Column I	Column II
(A) The sum of the factors of 8! which are odd and are the form $3\lambda +2, \lambda \in N,$ is	(p) $384$
(B) The number of divisors of $n=2^7. 3^5. 5^3$ which are the form $4\lambda +1, \lambda \in N$, is	(q) $240$
(C) Total number of divisors of $n=2^5. 3^4. 5^{10}.7^4$ which are the form $4\lambda +2, \lambda \ge 1,$ is	(r) $11$
(D) Total number of divisors of $n=3^5.5^7.7^9$ which are the form $4\lambda +1, \lambda \ge 0$, is	(s) $40$

(A) Here, $8!=2^7. 3^2. 5^1.7^1$

So, the factors may be $1,5,7,35$ of which $5$ and $35$ are of the form $3\lambda +2$.

$\therefore$Sum is $40$.

(B) Number of odd numbers $=(5+1)(3+1)=24$

Required number $=12,$ but $1$ is included.

$\therefore$ Required number of numbers $=12-1=11$ of the form $4\lambda +1$

(C) Here, $4\lambda +2= 2(2\lambda +1)$

$\therefore$ Total divisors $1.5.11.7-1=384$

[$\because$ one is subtracted because there will be case when selected powers $3,5$ and $7$ are zero]

(D) Here, any positive integer power of $5$ will be in the form $4\lambda +1$ when even powers of $3$ and 7 will be in the form of $4\lambda +1$ and odd powers of $3$ and 7 will be in the form of $4\lambda -1.$

$\therefore$ Required divisors $=8(3.5+3.5)=240$.