ISI2021-MMA: 12

Question

The number of distinct even divisors of $$\prod_{k=1}^{5} k!$$ is

• A. $24$
• B. $32$
• C. $64$
• D. $72$

Answer 1

Answer : (C) 64

$\prod_{k=1}^{5} k! = 1! *2!*3!*4!*5!$
$ = 1*2*1*3*2*1*4*3*2*1*5*3*2*1 = 2^8 *3^3 *5$

$\text{Total number of divisors } = (8+1)*(3+1)*(1+1) = 72$

$\text{total number of odd divisors} = (3+1)*(1+1) = 8$

$\text{total number of even divisors} = \text{total number of divisors} – \text{total number of odd divisors}$

$72 – 8 = 64$