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Thread: N!

  1. #1
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    N!

    For a positive integer,$\displaystyle N$, we define N! (read N factorial) to
    be the product of all the integers from 1 to N. Thus:
    $\displaystyle N! = 1 . 2 . 3 .... . N$
    It is clear that N! will end in a zero if N is larger or equal to
    5. For big values of N, N! will end in many, many zeros.
    How many zeros will $\displaystyle 75!$ end in?
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  2. #2
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    This is known as the trailing zeros problem.
    This function will answer the question for any $\displaystyle N<5^{10}$,
    $\displaystyle Z(N) = \sum\limits_{k = 1}^{10} {\left\lfloor {\frac{N}
    {{5^k }}} \right\rfloor } $
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  3. #3
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    Hello, perash!

    For a positive integer,$\displaystyle N$, we define N! (read N factorial)
    to be the product of all the integers from 1 to N.
    Thus: .$\displaystyle N! = 1\cdot2\cdot3 \cdots N$

    It is clear that $\displaystyle N!$ will end in a zero if $\displaystyle N \geq 5$.
    For large values of $\displaystyle N,\;N!$ will end in many, many zeros.

    How many zeros will $\displaystyle 75!$ end in?

    Plato is absolutely correct . . . Here's the basis of that formula . . .


    Every factor of 5 (paired with an even factor) will produce a final zero.

    The question becomes: how many 5's are in the prime factorization of $\displaystyle 75!$ ?


    Every fifth number has a factor of 5.
    . . Hence, there are: .$\displaystyle \frac{75}{5} \:=\:15$ factors of 5.

    But every twenty-fifth number has a factor of $\displaystyle 5^2 = 25.$
    . . Each of them contributes one more 5 to the total.
    And there are: .$\displaystyle \frac{75}{25} \:=\:3$ of them.

    Hence, $\displaystyle 75!$ contains: $\displaystyle 15 + 3 \:=\: 18$ factors of 5


    Therefore, $\displaystyle 75!$ ends in eighteen zeros.

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