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

**References**:**discrete math, how many zeroes in 125!***From:*trafh@aol.com (Timothy Anderson)