Divisibility #2

Consider $k^3 \leq n < (k+1)^3$ where $k$ is a positive integer

So, $\lfloor \sqrt[3]{n} \rfloor =k$

Therefore every integer within this range can be expressed as $n=k^3+r$ where $0 \leq r \leq 3k^2+3k$

When $\lfloor \sqrt[3]{n} \rfloor | n$ that is $k | k^3+r$ so, $k|r$ .Now $r$ will be $0,k,2k,3k,.....,3k^2+3k$ . So, total number of such $r$ is $3(k+1)+1=3k+4$

As $12^3 < 2017 <13^3$

So, total solutions upto $12^3-1$ is $\displaystyle \sum_{k=1}^{11} 3k+4=242$

Again the last part of the set is $12^3,12^3+12,.....2016$ there are $25$ more solutions.

So, total number of solutions are $242+25=267$