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MathGroup Archive 1998

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Re: discrete math, how many zeroes in 125!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg13496] Re: discrete math, how many zeroes in 125!
  • From: scottb (Scott Brown)
  • Date: Fri, 31 Jul 1998 04:33:15 -0400
  • Organization: Wolfram Research, Inc.
  • References: <6pebvk$i76@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

>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

  I'm not sure if this solution was offered yet.
  You want the number of trailing zeroes, I think, since you
  referred to 2 and 5.

  Here's a way to count the factors without atually factoring.
  Does this get you what you need?

(* Assuming n Prime,
 * the number of factors of n in M! is the number of multiples of n
 * that are less than M, plus the number of multiples of n^2 that
 * are less than M, plus (etc.), up to the number of multiples less
 * than M of the largest power of n that is less than M.
 *)
multiplesPreceding[ n_Integer?PrimeQ, M_Integer ] :=
    Plus @@ Table[ 
        Floor[ M/(n^k) ], 
        { k, 1, Floor[Log[n,M]] }
        ];



------------------------------------------------------------------------
| Scott Brown        | "I may be speaking from Wolfram Research, Inc.,
| | scottb at wolfram.com | but that doesn't mean I'm speaking for them." 
|
------------------------------------------------------------------------


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