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Math Help - N!

  1. #1
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    N!

    For a positive integer,  N, we define N! (read N factorial) to
    be the product of all the integers from 1 to N. Thus:
    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 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 N<5^{10},
    Z(N) = \sum\limits_{k = 1}^{10} {\left\lfloor {\frac{N}<br />
{{5^k }}} \right\rfloor }
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  3. #3
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    Hello, perash!

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

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

    How many zeros will 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 75! ?


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

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

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


    Therefore, 75! ends in eighteen zeros.

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