If $\sum_{n=1}^{k}\left[\frac{1}{3}+\frac{n}{90}\right]=21,$ where $[x]$ denotes the integral part of $x,$ then $k=$
(a) $84$ (b) $80$ (c) $85$ (d) none of these
$21=\sum_{n=1}^{k}\left[\frac{1}{3}+\frac{n}{90}\right],$
$=\left[\frac{1}{3}+\frac{1}{90}\right]+\left[\frac{1}{3}+\frac{2}{90}\right]+\cdots+\left[\frac{1}{3}+\frac{59}{90}\right]+\left[\frac{1}{3}+\frac{60}{90}\right]+\left[\frac{1}{3}+\frac{61}{90}\right]+\cdots+\left[\frac{1}{3}+\frac{k}{90}\right]$
How to proceed next?Any Hint?