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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- December 15th 2008, 11:08 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? - December 15th 2008, 12:54 PMSoroban
Hello, perplexed!

Quote:

is a divisor of the number .

What's the greatest 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 ?

The answer is: There are many zeros as there are factors-of-5 in

In , every fifth number is a multiple of 5.

. . There are: . multiples of 5.

But every 25th number is a multiple of ,

. . each of which contributes an addtional factor of 5.

There are: . multiples of 25.

And every 125th number is a multiple of ,

. . each of which contributes another factor of 5.

There are: . multiples of 125.

And every 625th number is a multiple of ,

. . each of which contributes yet another factor of 5.

There are: . multiples of 625.

Hence, contains: . factors of 5.

. . And ends in 500 zeros.

Therefore: .

- December 15th 2008, 03:05 PMperplexed
Thank you very much :-)

- December 15th 2008, 03:30 PMGreengoblin
- December 15th 2008, 03:58 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 .

Now . In any factorial like 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. - December 15th 2008, 04:09 PMGreengoblin
Ok, thanks!