 and itself (namely 13 and 78), it is a composite number.

Expression 2: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$

We can factor out the common term, 5, from the expression:

$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$

$= 5 \times (1008 + 1)$

$= 5 \times 1009$

The expression is a product of two integers, 5 and 1009. Since the number has factors other than 1 and itself (namely 5 and 1009), it is a composite number.

Conclusion: Both expressions can be factored into products of integers greater than 1, proving that they are composite numbers.

Given:

Time taken by Sonia to complete one round = 18 minutes.

Time taken by Ravi to complete one round = 12 minutes.

They start at the same point, at the same time, and go in the same direction.

To Find:

The time after which they will meet again at the starting point.

Solution:

To find when they will meet again at the starting point, we need to find the smallest time that is a multiple of both 18 and 12. This is the Least Common Multiple (LCM) of 18 and 12.

We will use the prime factorization method to find the LCM.

Prime factorization of 18:

$18 = 2 \times 9 = 2 \times 3^2$

Prime factorization of 12:

$12 = 4 \times 3 = 2^2 \times 3$

The LCM is the product of the highest powers of all prime factors involved.

Highest power of 2 is $2^2$.

Highest power of 3 is $3^2$.

LCM(18, 12) = $2^2 \times 3^2 = 4 \times 9 = 36$.

Final Answer: They will meet again at the starting point after 36 minutes.

To Prove:

$\sqrt{3}$ is an irrational number.

Proof by Contradiction:

Let us assume the opposite, that $\sqrt{3}$ is a rational number.

If it is rational, it can be expressed as a fraction $\frac{a}{b}$ in its simplest form, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are co-prime (their highest common factor is 1).

Squaring both sides of the equation:

$3 = \left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}$

Rearranging the equation gives:

This equation shows that $a^2$ is a multiple of 3. By a theorem, if a prime number (like 3) divides the square of an integer, it must also divide the integer itself. Therefore, 3 divides $a$.

We can write $a = 3c$ for some integer $c$.

Now, substitute this value of $a$ back into equation (i):

$(3c)^2 = 3b^2$

$9c^2 = 3b^2$

Divide both sides by 3:

This shows that $b^2$ is a multiple of 3. Following the same logic, if 3 divides $b^2$, then 3 must also divide $b$.

So, we have concluded that 3 is a factor of both $a$ and $b$.

This means that $a$ and $b$ have a common factor of 3. This contradicts our initial assumption that $a$ and $b$ are co-prime (have no common factors other than 1).

Since our assumption leads to a contradiction, the assumption must be false.

Conclusion:

Therefore, $\sqrt{3}$ cannot be expressed as a rational number, and it is irrational.

Hence Proved.

To Prove:

$5 - \sqrt{3}$ is an irrational number.

Proof by Contradiction:

Let us assume the opposite, that $5 - \sqrt{3}$ is a rational number.

If it is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.

Now, we rearrange the equation to isolate the term $\sqrt{3}$:

$5 - \frac{a}{b} = \sqrt{3}$

Combine the terms on the left side into a single fraction:

Let's analyze the right side of the equation, $\frac{5b - a}{b}$.

Since $a$ and $b$ are integers, $5b$ is an integer and $5b-a$ is also an integer. As $b \neq 0$, the fraction $\frac{5b - a}{b}$ is a ratio of two integers with a non-zero denominator. By definition, this means $\frac{5b - a}{b}$ is a rational number.

Our equation now states that $\sqrt{3}$ is equal to a rational number. This implies that $\sqrt{3}$ is rational.

However, we know for a fact that $\sqrt{3}$ is an irrational number.

This is a contradiction. The rational can never be equal to the irrational.

This contradiction arose because of our incorrect initial assumption that $5 - \sqrt{3}$ is rational.

Conclusion:

Therefore, our assumption is false, and $5 - \sqrt{3}$ must be an irrational number.

Hence Proved.

To Prove:

$3\sqrt{2}$ is an irrational number.

Proof by Contradiction:

Let us assume the opposite, that $3\sqrt{2}$ is a rational number.

If it is rational, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.

