HMMT November

First, we can use $3 x-2 y=n$ to get $x=\frac{n+2 y}{3}$. Thus $|x+y|=\left|\frac{n+5 y}{3}\right|$. Given a certain $n$, the only restriction on $y$ is that $3|n+2 y \Longleftrightarrow 3| n+5 y$. Hence the set of possible $x+y$ equals the set of integers of the form $\frac{n+5 y}{3}$, which in turn equals the set of integers congruent to $3^{-1} n \equiv 2 n(\bmod 5)$. (Prove this!) Thus $f(n)=|x+y|$ is minimized when $x+y$ equals the least absolute remainder $(2 n)_{5}$ when $2 n$ is divided by 5, i.e. the number between -2 and 2 (inclusive) congruent to $2 n$ modulo 5. We immediately find $f(n)=f(n+5 m)$ for all integers $m$, and the following initial values of $f: f(0)=\left|(0)_{5}\right|=0$; $f(1)=\left|(2)_{5}\right|=2 ; f(2)=\left|(4)_{5}\right|=1 ; f(3)=\left|(6)_{5}\right|=1$; and $f(4)=\left|(8)_{5}\right|=2$. Since $2013=403 \cdot 5-2$, it follows that $f(0)+f(1)+\cdots+f(2013)=403[f(0)+f(1)+\cdots+f(4)]-f(2014)=403 \cdot 6-2=2416$