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  1. #1
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    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!
    thank you in advance!
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  2. #2
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    Hello,
    Quote Originally Posted by z1llch View Post
    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!
    Your problem is equivalent to finding the greatest n such that 10^n divides 2008!

    Read this thread : http://www.mathhelpforum.com/math-he...1412-mods.html
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    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.
    Last edited by z1llch; June 12th 2009 at 09:10 PM.
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  4. #4
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    Quote Originally Posted by z1llch View Post
    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.
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  5. #5
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    Haha thanks. then i think something's wrong with the answer then
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