Re: discrete math, how many zeroes in 125!

*To*: mathgroup at smc.vnet.net*Subject*: [mg13484] Re: [mg13418] discrete math, how many zeroes in 125!*From*: BobHanlon at aol.com*Date*: Sun, 26 Jul 1998 02:33:43 -0400*Sender*: owner-wri-mathgroup at wolfram.com

zeroes[(n_Integer)?NonNegative] := Plus @@ Cases[Flatten[FactorInteger /@ Table[k, {k, 5, n, 5}], 1], {5, m_} :> m]; zeroes2[(n_Integer)?NonNegative] := Cases[FactorInteger[n!], {5, m_} :> m][[1]]; SetOptions[NumberForm, DigitBlock -> 5, NumberSeparator -> " "]; n = 125; zeroes[n] NumberForm[n!] To compare timing: TableForm[Table[{n=125k, Timing[zeroes[n]], Timing[zeroes2[n]]}, {k, 1, 15, 2}]] Bob Hanlon In a message dated 7/23/98 5:58:55 AM, trafh at AOL.com 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