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Math Help - quick question

  1. #1
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    quick question

    the problem is as follows, i have NO idea how to approach it or do it. any help is appreciated.

    find all positive integers n such that n! ends with exactly 74 zeroes in decimal notation
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  2. #2
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    Hello, silentbob!

    This takes a bit of brainwork . . .


    Find all positive integers n such that n! ends with exactly 74 zeroes.
    We know that: . n! \;=\;1\cdot2\cdot3\cdot4 \cdots n

    A final zero is created by a factor of 5 (and an even number).
    The question becomes: how many factors of 5 are contained in n! ?

    We know that every 5th number is a multiple of 5.

    Also that every 25th number is a multiple of 5 = 25,
    . . each of which contributes another 5.

    And that every 125th number is a multiple of 5 = 125,
    . . each of which contributes yet another 5.


    By trial-and error, we find that 299! ends in 72 zeros,

    . . but 300! ends in 74 zeros.

    . . \begin{array}{ccc}\left[\dfrac{300}{5}\right] &=& 60 \\ \\[-3mm]<br /> <br />
\left[\dfrac{300}{25}\right] &=& 12 \\ \\[-3mm]<br /> <br />
\left[\dfrac{300}{125}\right] &=& 2 \\ \\[-3mm] \hline \\[-3mm]<br />
\text{Total:} & & 74 \end{array}


    And the same is true for 301!,\:302!,\:303!,\:304!
    . . But 305! ends in 75 zeros.


    Therefore: . n \;=\;300,\:301,\:302,\:304

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