ardz24 wrote:How many zeros are at the end of 380! ?

(A) 90
(B) 91
(C) 94
(D) None
(E) 95

This problem is about TRAILING 0's: the number of 0's at the end of a large product.

380! = 380*379*378*....*3*2*1.

Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 380! will yield a 0 at the end of the integer representation of 380!.
The prime-factorization of 380! includes FAR MORE 2'S than 5's.
Thus, the number of 0's depends on the NUMBER OF 5's contained within 380!.

To count the number of 5's, simply divide increasing POWERS OF 5 into 380.

Every multiple of 5 within 380! provides at least one 5:
380/5 = 76 --> 76 5's.
Every multiple of 5² within 380! provides a SECOND 5:
380/5² = 15 --> 15 more 5's.
Every multiple of 5³ within 380! provides a THIRD 5:
380/5³ = 3 --> 3 more 5's.
Thus, the total number of 5's contained within 380! =76+15+3 = 94.

Since each of these 94 5's can serve to produce a trailing zero, the total number of trailing zeros = 94.

The correct answer is C.