Now, we rearrange the equation to isolate the term $\sqrt{2}$:

Divide both sides by 3:

Let's analyze the right side of the equation, $\frac{a}{3b}$.

Since $a$ and $b$ are integers, $3b$ is also an integer. Since $b \neq 0$, it follows that $3b \neq 0$.

The expression $\frac{a}{3b}$ is a ratio of two integers with a non-zero denominator. By definition, this means $\frac{a}{3b}$ is a rational number.

Our equation now states that $\sqrt{2}$ is equal to a rational number. This implies that $\sqrt{2}$ is rational.

However, it is a known mathematical fact that $\sqrt{2}$ is an irrational number.

This leads to a contradiction, as an irrational number ($\sqrt{2}$) cannot be equal to a rational number ($\frac{a}{3b}$).

This contradiction has arisen from our incorrect initial assumption that $3\sqrt{2}$ is rational.

Conclusion:

Therefore, our assumption is false, and $3\sqrt{2}$ must be an irrational number.

Hence Proved.

To Prove:

$\sqrt{5}$ is an irrational number.

Proof by Contradiction:

Let us assume the contrary, that $\sqrt{5}$ is a rational number.

If it is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are co-prime (i.e., they have no common factors other than 1).

Squaring both sides of the equation, we get:

$(\sqrt{5})^2 = \left(\frac{a}{b}\right)^2$

$5 = \frac{a^2}{b^2}$

Rearranging the equation gives:

From equation (i), we can see that $a^2$ is divisible by 5. By a theorem, if a prime number (like 5) divides the square of an integer, it must also divide the integer itself. Therefore, 5 divides $a$.

So, we can write $a = 5c$ for some integer $c$.

Substitute $a = 5c$ into equation (i):

$(5c)^2 = 5b^2$

$25c^2 = 5b^2$

Divide both sides by 5:

$5c^2 = b^2$

This shows that $b^2$ is divisible by 5. Following the same logic, if 5 divides $b^2$, then 5 must also divide $b$.

Thus, we have established that 5 is a common factor of both $a$ and $b$.

This contradicts our initial assumption that $a$ and $b$ are co-prime (their HCF is 1).

This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.

Conclusion:

Therefore, our assumption is false. $\sqrt{5}$ cannot be a rational number.

Hence, $\sqrt{5}$ is an irrational number.

To Prove:

$3 + 2\sqrt{5}$ is an irrational number.

Proof by Contradiction:

Let us assume the contrary, that $3 + 2\sqrt{5}$ is a rational number.

If it is rational, it can be written in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.

Now, we rearrange the equation to isolate the irrational part ($\sqrt{5}$).

Subtract 3 from both sides:

$2\sqrt{5} = \frac{a}{b} - 3$

$2\sqrt{5} = \frac{a - 3b}{b}$

Divide both sides by 2:

Let's analyze the right-hand side (RHS) of the equation, $\frac{a - 3b}{2b}$.

Since $a$ and $b$ are integers, $a-3b$ and $2b$ are also integers. Since $b \neq 0$, it follows that $2b \neq 0$.

Therefore, the RHS, $\frac{a - 3b}{2b}$, is a ratio of two integers with a non-zero denominator, which by definition is a rational number.

Our equation now states that $\sqrt{5}$ (an irrational number) is equal to $\frac{a - 3b}{2b}$ (a rational number).

This is a contradiction, as an irrational number cannot be equal to a rational number.

This contradiction arose because of our incorrect initial assumption that $3 + 2\sqrt{5}$ is rational.

Conclusion:

Therefore, our assumption is false.

Hence, $3 + 2\sqrt{5}$ is an irrational number.

(i) Prove that $\frac{1}{\sqrt{2}}$ is irrational.

Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational. Then it can be written as $\frac{a}{b}$, where $a, b$ are co-prime integers and $b \neq 0$.

$\frac{1}{\sqrt{2}} = \frac{a}{b}$

Rearranging the equation, we get:

$\sqrt{2} = \frac{b}{a}$

Since $a$ and $b$ are integers, $\frac{b}{a}$ is a rational number. This implies that $\sqrt{2}$ is a rational number.

