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!

Printable View

- 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 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)