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