BTGmoderatorDC wrote:If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
A. 6
B. 12
C. 14
D. 42
E. 56
60! = 60*59*58*....*3*2*1.
Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 60! will yield a 0 at the end of the integer representation of 60!.
The prime-factorization of 60! is composed of FAR MORE 2'S than 5's.
Thus, the number of 0's depends on the NUMBER OF 5's contained within 60!.
To count the number of 5's, simply divide increasing POWERS OF 5 into 60.
Every multiple of 5 within 60! provides at least one 5:
60/5 = 12 --> twelves 5's.
Every multiple of 5² provides a SECOND 5:
60/5² = 2 --> two more 5's.
Thus, the total number of 5's contained within 60! = 12+2 = 14.
The correct answer is
C.
Another example:
If 200! is written out as an integer, with how many consecutive 0’s will that integer end?
Every multiple of 5 within 200! provides at least one 5:
200/5 = 40 --> forty 5's.
Every multiple of 5² provides a SECOND 5:
200/5² = 8 --> eight more 5's.
Every multiple of 5³ provides a THIRD 5:
200/5³ = 1 --> one more 5.
Thus, the total number of 5's contained within 200! = 40+8+1 = 49.
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