Let $k$ be an integer satisfying
$0\le k \le 9$. Show that $0\le 10k-k^2\le 25$
and that $k(k-1)(k+1)$ is divisible by $3$.
For each $3$-digit number $N$,
where $N\ge100$,
let $S$ be the sum of the hundreds digit,
the square of the tens digit
and the cube of the units digit.
Find the numbers $N$ such that $S=N$.
Hint:
write $N=100a+10b+c$
where $a$, $b$ and $c$ are the digits of $N$.
