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Let $s(n)$ denote the sum of the squares of the digits of $n$. For example, $s(14) = 1 ^ 2 + 4 ^ 2 = 17$ Determine all integers adding n for which $s(n) = n$ holds.

I bound it to $243$ due to $9^2 *4 < 1000$. And thus $n$ must have at most $3$ digits so $3*81 =243$. I have no idea what to do now.

Sergey6552
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    You could just check up to $243$. It will be quick, as it will become obvious that many cannot come close: e.g. since $4+81+81 < 200$ you will not have any three digit numbers starting with $2$. Clearly the only single digit cases are $s(1)=1$ and $s(0)=0$. Are there any two or three digit cases? – Henry Sep 26 '23 at 10:06

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