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Re: discrete math, how many zeroes in 125!


Tim:

If p is a prime, the number of factors of p in n! is equal to
Sum[Floor[n/(p^k)],{k,1,Infinity}].   The sum is finite since all but a
finite number of terms are zero.  Since there are plenty of 2's in
125!, you need only count the 5's.

Ken Levasseur
Math. Sci.
UMass Lowell

Timothy Anderson wrote:

> how can I solve this problem by counting the factors of 2 and 5 without
> doing each factor individually? thanks for any real quick help! Tim




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