# factorial problem.

• June 12th 2009, 08:19 AM
z1llch
factorial problem.
The number 2008! means the product of all the integers 1,2,3,4,...., 2007,2007. With how many zeroes does 2008! end?

any help would be so much appreciated!
• June 12th 2009, 08:22 AM
Moo
Hello,
Quote:

Originally Posted by z1llch
The number 2008! means the product of all the integers 1,2,3,4,...., 2007,2007. With how many zeroes does 2008! end?

any help would be so much appreciated!

Your problem is equivalent to finding the greatest n such that $10^n$ divides 2008!

• June 12th 2009, 08:51 AM
z1llch
i used that method to find the number of 5s in 2008!
i supposed that any number that ends with zero is divisible by 10, which is 2x5. meaning i have to calculate the number of parings of 2 and 5 with all the numbers from 1 to 2008. number of two > number of five. so i find the number of fives in 2008! and i got that there are 500 fives in 2008! concluding there are 500 zeroes in 2008! but the answer's 200 zeroes.
• June 12th 2009, 08:55 AM
Moo
Quote:

Originally Posted by z1llch
i use that method to find the number of 5s in 2008!
i've read from somewhere else that any number that ends with zero is divisible by 10, which is 2x5. meaning i have to calculate the number of parings of 2 and 5 with all the numbers from 1 to 2008. number of two > number of five. so i find the number of fives in 2008! and i got that there are 500 fives in 2008! concluding there are 500 zeroes in 2008! but the answer's 200 zeroes.

I find 500 zeroes as well... and checked it with maxima to get 500 too. (Wink)
• June 12th 2009, 08:57 AM
z1llch
Haha thanks. then i think something's wrong with the answer then (Giggle)