1005! / 10^n: A Number Puzzle

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 
b) your opinion of the grade level of the problem

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.

b) I cannot answer which grade level... it's hard to answer that one.
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/