BTGmoderatorDC wrote: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.
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