Leftovers

Stage: 4 Short Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Note that

${n^2 \over n+4}={n^2-16 \over n+4}+{16 \over n+4}={(n+4)(n-4) \over n+4}+{16 \over n+4}=n-4+{16 \over n+4}$.

So when $n> 12$, the remainder when $n^2$ is divided by $n+4$ is always $16$.

For $1 \leq n \leq 12$, the remainder when $n^2$ is divided by $n+4$ is shown in the table below.

$n\quad\quad\quad$ 1 2 2 4 5 6 7 8 9 10 11 12
$n+4$ 5 6 7 8 9 10 11 12 13 14 15 16
remainder 1 4 2 0 7 6 5 4 3 2 1 0


So there are $9$ different remainders, namely $0, 1, 2, 3, 4, 5, 6, 7, 16$.

This problem is taken from the UKMT Mathematical Challenges.
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