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 $(2n+1),n\in N$ is $11^2(133K)$ 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)$S=\left(3^0+3^1+3^2+3^3+3^4\right)\left(5^0+5^1+5^2\right)\left(11^0+11^1+11^2\right)$
$121\times 3\times 1330$
$11^2\times 133\times 30$
$\lambda =30$