BTGmoderatorDC wrote:How many trailing zeroes would be found in 63! upon expansion?

A. 6
B. 12
C. 14
D. 53
E. 57

OA C

Source: e-GMAT

The number of trailing zeros in 63! is the number of 2-and-5 pairs it contains since each such pair produces a trailing zero (notice that 2 x 5 = 10). However, since there are more factors of 2 than factors of 5, the number of trailing zeros really depends on the number of factors of 5 in 63!. To find the number of factors of 5 in n!, we can use the following trick: Divide n by 5, then divide the nonzero quotient (ignore any nonzero remainder) by 5, and continue this process until the quotient become 0. Lastly, add these nonzero quotients up, and the sum will be the number of factors of 5 in n!. Let's use this trick for 63!:

63/5 = 12 R 3 (ignore remainder 3)

12/5 = 2 R 2 (ignore remainder 2)

Since 2/5 = 0 R 2, we can stop. Therefore, the number of factors of 5 in 63! is 12 + 2 = 14, and hence there are 14 trailing zeros in 63!.

Answer: C