But this contradicts the fact that $\sqrt{2}$ is irrational. This contradiction has arisen because of our incorrect assumption.

Therefore, $\frac{1}{\sqrt{2}}$ is irrational.

(ii) Prove that $7\sqrt{5}$ is irrational.

Let us assume, to the contrary, that $7\sqrt{5}$ is rational. Then it can be written as $\frac{a}{b}$, where $a, b$ are co-prime integers and $b \neq 0$.

$7\sqrt{5} = \frac{a}{b}$

Rearranging the equation, we get:

$\sqrt{5} = \frac{a}{7b}$

Since $a$, $b$, and 7 are integers, $\frac{a}{7b}$ is a rational number. This implies that $\sqrt{5}$ is a rational number.

But this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction has arisen because of our incorrect assumption.

Therefore, $7\sqrt{5}$ is irrational.

(iii) Prove that $6 + \sqrt{2}$ is irrational.

Let us assume, to the contrary, that $6 + \sqrt{2}$ is rational. Then it can be written as $\frac{a}{b}$, where $a, b$ are co-prime integers and $b \neq 0$.

$6 + \sqrt{2} = \frac{a}{b}$

Rearranging the equation, we get:

$\sqrt{2} = \frac{a}{b} - 6$

$\sqrt{2} = \frac{a - 6b}{b}$

Since $a$ and $b$ are integers, $\frac{a - 6b}{b}$ is a rational number. This implies that $\sqrt{2}$ is a rational number.

But this contradicts the fact that $\sqrt{2}$ is irrational. This contradiction has arisen because of our incorrect assumption.

Therefore, $6 + \sqrt{2}$ is irrational.

Concept Used:

A rational number $\frac{p}{q}$ (in its simplest form) has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^m \cdot 5^n$, where $m$ and $n$ are non-negative integers.

If the prime factorization of the denominator $q$ has any prime factor other than 2 or 5, the decimal expansion will be non-terminating repeating.

(i) $\frac{13}{3125}$

The denominator is $3125$. Prime factorization of $3125 = 5^5$.

Since the denominator's prime factorization is of the form $2^0 \cdot 5^5$, the decimal expansion is terminating.

(ii) $\frac{17}{8}$

The denominator is $8$. Prime factorization of $8 = 2^3$.

Since the denominator's prime factorization is of the form $2^3 \cdot 5^0$, the decimal expansion is terminating.

(iii) $\frac{64}{455}$

The denominator is $455$. Prime factorization of $455 = 5 \times 7 \times 13$.

Since the denominator has prime factors (7 and 13) other than 2 and 5, the decimal expansion is non-terminating repeating.

(iv) $\frac{15}{1600}$

First, simplify the fraction: $\frac{15}{1600} = \frac{3 \times 5}{320 \times 5} = \frac{3}{320}$.

The denominator is $320$. Prime factorization of $320 = 32 \times 10 = 2^5 \times 2 \times 5 = 2^6 \times 5^1$.

Since the denominator's prime factorization is of the form $2^6 \cdot 5^1$, the decimal expansion is terminating.

(v) $\frac{29}{343}$

The denominator is $343$. Prime factorization of $343 = 7^3$.

Since the denominator has a prime factor (7) other than 2 and 5, the decimal expansion is non-terminating repeating.

(vi) $\frac{23}{2^{3}5^{2}}$

The denominator is already given in prime factor form as $2^3 \cdot 5^2$.

The prime factors are only 2 and 5. Therefore, the decimal expansion is terminating.

(vii) $\frac{129}{2^{2}5^{7}7^{5}}$

The prime factorization of the numerator is $129 = 3 \times 43$. There are no common factors to cancel with the denominator.

The denominator $2^2 \cdot 5^7 \cdot 7^5$ has a prime factor of 7, which is not 2 or 5.

Therefore, the decimal expansion is non-terminating repeating.

(viii) $\frac{6}{15}$

First, simplify the fraction: $\frac{6}{15} = \frac{2 \times 3}{5 \times 3} = \frac{2}{5}$.

