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Thread: Factorials & Perfect Squares

  1. #1
    Junior Member star_tenshi's Avatar
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    Factorials & Perfect Squares

    I've been racking my brain out for the last two days on this problem, but I don't even have the slightest clue on how to approach this problem. Could anybody point me in the right direction please?

    Q: Find the values of $\displaystyle n \geq 1$ for which $\displaystyle 1! + 2! + ... + n!$ is a perfect square. (Hint: Determine the possibilities for the units digit of a perfect square.)

    Using the hint:
    $\displaystyle 0^2 = 0$
    $\displaystyle 1^2 = 1$
    $\displaystyle 2^2 = 4$
    $\displaystyle 3^2 = 9$
    $\displaystyle 4^2 = 16$
    $\displaystyle 5^2 = 25$
    $\displaystyle 6^2 = 36$
    $\displaystyle 7^2 = 49$
    $\displaystyle 8^2 = 64$
    $\displaystyle 9^2 = 81$
    $\displaystyle 10^2 = 100$

    This leads to a pattern: 0 1 4 9 6 5 6 9 4 1 0

    Then I'm just lost....
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  2. #2
    MHF Contributor red_dog's Avatar
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    The unit digit of $\displaystyle 1!$ is 1.

    The unit digit of $\displaystyle 2!$ is 2.

    The unit digit of $\displaystyle 3!$ is 6.

    The unit digit of $\displaystyle 4!$ is 4.

    The unit digit of $\displaystyle n!, \ n\geq 5$ is 0.

    Now, we have:
    $\displaystyle 1!=1=1^2$

    $\displaystyle 1!+2!+3!=1+2+6=9=3^2$

    For $\displaystyle n\geq 4$ the unit digit of $\displaystyle 1!+2!+3!+\ldots+n!$ is 3 and it can't be a perfect square.
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  3. #3
    Junior Member star_tenshi's Avatar
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    Thanks a lot red_dog.

    After your hint, I completed the mathematical proof and finished off the problem.
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