5
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I have no idea how to start, except that I know the last two digits must be $24, 04, 84, 64$.

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  • $\begingroup$ It could also be $04$ or $84$, I presume. $\endgroup$ Dec 3, 2017 at 0:52
  • $\begingroup$ @астонвіллаолофмэллбэрг Yes sorry. $\endgroup$
    – Gerard L.
    Dec 3, 2017 at 0:56

1 Answer 1

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The remainder by $3$ is $2+4+6+8+0 \equiv 20 \equiv 2$ (using the divisibility rule mod $3$: it's the digit sum); but all squares have remainder $0$ or $1$ mod $3$.

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    $\begingroup$ I noticed you left “mod” outside of the MathJax; you can use a\equiv b \pmod x for $a\equiv b \pmod x$. $\endgroup$ Dec 3, 2017 at 8:18

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