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?

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- Dec 15th 2008, 10:08 AMperplexeda 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? - Dec 15th 2008, 11:54 AMSoroban
Hello, perplexed!

Quote:

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

- Dec 15th 2008, 02:05 PMperplexed
Thank you very much :-)

- Dec 15th 2008, 02:30 PMGreengoblin
- Dec 15th 2008, 02:58 PMPlato

This is a well known problem,*The trailing zeros problem*.

There is an easy answer to your question.

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.

Therefore, the number of trailing zeros depends upon the number of factors of five in the factorial. - Dec 15th 2008, 03:09 PMGreengoblin
Ok, thanks!