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!

thank you in advance!

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- Jun 12th 2009, 08:19 AMz1llchfactorial 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!*

thank you in advance! - Jun 12th 2009, 08:22 AMMoo
Hello,

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

Read this thread : http://www.mathhelpforum.com/math-he...1412-mods.html :) - Jun 12th 2009, 08:51 AMz1llch
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. - Jun 12th 2009, 08:55 AMMoo
- Jun 12th 2009, 08:57 AMz1llch
Haha thanks. then i think something's wrong with the answer then (Giggle)