Re: discrete math, how many zeroes in 125!

• To: mathgroup at smc.vnet.net
• Subject: [mg13518] Re: [mg13418] discrete math, how many zeroes in 125!
• From: Wouter Meeussen <eu000949 at pophost.eunet.be>
• Date: Fri, 31 Jul 1998 04:33:31 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```At 03:33 23-07-98 -0400, 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
>
>

Equivalently to
Ken Levasseur <Kenneth_Levasseur at uml.edu>
from UMass Lowell Mathematical Sciences
ref:  [mg13440] Re: [mg13418] discrete math, how many zeroes in 125!

and with due thanks to Richard Schroeppel who showed me this : quote:
" the exact power of p that divides ( n! )  is  " (n-Sum of the digits
of the base p representation of n)/(p-1) " end_quote

note that n_factorial need not be computed, giving a small but
significant (;-) advantage for moderate to large n.

try for instance :

In[1]:=n=2345;
Length[Last[Split[IntegerDigits[n!]]]]//Timing
Min[(n-Plus@@IntegerDigits[n,#])/(#-1)  &/@ {2,5}]//Timing

Out[1]=
{1.92 Second, 583}
Out[2]=
{0.05 Second, 583}
Dr. Wouter L. J. MEEUSSEN
w.meeussen.vdmcc at vandemoortele.be
eu000949 at pophost.eunet.be

```

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