(3)

Smallest 2-digit number = 10 and smallest composite number = 4
LCM of 10 and 4 = 20

(3)

A prime number has only 2 factors, i.e. the number itself and 1.

(2)

$$\begin{gathered} We have, 196 = 2² \times 7² \\ \Rightarrow Sum of exponents = 2 + 2 = 4 \end{gathered}$$

(3)

Given, $$\begin{gathered} a=p^3 q^4, b=p^2 q^3 \\ \therefore HCF (a, b)=p^2 q^3 and LCM (a, b)=p^3 q^4 \end{gathered}$$

Comparing with the HCF and LCM given in the question, we get

$$\begin{gathered} HCF (a, b) = p^m q^n=p^2 q^3 \\ \Rightarrow m=2,n=3 \\ and LCM (a,b) = p^r q^s=p^3 q^4 \\ \Rightarrow r=3,s=4 \\ \Rightarrow (m+n)(r+s)=(2+3)(3+4)=5\times 7=35 \end{gathered}$$

(3)

p and q (p > q) are prime numbers, HCF (p,q) = 1

Given, LCM (p,q) = 221

$$\begin{gathered} \Rightarrow LCM (p,q)\times HCF (p,q)=p\times q \\ \Rightarrow 1\times 221=p\times q\Rightarrow p\times q=221 \\ So, p=17,q=13(\because p>q) \\ 3p-q=3\times 17-13=51-13=38 \end{gathered}$$

(3)

$$\begin{gathered} Given LCM (x, 18) = 36 and HCF (x, 18) = 2 \\ \therefore LCM (x, 18) \times HCF (x, 18) = x \times 18 \\ \Rightarrow 36 \times 2 = x \times 18 \\ \Rightarrow x = (36 \times 2) ÷ 18 = 4 \end{gathered}$$

(3)

HCF = 81, let the two numbers be 81x and 81y.
According to the question,

81 x +81 y  = 1215
x + y = 1215

The gives 4 co-prime pairs (1, 14), (2, 13), (4, 11), (7, 8)

(2)

We have,

 $$\begin{gathered} p=x^2{y}^2 \\ q=x{y}^3 \\ \Rightarrow HCF(p,q)=x{y}^2 \end{gathered}$$

(3)

$$\begin{gathered} We have, HCF = (n – 1) and LCM = (n² – 1)(n² – 4) \\ We know that product of two numbers = LCM \times HCF \\ So product of two numbers = (n – 1)(n² – 1)(n² – 4) \\ = (n – 1)(n – 1)(n + 1)(n² – 4) \\ = (n + 1)(n – 1)²(n² – 4) \end{gathered}$$

(4)

$$\begin{gathered} We know that LCM (a, b) \times HCF (a, b) = a \times b \\ \Rightarrow LCM (P, 18) \times HCF (P, 18) = P \times 18 \\ \Rightarrow 36 \times 2 = P \times 18 \Rightarrow 72 = 18P \\ \Rightarrow P = 72 ÷ 18 = 4 \end{gathered}$$