Ans: $945 = 3 \times 315 = 3^2 \times 105 = 3^3 \times 35 = 3^3 \times 5 \times 7$. The exponents are 3, 1, and 1. Sum of exponents $= 3 + 1 + 1 = \mathbf{5}$.
7.
Ans: The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes in a unique way (except for the order of factors). For $6^n$ to end with 0, it must be divisible by 10, i.e., it must have both 2 and 5 as prime factors. But $6^n = (2 \times 3)^n = 2^n \times 3^n$. By the uniqueness of prime factorization, 5 is not a prime factor of $6^n$ for any $n$. Hence, $6^n$ cannot end with digit 0.
8.
Ans: $4^n = (2^2)^n = 2^{2n}$. The only prime factor of $4^n$ is 2. For a number to end with 0, it must contain both 2 and 5 as prime factors. Since 5 is absent, $4^n$ can never end with digit 0 for any natural number $n$.
9.
Ans: $7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13 \times 78 = 13 \times 2 \times 3 \times 13$. Since the expression equals a product of two integers each greater than 1, it has factors other than 1 and itself. Hence it is a composite number.
10.
Ans: $7! + 5 = 5040 + 5 = 5(1008 + 1) = 5 \times 1009$. Since it is a product of two numbers (5 and 1009) both greater than 1, it is a composite number.
Ans:No. The LCM of two numbers must always be a multiple of their HCF. Here, $\dfrac{380}{18} = 21.1\overline{1}$, which is not an integer. So 380 is not a multiple of 18. Therefore, two numbers cannot have HCF = 18 and LCM = 380.
22.
Ans: Since $p$ and $q$ are distinct primes, they have no common factors other than 1. Therefore, $\text{HCF}(p, q) = \mathbf{1}$ and $\text{LCM}(p, q) = \mathbf{pq}$.
23.
Ans: For co-prime numbers, $\text{HCF} = 1$. Using $\text{HCF} \times \text{LCM} = a \times b$: $1 \times 4875 = 75 \times b \implies b = \dfrac{4875}{75} = \mathbf{65}$.
24.
Ans: The smallest number divisible by both is their LCM. $306 = 2 \times 3^2 \times 17$, $657 = 3^2 \times 73$. $\text{LCM} = 2 \times 3^2 \times 17 \times 73 = \mathbf{22338}$.
Ans: For minimum area (minimum number of stacks), each stack must have the maximum number of barfis, which equals $\text{HCF}(420, 130)$. $420 = 2^2 \times 3 \times 5 \times 7$, $130 = 2 \times 5 \times 13$. $\text{HCF} = 2 \times 5 = \mathbf{10}$ barfis per stack.
36.
Ans: Height of each stack is equal to the maximum number of books per stack $= \text{HCF}(96, 240, 336)$.
$840 \times 130 = 109200$ and $840 \times 131 = 110040$. Both are near 110000 but 110040 $> 100000$ and is the nearest multiple greater than 100000 that is $> 110000$ is ambiguous; taking nearest to 110000: $|110000 - 109200| = 800$ and $|110040 - 110000| = 40$. Hence the required number is $\mathbf{110040}$.
48.
Ans: Length $= 1872$ cm, Breadth $= 1320$ cm. Size of largest square tile $= \text{HCF}(1872, 1320)$.
Ans:Proof: Assume $\sqrt{2}$ is rational. So $\sqrt{2} = \dfrac{p}{q}$, where $p, q$ are co-prime integers, $q \neq 0$. Squaring: $2 = \dfrac{p^2}{q^2} \implies p^2 = 2q^2$. So $2 | p^2 \implies 2 | p$. Let $p = 2c$. Then $(2c)^2 = 2q^2 \implies 4c^2 = 2q^2 \implies q^2 = 2c^2$. So $2 | q$. Both $p$ and $q$ are divisible by 2, contradicting that they are co-prime. Hence $\sqrt{2}$ is irrational. $\blacksquare$
50.
Ans: Assume $\sqrt{3} = \dfrac{a}{b}$ ($a, b$ co-prime). Squaring: $a^2 = 3b^2$. So $3 | a \implies a = 3c$. Then $9c^2 = 3b^2 \implies b^2 = 3c^2 \implies 3 | b$. This contradicts that $a$ and $b$ are co-prime. Hence $\sqrt{3}$ is irrational. $\blacksquare$
51.
Ans: Assume $\sqrt{5} = \dfrac{p}{q}$ ($p, q$ co-prime). $p^2 = 5q^2 \implies 5 | p \implies p = 5k$. Then $25k^2 = 5q^2 \implies q^2 = 5k^2 \implies 5 | q$. Contradiction. Hence $\sqrt{5}$ is irrational. $\blacksquare$
52.
Ans: Assume $\sqrt{7} = \dfrac{a}{b}$ ($a, b$ co-prime). $a^2 = 7b^2 \implies 7 | a \implies a = 7m$. Then $49m^2 = 7b^2 \implies b^2 = 7m^2 \implies 7 | b$. Contradiction. Hence $\sqrt{7}$ is irrational. $\blacksquare$
53.
