Bunuel wrote:

Let a be a positive integer. If n is divisible by 2^a and n is also divisible by 3^(2a), then it is possible that n is NOT divisible by

A. 6

B. 3 × 2^a

C. 2 × 3^(2a)

D. 6^a

E. 6^(2a)

Kudos for a correct solution.

Easiest method is to assume a value of a = 1

Given n is divisible by 2^2 ---> n = 2^1*p ---> n =2p

Also, n is also divisible by 3^(2a) ---> n = 3^(2a) * q = 3^2*q = 9q

Lets look at a few numbers that are multiples of both 2 and 9 are 18,36,72....

Thus looking at the options , A-D divide the numbers 18,36,72.... while E (=6^(2) = 36 ) does not divide 18. Thus E is the correct answer.