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)
- discrete math, how many zeroes in 125!