Product of Numbers 1-100

Date: 02/23/2002 at 21:24:04
From: Katie 
Subject: Product of numbers 1-100

I was wondering how to find out how many zeros will be at the end of 
the product of all the numbers from 1 to 100 without multiplying them 
all out.  I know there must be some logical way to figure this out, 
but I'm stuck. Any and all help would be appreciated.  Thank you in 
advance.  :)

Katie

Date: 02/24/2002 at 01:56:33
From: Doctor Twe
Subject: Re: Product of numbers 1-100

Hi Katie - thanks for writing to Dr. Math.

Well, each zero at the end of the number means that the number has 10 
as a factor. 10, 20, 30, ... have 10 as a factor exactly once; 100, 
200, 300, ... have 10 as a factor exactly twice; and so on.

Ten's prime factors are 2 and 5. So if we can count how many times 
2 and 5 are factors of each of the numbers multiplied to get your 
product, the lesser of these values (since we need one of each to make 
a 10) will be the number of zeroes at the end of the number.

Counting the number of 2's and 5's isn't that easy, however. Every 
even number has at least one factor of 2, but many have more than one. 
Forty, for example, has three factors of 2 (40 = 2*2*2*5). Similarly, 
every multiple of 5 will have at least one factor of 5, but some will 
have two. Since 5*5*5 = 125 > 100, we don't have to worry about any 
having more than two factors of 5.

A final thought: it seems pretty clear to me that there will be many 
more 2's than 5's, and we're only interested in the lesser total, so 
I'd start by counting the number of 5 factors in the product. You can 
probably pretty quickly demonstrate that there are at least that many 
2 factors as well.

I hope this helps. If you have any more questions, write back.

- Doctor TWE, The Math Forum
  http://mathforum.com/dr.math/   

Date: 02/25/2002 at 23:42:37
From: Katie 
Subject: Product of numbers 1-100

First of all, thank you for the help and the tips. I appreciate your 
quick response. However, I'm not sure I understand what you meant.  
100 = 2*2*5*5, right? You said something about the lesser of these 
values would be the number or zeros at the end of the product of the 
numbers from 1-100. So then, since there are two 2's and two 5's, 
would there be two zeros at the end of the number? 

Maybe I'm thinking too hard about this one, but I'm just confusing 
myself more as I try to rationalize... please help if you get a 
chance.

Katie

Date: 02/26/2002 at 00:51:00
From: Doctor Jubal
Subject: Re: Product of numbers 1-100

Hi Katie,

Thanks for writing Dr. Math.

Here are some hints.  

The number of zeroes at the end of a number is the number of times 
that number is divisible by 10.  Dividing by 10 is the same as 
dividing by 2 and 5, so the number of times you can divide by 10 is 
the lesser of the number of 2's and 5's in a number's prime 
factorization.

Now, since in any long list of consective numbers, there will be more 
even numbers than numbers divisible by 5, there will be more 2's in 
their product's prime factorization than 5's, so the "limiting factor" 
is going to be 5.

So what you're really asking is, how many 5's will appear in this 
number's prime factorization? 

Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

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

Date: 02/26/2002 at 00:51:20
From: Doctor Twe
Subject: Re: Product of numbers 1-100

Hi Katie. Thanks for writing back.

You have the right idea about the lesser of the two numbers. With two 
2's and two 5's, you'd have the minimum of (2,2) = 2 zeroes at the 
end. If you had 2*2*2*2*2*2*5*5*5, you'd have the minimum of (6,3) = 3 
zeroes. Just to verify this, 2*2*2*2*2*2*5*5*5 = 8000, which indeed 
has 3 zeroes at the end.

But the number you want to count is:

     1*2*3*4*5*6*7*8* ... *100

So you'd have to break it into its prime factors:

     (1)*(2)*(3)*(2*2)*(5)*(2*3)*(7)*(2*2*2)* ... *(2*2*5*5)

and count all the 2's and 5's in _that_.

Fortunately, we can show that there are fewer 5's than 2's (I've 
listed three 5's and nine 2's in the part I did above, but of course I 
didn't list the factors 9 to 99), so we only have to count the 5's. 

I hope this helps. If you have any more questions or comments, write 
back again.

- Doctor TWE, The Math Forum
  http://mathforum.com/dr.math/