Hello, perplexed!

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: .