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Let $n$ be a positive integer greater than $3$.

(a) Determine if it is possible that the digits of $n^2$ are five different even digits.

(b) Determine all $n$ such that the digits of $n^2$ are different odd digits.


(b) If $n$ is odd then $n = 2k+1$, then we have $n^2 = 4k^2 +4k+1$ and hence $n^2 = 1 \mod 4$. Thus we have $n = 1,3 \mod 4$.

Thus $n$ always ends in either of $1,3,5,7$ or $9$.

But in that case the digit in ten's place of $n^2$ is always even, hence no such $n$ is possible.


I am stuck with part (a). Any hints are welcome.

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    $\begingroup$ For $a$...have you tried a search? Note: I don't really understand what the rules are. Do you mean $n^2$ must be $5$ digits long and the digits are all distinct and even? Like $20468$ were it a perfect square? If so, just search. $\endgroup$
    – lulu
    Feb 26, 2023 at 0:23
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    $\begingroup$ For (a) you could say any such number would be $\equiv 2+4+6+8+0 \equiv 2 \pmod 3$ while squares are only $\equiv 0 \text{ or } 1 \pmod 3$ $\endgroup$
    – Henry
    Feb 26, 2023 at 0:41
  • $\begingroup$ @Henry: That's a great solution – worthy of an answer, I'd say :-) $\endgroup$
    – joriki
    Feb 26, 2023 at 0:42

1 Answer 1

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From comment:

For (a) you could say any such number with the digits $2,4,6,8,0$ in some order

would be $\equiv 2+4+6+8+0 \equiv 2 \pmod 3$

while squares are only $\equiv 0 \text{ or } 1 \pmod 3$

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  • $\begingroup$ If the number has fewer than 5 digits then? $\endgroup$
    – user5210
    Feb 26, 2023 at 0:56
  • $\begingroup$ @user5210: It is possible, for example $8^2=64\equiv 1 \pmod 3$ or $78^2=6084\equiv 0 \pmod 3$ $\endgroup$
    – Henry
    Feb 26, 2023 at 1:49

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