Bunuel wrote:

If a!/6^5 is an integer and a is a positive integer, what is the minimum possible value of a?

A. 30

B. 18

C. 16

D. 15

E. 12

Since we're told that a!/6^5 is an integer we'll look for a solution dealing with integer properties.

This is a Logical approach.

Saying that a!/6^5 is an integer is the same as saying that a! is divisible by 6^5

This means it has 5 6's in its factorization meaning 5 3's and 5 2's.

Since the first 5 factors of 2 are in are 2,(2*2)=4,6,(2*2*2)=8 and the first 5 factors of 3 are 3,6,(3*3)=9,12

then any number larger or equal to 12! willl have 5 3's and 5 2's in its factorization, as required.

(E) is our answer.

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