Question

Let $\left[x\right]$ represents the greatest integer function less than or equal to $x$ and $S$ be the sum of all integers $n,$ such that $1\le n\le 1998$ and that $60$ divides $n^3+30n^2+100n.$ Determine the value of $\left[\frac{S}{1000}\right].$
(correct answer + 5, wrong answer 0)

Answer 1

Note that $60=3\times 4\times 5$

$I.$ $4|100n$
So, $4$ should divide $n^3+30n^2$
i.e., $4$ should divide $n^2(n+30)$
$\Rightarrow n$ is even.
$\Rightarrow 2|n$

$II.$ $5|30n^2+100n$
So, $5$ should divide $n^3$
$\Rightarrow 5|n$

$III.$ $3|30n^2$
So, $3$ should divide $n^3+100n$
i.e., $n^3+100n\equiv 0\left(\text{mod}3\right)$
$\Rightarrow n^3+n\equiv 0\left(\text{mod}3\right)$
$\Rightarrow n(n^2+1)\equiv 0\left(\text{mod}3\right)$

If $n\equiv \pm 1\left(\text{mod}3\right),$
then $n^2\equiv 1\left(\text{mod}3\right)$
$\Rightarrow n^2+1\equiv 2\left(\text{mod}3\right)$
So, neither of $n$ and $n^2+1$ are divisible by $3$

If $n\equiv 0\left(\text{mod}3\right),$
then $n(n^2+1)\equiv 0\left(\text{mod}3\right)$
So, $n$ should be a multiple of $3$
i.e., $3|n$

From $I,II$ and $III,$
$n$ must be a multiple of $2\times 3\times 5=30$
So, we have to find $n$ such that $1\le n\le 1998$ and $n$ should be a multiple of $30.$
$\therefore S=30+60+\cdots +1980$
$\Rightarrow S=30(1+2+\cdots +66)$
$=30\times \frac{66\times 67}{2}$
$=66330$
$\left[\frac{S}{1000}\right]=66$