HMMT November

Solution:

Answer: $2416$

First, we can use $3x - 2y = n$ to get $x = \frac{n + 2y}{3}$. Thus $|x + y| = \left| \frac{n + 5y}{3} \right|$. Given a certain $n$, the only restriction on $y$ is that $3 \mid n + 2y \Longleftrightarrow 3 \mid n + 5y$. Hence the set of possible $x + y$ equals the set of integers of the form $\frac{n + 5y}{3}$, which in turn equals the set of integers congruent to $3^{-1} n \equiv 2n \pmod{5}$. (Prove this!)

Thus $f(n) = |x + y|$ is minimized when $x + y$ equals the least absolute remainder $(2n)_5$ when $2n$ is divided by $5$, i.e., the number between $-2$ and $2$ (inclusive) congruent to $2n$ modulo $5$. We immediately find $f(n) = f(n + 5m)$ for all integers $m$, and the following initial values of $f$:

$f(0) = |(0)_5| = 0$

$f(1) = |(2)_5| = 2$

$f(2) = |(4)_5| = 1$

$f(3) = |(6)_5| = 1$

$f(4) = |(8)_5| = 2$

Since $2013 = 403 \cdot 5 - 2$, it follows that

$$f(0) + f(1) + \cdots + f(2013) = 403 [f(0) + f(1) + \cdots + f(4)] - f(2014) = 403 \cdot 6 - 2 = 2416.$$