(3)    $a^3{b}^2$

(1)

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

$6^n={(2\times 3)}^n=2^n\times 3^n$ , Its prime factors do not contain 5 i.e., of the form $2^m\times 5^n$,

where m, n are natural numbers. Hence, $6^n$ never ends with the digit zero.

(3)

$a=p^3{q}^4 \text{and} b=p^2{q}^3$

          $\text{HCF}(a,b)=p^2{q}^3$          ...(i)

and   $\text{LCM}(a,b)=p^3{q}^4$           ...(ii)

But gives: $\text{HCF}(a,b)=p^m{q}^n$ and $\text{LCM}(a,b)=p^r{q}^s$

From eq. (i), $p^m{q}^n=p^2{q}^3$

So, $m=2$ and $n=3$

From eq. (ii), $p^r{q}^s=p^3{q}^4$

So, $r=3$ and $s=4$

$\therefore$ $(m+n)(r+s)=(2+3)(3+4)=35.$

(4)

Given, $p=18a^2{b}^4$ and $q=20a^3{b}^2$

$\text{LCM}(p,q)=\text{LCM}(18a^2{b}^4,20a^3{b}^2)=180a^3{b}^4$

(2)

Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)

(3)     $p^5{q}^4$

(1)

Given, HCF = 40 and LCM = 252 × k

We know that, LCM × HCF = Product of two number

$\Rightarrow 40\times 252\times k=2520\times 6600$

$\Rightarrow k=\frac{2520\times 6600}{40\times 252}$

$\Rightarrow k=1650$

(1)

2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7 
$= 2³ \times 3² \times 5 \times 7$
$Comparing with 2520 = 2³ \times 3ᵃ \times b \times 7$
$$\begin{gathered} We get, a = 2, b = 5 \\ \therefore a + 2b = 2 + 2 \times 5 = 2 + 10 = 12 \end{gathered}$$

(4)

We have, the smallest prime number = 2

and the smallest odd composite number = 9

∴ LCM of the smallest prime number and the smallest odd composite number

= LCM (2, 9) = 18

(3)

If LCM (a,b,c) = 3780
By prime factorisation of 3780

$$\begin{gathered} 3780=2^2\times 3^3\times 5\times 7 \\ LCM (a,b,c)=2^2\times 3^3\times 5\times 7 \left(i\right) \\ a=2^2\times 3^x;b=2^2\times 3\times 5; c=2^2\times 3\times 7 \\ LCM (a,b,c)=2^2\times 3^x\times 5\times 7 \left(ii\right) \\ Comparing \left(i\right) and \left(ii\right), we get 𝑥 = 3 \end{gathered}$$