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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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