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### 1005! / 10^n: A Number Puzzle

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Date: 9/11/96 at 17:38:17
From: Anonymous
Subject: 1005! / 10^n: A Number Puzzle

Find the largest positive integer, n, such that 1005! is divisible by
10 to the power of n.

I would like to know:
a) how one would go about solving such a problem and
```

```
Date: 9/12/96 at 14:39:6
From: Doctor Ceeks
Subject: Re: 1005! / 10^n: A Number Puzzle

Hi,

a) The problem is about prime factorization.

Given any number, it can be factored as 2^X 5^Y M, where M has no
factors of 2 or 5.  Since 10 = 2*5, the number of tens which divide
the number is the minimum of X and Y...each time you divide by
10, you eliminate a factor of 2 and a factor of 5, and you can
keep dividing until you run out of 2's or 5's (you have to have
both).

So, we have to figure out how many factors of 2 and how many factors
of 5 there are in 1005!=1005*1004*1003*...*3*2*1

1005/5 = 201 of the numbers in the above product are divisible by 5.

[1005/25]=40 are divisible by 25.  ([x]=greatest integer < or = to x)

[1005/125]=8 are divisible by 125=5^3.

[1005/625]=1 is divisible by 625=5^4.

[1005/3025]=0 are divisible by 3025=5^5.

We have to make sure we count the number of factors of five properly.
201 contribute at least 1 five, and of those, 40 give at least 2,
8 at least 3, and 1 gives 4.  So the total number of factors of five
in 1005! is 201+40+8+1=250.

You do the same computation for factors of 2, and get some number.
It will be bigger than 250.

So the answer is: n = 250.

Another way of saying it is that 1005! is a number ending with 250
zeroes.

Different people learn different things at different times.  So if
grade level is supposed to measure what most kids are supposed to
be able to do by a certain age, then I don't have the data for this
particular problem.

-Doctor Ceeks,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents
Middle School Exponents
Middle School Factoring Numbers

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