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MathGroup Archive 1998

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Re: discrete math, how many zeroes in 125!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg13440] Re: [mg13418] discrete math, how many zeroes in 125!
  • From: Ken Levasseur <Kenneth_Levasseur at uml.edu>
  • Date: Fri, 24 Jul 1998 01:45:30 -0400
  • Organization: UMass Lowell Mathematical Sciences
  • References: <199807230733.DAA05535@smc.vnet.net.>
  • Sender: owner-wri-mathgroup at wolfram.com

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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