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Math Help - a factorial problem

  1. #1
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    a factorial problem

    10^n is a divisor of the number 2008!. What's the greatest n that satisfies this condition?

    According to my calculations it's 502, but I doubt this is the right result. Could you help me out with this one?
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  2. #2
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    Hello, perplexed!

    10^n is a divisor of the number 2008!.
    What's the greatest n 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 2008! ?

    The answer is: There are many zeros as there are factors-of-5 in 2008!


    In 2008! \:=\:1\cdot2\cdot3\:\hdots\:2008, every fifth number is a multiple of 5.
    . . There are: . \left[\frac{2008}{5}\right] \:=\:401 multiples of 5.

    But every 25th number is a multiple of 5^2,
    . . each of which contributes an addtional factor of 5.
    There are: . \left[\frac{2008}{25}\right] \:=\:80 multiples of 25.

    And every 125th number is a multiple of 5^3,
    . . each of which contributes another factor of 5.
    There are: . \left[\frac{2008}{125}\right] \:=\:16 multiples of 125.

    And every 625th number is a multiple of 5^4,
    . . each of which contributes yet another factor of 5.
    There are: . \left[\frac{2008}{625}\right] \:=\:3 multiples of 625.


    Hence, 2008! contains: . 401 + 80 + 16 + 3 \:=\:500 factors of 5.
    . . And 2008! ends in 500 zeros.

    Therefore: . 10^{500}\text{ divides }2008! \quad\hdots\quad\text{The greatest }n\text{ is } 500.

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  3. #3
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    Thank you very much :-)
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  4. #4
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    Quote Originally Posted by Soroban View Post

    The answer is: There are many zeros as there are factors-of-5 in 2008!
    I was looking at this for ages, and was thinking along the same line - how many zero's are at the end - but why is it equal to the number of factors of five in 2008! ?
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  5. #5
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    Quote Originally Posted by Greengoblin View Post
    how many zero's are at the end - but why is it equal to the number of factors of five in 2008! ?

    This is a well known problem, The trailing zeros problem.
    There is an easy answer to your question.
    Any number ending is exactly n zeros is a multiple of 10^n.
    Now 10^n = (2^n)(5^n). In any factorial like 2008! there are many factors of two but less factors of five.
    Therefore, the number of trailing zeros depends upon the number of factors of five in the factorial.
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  6. #6
    Member Greengoblin's Avatar
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    Ok, thanks!
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