Question

Consider $M =2^4 3^4 5^2 7^2 11^2$ and match the column.

Column I		Column II
P	Number of ways in which $M$ can be resolved as the product of 2 divisors is	1	30
Q	Number of divisors of M which are divisible by 60 is	2	216
R	Sum of all the divisors of $M$ of the form $(2 n+1), n \in N$ is $11^2(133 K )$ then $K$ is equal to	3	338
S	Sum of all the odd divisors which are divisible by ' 5 ' but not ' 7 ' is $11^2(133 \lambda)$ then $\lambda$ is equal to	4	1767

• A. P-3, Q-2, R-4, S-1
• B. P-2, Q-3, R-4, S-1
• C. P-3, Q-2, R-1, S-4
• D. P-2, Q-3, R-1, S-4

Answer 1

Answer: (A)

(P) Total divisors $=675$
$\therefore \frac{674}{2}+1=338$
(Q) number should be divisible by 4,3,5
at least two 2 's can be taken in 3 ways
at least one 3 's can be taken in 4 ways
at least one 5 's can be taken in 2 ways
7 's can be taken in 3 ways
11 's can be taken in 3 ways
(R)$\therefore \text { no. of divisors }=216$
$\text { sum }=\left(3^0+3^1+3^2+3^3+3^4\right)\left(5^0+5^1+5^2\right)\left(7^0+7^1+7^2\right)\left(11^0+11^1+11^2\right) $
$ =\frac{3^5-1}{3-1} \times \frac{5^3-1}{5-1} \times \frac{7^3-1}{7-1} \times \frac{11^3-1}{11-1}=11^2 \times 31 \times 57 \times 133$
$= K =31 \times 57$
(S)$S=\left(3^0+3^1+3^2+3^3+3^4\right)\left(5^0+5^1+5^2\right)\left(11^0+11^1+11^2\right) $
$121 \times 3 \times 1330 $
$11^2 \times 133 \times 30$
$\lambda=30$