varun289 wrote:If k is the product of the integers from 1 to 20, inclusive, what is the greatest integer n for which 4^n is a factor of k?
(A) 5
(B) 7
(C) 9
(D) 10
(E) 12

Since 4 = 2², we need to count how many 2's are contained within the prime-factorization of 20!.
To determine the number of 2's, count how many times EACH POWER OF 2 can divide into 20.

Every multiple of 2 that can divide into 20 provides at least one 2:
20/2¹ = 10..

Every multiple of 2² that can divide into 20 adds a second 2:
20/2² = 20/4 = 5.

Every multiple of 2³ that can divide into 20 adds a third 2:
20/2³ = 20/8 = 2.

Every multiple of 2� that can divide into 20 adds a fourth 2:
20/2� = 20/16 = 1.

Total number of 2's included in the prime-factorizaion of 20! = 10+5+2+1 = 18.
Since there are 18 2's, the number of times that 4=2² can divide into 20! = 9.

The correct answer is C.

Take out all the even numbers between 1-20
do factorization of those numbers u'll get 2^18
which is equal to 4^9