The denominator is 5. Its prime factorization is $5^1$ (or $2^0 \cdot 5^1$).

Therefore, the decimal expansion is terminating.

(ix) $\frac{35}{50}$

First, simplify the fraction: $\frac{35}{50} = \frac{7 \times 5}{10 \times 5} = \frac{7}{10}$.

The denominator is $10 = 2^1 \times 5^1$.

Since the denominator's prime factors are only 2 and 5, the decimal expansion is terminating.

(x) $\frac{77}{210}$

First, simplify the fraction: $\frac{77}{210} = \frac{11 \times 7}{30 \times 7} = \frac{11}{30}$.

The denominator is $30$. Prime factorization of $30 = 2 \times 3 \times 5$.

Since the denominator has a prime factor (3) other than 2 and 5, the decimal expansion is non-terminating repeating.

The rational numbers from Question 1 with terminating decimal expansions are (i), (ii), (iv), (vi), (viii), and (ix).

(i) $\frac{13}{3125}$

We have $3125 = 5^5$. To make the denominator a power of 10, we need to multiply by $2^5$.

$\frac{13}{3125} = \frac{13}{5^5} = \frac{13 \times 2^5}{5^5 \times 2^5} = \frac{13 \times 32}{(5 \times 2)^5} = \frac{416}{10^5} = \frac{416}{100000} = 0.00416$.

Decimal expansion: 0.00416

(ii) $\frac{17}{8}$

We have $8 = 2^3$. To make the denominator a power of 10, we need to multiply by $5^3$.

$\frac{17}{8} = \frac{17}{2^3} = \frac{17 \times 5^3}{2^3 \times 5^3} = \frac{17 \times 125}{(2 \times 5)^3} = \frac{2125}{10^3} = 2.125$.

Decimal expansion: 2.125

(iv) $\frac{15}{1600}$

Simplified form is $\frac{3}{320}$. We have $320 = 2^6 \times 5^1$. We need to multiply by $5^5$.

$\frac{3}{320} = \frac{3}{2^6 \times 5^1} = \frac{3 \times 5^5}{2^6 \times 5^6} = \frac{3 \times 3125}{10^6} = \frac{9375}{1000000} = 0.009375$.

Decimal expansion: 0.009375

(vi) $\frac{23}{2^{3}5^{2}}$

We have $2^3 \cdot 5^2$. We need to multiply by $5^1$ to get $2^3 \cdot 5^3 = 10^3$.

$\frac{23}{2^3 \times 5^2} = \frac{23 \times 5}{2^3 \times 5^3} = \frac{115}{1000} = 0.115$.

Decimal expansion: 0.115

(viii) $\frac{6}{15}$

Simplified form is $\frac{2}{5}$. We need to multiply by 2.

$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10} = 0.4$.

Decimal expansion: 0.4

(ix) $\frac{35}{50}$

Simplified form is $\frac{7}{10}$.

$\frac{7}{10} = 0.7$.

Decimal expansion: 0.7

(i) 43.123456789

The decimal expansion is terminating. A number with a terminating decimal expansion is always rational.

When a rational number has a terminating decimal expansion, its denominator $q$ (in the simplest form $\frac{p}{q}$) must have a prime factorization of the form $2^m \cdot 5^n$, where $m$ and $n$ are non-negative integers.

Conclusion: The number is rational. The prime factors of $q$ will be either 2 or 5 or both.

(ii) 0.120120012000120000...

The decimal expansion is non-terminating and non-repeating (the number of zeroes increases each time). A number with a non-terminating, non-repeating decimal expansion is irrational.

Conclusion: The number is irrational. It cannot be expressed in the form $\frac{p}{q}$.

(iii) 43.$\overline{123456789}$

The decimal expansion is non-terminating and repeating (the block of digits 123456789 repeats indefinitely). A number with a non-terminating, repeating decimal expansion is always rational.

When a rational number has a non-terminating repeating decimal expansion, its denominator $q$ (in the simplest form $\frac{p}{q}$) must have at least one prime factor other than 2 or 5.

Conclusion: The number is rational. The prime factors of $q$ will have at least one factor other than 2 or 5.