Smallest Value of N!; Factorial Table

Date: 11/07/2001 at 19:46:26
From: Melissa
Subject: How to find N

The symbol 3! means 3*2*1, which equals 6. Similarly, 5! means 
5*4*3*2*1, which equals 120, and so on. 

Suppose that N! ends in exactly 3 zeros after fully multiplying it 
out.  What is the smallest value that N can have?

I have tried doing this on a calculator but it would have taken 
forever by hand. Is there any pattern or way to make this faster?

Date: 11/07/2001 at 20:20:38
From: Doctor Paul
Subject: Re: How to find N

Think of what generates a zero at the end of a number.

Pick a number - say 3.

What do you have to do to 3 to add a zero to the end of it?

I think you'll agree that the answer is: multiply it by ten.

So certainly 30! will have at least three zeros at the end of it 
because you multiply by 10 three times in the computation of 30!

But are there any other ways to get 10?  Yes = 10 = 2*5

Every even number (every other number) contains a 2 - but the fives 
are much more rare.

So it seems that 5! will end in one zero, 10! will end in 2 zeros and 
15! will be the first number to end in 3 zeros. Similarly, 20! ends in 
4 zeros but 25! ends in *six* zeros!  Why the jump?  Because 25 
contains two fives!

Here's a factorial table from 1! to 30! :

01! = 1
02! = 2
03! = 6
04! = 24
05! = 120
06! = 720
07! = 5040
08! = 40320
09! = 362880
10! = 3628800
11! = 39916800
12! = 479001600
13! = 6227020800
14! = 87178291200
15! = 1307674368000
16! = 20922789888000
17! = 355687428096000
18! = 6402373705728000
19! = 121645100408832000
20! = 2432902008176640000
21! = 51090942171709440000
22! = 1124000727777607680000
23! = 25852016738884976640000
24! = 620448401733239439360000
25! = 15511210043330985984000000
26! = 403291461126605635584000000
27! = 10888869450418352160768000000
28! = 304888344611713860501504000000
29! = 8841761993739701954543616000000
30! = 265252859812191058636308480000000                                           
   

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/