Question

If $\sum _{n=1}^k[\frac{1}{3}+\frac{n}{90}]=21$, where $\left[x\right]$ denotes the integral part of $x$, then $k=$

Answer 1

$\sum _{n=1}^k[\frac{1}{3}+\frac{n}{90}]=21$
$\Rightarrow [\frac{1}{3}+\frac{1}{90}]+[\frac{1}{3}+\frac{2}{90}]+[\frac{1}{3}+\frac{3}{90}]...[\frac{1}{3}+\frac{60}{90}]...[\frac{1}{3}+\frac{k}{90}]=21$
$\Rightarrow (0+0+...$ to $59$ terms $)+(1+1+...$ to $(k-59)$ terms $)=21$
$\therefore 21=k-59\Rightarrow k=80$