2023/24 AHSMC Part 1 Problem 7

For every integer $m$, either $m$ or $m+5$ is even, so $m(m+5)$ is always even. To have $m(m+5)$ divisible by $6$ we need only divisibility by $3$. Either $3 \mid m$, and there are $\lfloor 2023/3 \rfloor = 674$ such $m$ between $1$ and $2023$; or $3 \mid (m+5)$, i.e. $m = 1,4,7,\ldots,2023$, of which there are $675$. The total is $674+675 = 1349$. The answer is (D).

Alternatively, writing $m = 6k+t$ with $0 \le t \le 5$, the quantity $m(m+5)/6$ is an integer precisely when $t \in \{0,1,3,4\}$. Thus there are $\frac{4}{6}(2022)+1 = 1349$ such values.