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Math Help - Last digit of perfect squre

  1. #1
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    Last digit of perfect squre

    Show that the last decimal digit of a perfect square cannot be 2,3,7, or 8.

    Any advice? I don't know where to begin to show this...
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  2. #2
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    Nevermind... Figured it out..
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  3. #3
    Super Member Bacterius's Avatar
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    Hello,
    Although you figured it out, just look to this proof (if yours is different of course) :

    Here is a little table containing the last digit of any number n, and the last digit of its square (thanks to Soroban for the array template) :

    \left|\begin{array}{c|cccccccccc}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ n^2 & 0 & 1 & 4 & 9 & 6 & 5 & 6 & 9 & 4 & 1 \end{array}\right|

    The last digits 2, 3, 7 and 8 are missing. Therefore, a perfect square cannot have 2, 3, 7 or 8 as a last digit.
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