Ans: Assume $\sqrt{11} = \dfrac{a}{b}$ ($a, b$ co-prime). $a^2 = 11b^2 \implies 11 | a \implies a = 11k$. Then $121k^2 = 11b^2 \implies b^2 = 11k^2 \implies 11 | b$. Contradiction. Hence $\sqrt{11}$ is irrational. $\blacksquare$
Since $a, b$ are integers, $\dfrac{a-3b}{2b}$ is rational. This means $\sqrt{5}$ is rational – a contradiction. Hence $3 + 2\sqrt{5}$ is irrational. $\blacksquare$
55.
Ans: Assume $\dfrac{1}{\sqrt{2}}$ is rational $= \dfrac{a}{b}$ (integers, $a, b \neq 0$). Then $\sqrt{2} = \dfrac{b}{a}$, which is rational. This contradicts the fact that $\sqrt{2}$ is irrational. Hence $\dfrac{1}{\sqrt{2}}$ is irrational. $\blacksquare$
56.
Ans: Assume $7\sqrt{5}$ is rational $= \dfrac{a}{b}$. Then $\sqrt{5} = \dfrac{a}{7b}$, which is rational. This contradicts the given fact that $\sqrt{5}$ is irrational. Hence $7\sqrt{5}$ is irrational. $\blacksquare$
57.
Ans: Assume $6 + \sqrt{2}$ is rational $= \dfrac{a}{b}$. Then $\sqrt{2} = \dfrac{a}{b} - 6 = \dfrac{a - 6b}{b}$, which is rational – contradicting that $\sqrt{2}$ is irrational. Hence $6 + \sqrt{2}$ is irrational. $\blacksquare$
58.
Ans: Assume $5 - \sqrt{3}$ is rational $= \dfrac{p}{q}$. Then $\sqrt{3} = 5 - \dfrac{p}{q} = \dfrac{5q - p}{q}$, which is rational – contradicting that $\sqrt{3}$ is irrational. Hence $5 - \sqrt{3}$ is irrational. $\blacksquare$
Since the right side is rational, $\sqrt{6}$ must be rational. But $6 = 2 \times 3$ (not a perfect square), so $\sqrt{6}$ is irrational – a contradiction. Hence $\sqrt{2} + \sqrt{3}$ is irrational. $\blacksquare$
60.
Ans:Yes, always irrational. Let the non-zero rational be $\dfrac{p}{q} \neq 0$ and the irrational be $x$. Assume the product $\dfrac{p}{q} \cdot x$ is rational, say $r$. Then $x = \dfrac{rq}{p}$. Since $r, q, p$ are integers ($p \neq 0$), $\dfrac{rq}{p}$ is rational. This contradicts that $x$ is irrational. Hence the product is always irrational. $\blacksquare$
Ans:Option (a) – Both A and R are true and R is the correct explanation of A.
Using the formula in R: $\text{LCM} = \dfrac{\text{Product}}{\text{HCF}} = \dfrac{150}{5} = 30$. So A is correct and R correctly justifies it.
62.
Ans:Option (d) – A is false but R is true.
$6^n = (2 \times 3)^n = 2^n \times 3^n$. Since 5 is NOT a prime factor of $6^n$, it can never end with digit 0. So A is false. However, R is a true and correct statement about numbers ending in 0.
63.
Ans:Option (a) – Both A and R are true and R is the correct explanation of A.
The theorem in R (if prime $p$ divides $a^2$, then $p$ divides $a$) is the key lemma used in the proof that $\sqrt{2}$ is irrational, making R the correct explanation for A.
64.
Ans:Option (c) – A is true but R is false.
$\text{LCM}(12, 21, 15) = 2^2 \times 3 \times 5 \times 7 = 420$ is correct, so A is true. But R says "lowest powers of common prime factors" – that is the definition of HCF, not LCM. LCM uses the highest powers of all prime factors. So R is false.
65.
Ans:Option (a) – Both A and R are true and R is the correct explanation of A.
$5 \times 11 \times 13 + 13 = 13(5 \times 11 + 1) = 13 \times 56 = 13 \times 8 \times 7$. It has factors other than 1 and itself, so it is composite as per the definition in R.
66.
Ans:Option (d) – A is false but R is true.
$\dfrac{350}{12} = 29.1\overline{6}$, which is not an integer. So 350 is not a multiple of 12, which means they cannot be LCM and HCF of the same pair of numbers. A is false. R is a correct statement – LCM is always a multiple of HCF.
67.
Ans:Option (c) – A is true but R is false.
A is a correct theorem about terminating decimals. But R is false – not all rational numbers are terminating decimals (e.g., $\dfrac{1}{3} = 0.\overline{3}$ is a non-terminating repeating decimal). So R is incorrect.
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