Check this: everything-about-factorials-on-the-gmat-85592.html other examples: facorial-ps-105746.html#p827453

1. Trailing zeros:
Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}, where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32, 5^2=25 is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

Back to the original question:
If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which 10^p is a factor of m?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

Given: m=40!. So we should find the # of trailing zeros in 40!, as it'll be the greatest value of p for which 40!/10^p will be an integer.

40! has \frac{40}{5}+\frac{40}{5^2}=8+1=9 trailing zeros, which means that 40! ends with 9 zeros so p=9 is the greatest integer for which 10^p (10^9) is a factor of 40!.

Answer: C.

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The factors of 10 are 1, 2, 5, and 10.

To find what k is, find how many times you can make 10 with the numbers in 1-40.

First, find the 10's as they are going to be most restrictive. Only 10, 20, 30 and 40 have 10 as a factor. 4 total.

Next find 5's. 5, 15, 25 and 35 each have a 5, with 25 having 2. 5 total.

You dont need to find the twos as you know there are more than 5's.

Add the two you get 9.

This is just an amazing way to solve it. so simple its crazy.
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For a detailed discussion, check out: http://www.veritasprep.com/blog/2011/06 ... actorials/

As for your question, forget 40!. Let's look at 10!

What is the maximum value of p such that 10^p is a factor of 10!
To make a 10, you need a 2 and a 5. In 10!, will you have more 2s or more 5s?

1*2*3*4*5*6*7*8*9*10 - Every alternate number has at least one 2 in it. Every fifth number has at least one 5 in it. You certainly have more 2s than 5s. If you have eight 2s and two 5s, how many 10s can you make? You certainly need a 5 to make a 10. Since you have only two 5s, you can make only two 10s. It doesn't matter how many 2s are extra. You cannot make a 10 with only 2s.
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