       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][];

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

```

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