How many trailing zeroes would be found in 63! upon

[BTGmoderatorDC]

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

[GMATGuruNY]

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

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

TRAILING 0's = the number of 0's at the end of a large product.

63! = 63*62*61*....*3*2*1.

Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 63! will yield a 0 at the end of the integer representation of 63!.
The prime-factorization of 63! includes FAR MORE 2'S than 5's.
Thus, the number of 0's depends on the NUMBER OF 5's contained within 63!.

To count the number of 5's, simply divide increasing POWERS OF 5 into 63.

Every multiple of 5 within 63! provides at least one 5:
63/5 = 12 --> twelve 5's.
Every multiple of 5² within 63! provides a SECOND 5:
63/5² = 2 --> two more 5's.
Thus, the total number of 5's contained within 63! = 12+2 = 14.

Since each of these 14 5's can serve to produce a trailing zero, the total number of trailing zeros = 14.

The correct answer is C.

[Scott@TargetTestPrep]

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