1. a factorial problem

10^n is a divisor of the number 2008!. What's the greatest n that satisfies this condition?

According to my calculations it's 502, but I doubt this is the right result. Could you help me out with this one?

2. Hello, perplexed!

$\displaystyle 10^n$ is a divisor of the number $\displaystyle 2008!$.
What's the greatest $\displaystyle n$ that satisfies this condition?

According to my calculations it's 502. . . . . You're close!

The question becomes: How many zeros at the far right of $\displaystyle 2008!$ ?

The answer is: There are many zeros as there are factors-of-5 in $\displaystyle 2008!$

In $\displaystyle 2008! \:=\:1\cdot2\cdot3\:\hdots\:2008$, every fifth number is a multiple of 5.
. . There are: .$\displaystyle \left[\frac{2008}{5}\right] \:=\:401$ multiples of 5.

But every 25th number is a multiple of $\displaystyle 5^2$,
. . each of which contributes an addtional factor of 5.
There are: .$\displaystyle \left[\frac{2008}{25}\right] \:=\:80$ multiples of 25.

And every 125th number is a multiple of $\displaystyle 5^3$,
. . each of which contributes another factor of 5.
There are: .$\displaystyle \left[\frac{2008}{125}\right] \:=\:16$ multiples of 125.

And every 625th number is a multiple of $\displaystyle 5^4$,
. . each of which contributes yet another factor of 5.
There are: .$\displaystyle \left[\frac{2008}{625}\right] \:=\:3$ multiples of 625.

Hence, $\displaystyle 2008!$ contains: .$\displaystyle 401 + 80 + 16 + 3 \:=\:500$ factors of 5.
. . And $\displaystyle 2008!$ ends in 500 zeros.

Therefore: .$\displaystyle 10^{500}\text{ divides }2008! \quad\hdots\quad\text{The greatest }n\text{ is } 500.$

3. Thank you very much :-)

4. Originally Posted by Soroban

The answer is: There are many zeros as there are factors-of-5 in $\displaystyle 2008!$
I was looking at this for ages, and was thinking along the same line - how many zero's are at the end - but why is it equal to the number of factors of five in 2008! ?

5. Originally Posted by Greengoblin
how many zero's are at the end - but why is it equal to the number of factors of five in 2008! ?

This is a well known problem, The trailing zeros problem.
Any number ending is exactly n zeros is a multiple of $\displaystyle 10^n$.
Now $\displaystyle 10^n = (2^n)(5^n)$. In any factorial like $\displaystyle 2008!$ there are many factors of two but less factors of five.