100 Factorial in Base 6: How Many Zeros?

Date: 10/24/2001 at 15:27:22
From: Krista 
Subject: Factorial in base 6

How many zeros are at the end of 100! in base 6? I have tried to 
figure it out by writing out the entire list of base 6 numbers up to 
100. I come up with 7 just using the numbers with zeroes in the end, 
but I have a feeling there are more. Could you please help? Thanks!

Date: 10/29/2001 at 11:14:06
From: Doctor Greenie
Subject: Re: Factorial in base 6

Hi, Krista -

In base 10, you get a zero at the end of a product if one of the 
factors contains a prime factor of 5 and another of the factors 
contains a prime factor of 2. The prime factors of 2 and 5 multiplied 
together make 10 and produce a 0 at the end of the product.

If you want to (for example) find the number of 0's at the end of 
1000! in base 10, you need to find the total number of factors of 5 
and the total number of factors of 2 in all the numbers from 1 to 
1000; the smaller of those two numbers will be the number of 0's at 
the end of 1000!

Here is how you would determine the number of 0's at the end of 1000! 
in base 10:

1) 1 out of every 5 numbers 1 to 1000 inclusive (200 of them) 
   contains at least one factor of 5.

2) Of the 200 numbers 1 to 1000 inclusive that contain at least one 
   factor of 5, 1 out of every 5 (40 of them) contains a second 
   factor of 5.

3) Of the 40 numbers 1 to 1000 inclusive that contain a second 
   factor of 5, 1 out of every 5 (8 of them) contains a third factor 
   of 5.

4) Of the 8 numbers 1 to 1000 inclusive that contain a third factor 
   of 5, 1 out of every 5 (1 of them) contains a fourth factor of 5.

The total number of factors of 5 contained in the numbers 1 to 1000 
inclusive is then

   200 + 40 + 8 + 1 = 249

Performing the arithmetic without writing down all the words to 
describe what we are doing, the actual determination of the number of 
factors of 5 contained in the numbers 1 to 1000 inclusive looks like 
this:

   1000/5 = 200
    200/5 =  40
     40/5 =   8
      8/5 =   1
      1/5 =   0
           ----
            249

We know now that there are 249 factors of 5 contained among the 
numbers 1 to 1000 inclusive. There will be 249 0's at the end of 1000! 
if there are at least 249 factors of 2 contained among the numbers 
1 to 1000 inclusive. We know that there are, because each even number 
(500 of them from 1 to 1000 inclusive) contains at least one factor 
of 2. 

However, to practice the method we use to solve this kind of problem, 
let's go ahead and find the total number of factors of 2 contained in 
the numbers 1 to 1000 inclusive. We use the shorthand form of the 
calculation as demonstrated immediately preceding this paragraph:

   1000/2 = 500
    500/2 = 250
    250/2 = 125
    125/2 =  62
     62/2 =  31
     31/2 =  15
     15/2 =   7
      7/2 =   3
      3/2 =   1
      1/2 =   0
          -----
            994

And now on to your problem.

First, I am assuming that the 100 in 100! is a base-10 number, and not 
a base-6 number. (If it was supposed to be a base-6 number equivalent 
to 36 base 10, then simply replace my starting point 100 in the 
calculations below with 36.)

To get a 0 at the end of a product in base 6, you need prime factors 
that, when multiplied together, give you 6 - so you need a factor of 2 
and a factor of 3. To find the number of 0's at the end of 100! in 
base 6, then, you need to determine the total number of factors of 2 
and the total number of factors of 3 among the numbers 1 to 100 
inclusive.

number factors of 2 in 100!....

   100/2 = 50
    50/2 = 25
    25/2 = 12
    12/2 =  6
     6/2 =  3
     3/2 =  1
     1/2 =  0
        -----
           97

number factors of 3 in 100!....

   100/3 = 33
    33/3 = 11
    11/3 =  3
     3/3 =  1
     1/3 =  0
         ----
           48

With 48 factors of 3 and 97 factors of 2 among the numbers 1 to 100 
inclusive, there will be 48 0's at the end of the base-6 
representation of 100!

I hope this helps. Write back if you have any further questions on 
this topic.

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