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...
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.