Bunuel wrote:

If 20! × 20!/20^n is an integer, what is the largest possible value of n?

A. 20

B. 16

C. 8

D. 4

E. 1

20! × 20!/20^n is an integer.

Or 20! × 20!/(2^2n*5^n)

This means we need to find the power of 5 in the numerator. We can safely say that the number of powers of 5 will be lower than the number of powers of 4 in the numerator.

Largest power of 5 in 20! = [20/5] + [20/25] = 4

Since there are 2 20!'s, power of 4 in the numerator = 2*4 = 8

Hence the largest value of b for which 20! × 20!/20^n is integer = 8

Correct Option : C