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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- Feb 15th 2010, 11:00 PMjzelltLast 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... - Feb 15th 2010, 11:16 PMjzellt
Nevermind... Figured it out..

- Feb 16th 2010, 12:47 AMBacterius
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 $\displaystyle n$, and the last digit of its square (thanks to Soroban for the array template) :

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

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