Naturally NMTC - Part 3

Given expression = $\displaystyle 81n +\sum_{k=20} ^{100} k = 81n +4860$

As we want this to be a square, we can cancel out factors that are squares.

$81n +4860=81 (n+60)$ and thus if $81n +4860$ is a square, then surely $n+60$ is a square.

The smallest natural number that makes $n+60$ , a perfect square, is $n=\boxed{4}$ which will make $n+60 =8^2$

Nmtc ques. Of yesterday's paper.