2022/23 AHSMC Part 1 Problem 16

From $5x + 3y = n$ one obtains $y = \dfrac{n-5x}{3} = \dfrac{n+x}{3} - 2x$, and since $x$ and $y$ are integers we must have $n + x = 3k$, where $k$ is an integer. Thus $x = 3k - n$, $y = k - 2(3k - n) = 2n - 5k$ and hence $$f(n) = \min |x+2y| = \min \bigl|(3k-n) + 2(2n-5k)\bigr| = \min |3n - 7k|,$$ where $n \in \mathbb{Z}_{\ge 0}$ and $k$ is any integer. We remark that $f(n+7) = \min |3n - 7(k-3)| = f(n)$ so $f$ is periodic of principal period of $7$. On the other hand, $$f(0) = 0,\ f(1) = 3,\ f(2) = 1,\ f(3) = 2,\ f(4) = 2,\ f(5) = 1,\ f(6) = 3,$$ values which are obtained using $k = 0$, $k = 0$, $k = 1$, $k = 1$, $k = 2$, $k = 2$ and $k = 3$ respectively. We conclude that $$f(0) + f(1) + \cdots + f(100) = 14 \cdot \bigl(f(0) + f(1) + \cdots + f(6)\bigr) + f(0) + f(1) + f(2) = 14 \cdot 12 + 4 = 172.$$ The answer is (A).