Prove that a 5 digit even number consisting of distinct even digits can't be a perfect square.

Question

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

It could also be $04$ or $84$, I presume. – Sarvesh Ravichandran Iyer Dec 03 '17 at 00:52 — Sarvesh Ravichandran Iyer, Dec 03 '17 at 00:52
@астонвіллаолофмэллбэрг Yes sorry. – Gerard L. Dec 03 '17 at 00:56 — Gerard L., Dec 03 '17 at 00:56

score 16 · Accepted Answer · answered Dec 03 '17 at 00:56

16

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 $3$.

answered Dec 03 '17 at 00:56

user16394

781

1

I noticed you left “mod” outside of the MathJax; you can use a\equiv b \pmod x for $a\equiv b \pmod x$. – gen-ℤ ready to perish Dec 03 '17 at 08:18

Prove that a 5 digit even number consisting of distinct even digits can't be a perfect square.

1 Answers1