2016 AMC 10A Problem 14

A way to write $2016$ is a pair of nonnegative integers $(a, b)$ with
$$2a + 3b = 2016,$$
where $a$ is the number of twos and $b$ the number of threes.

Since $2a$ and $2016$ are even, $3b$ must be even, so $b$ is even. Write $b = 2k$. Then $2a + 6k = 2016$, i.e. $a = 1008 - 3k$. This is a nonnegative integer exactly when $0 \le k \le 336$.

Each $k$ in that range gives one valid pair, so the number of ways is $336 - 0 + 1 = 337$.

The answer is $\boxed{\textbf{(C)}\ 337}$.

For a two-variable linear equation, once one variable is pinned to a residue class (here, $b$ even), the solutions form an arithmetic progression, and counting them is just counting the terms of that progression within the allowed bounds. Parity is the fastest way to find the residue condition; the example in the problem ($404$ threes) is a hint that $b$ is even.

Count values of $b$ directly: $b$ ranges over even numbers from $0$ to $672$ (since $3b \le 2016$ means $b \le 672$). Even numbers from $0$ to $672$ inclusive: $\frac{672}{2} + 1 = 337$.

• Forgetting the solution with zero threes (or zero twos) and answering 336.
• Counting all $b$ from $0$ to $672$ without the parity restriction, giving $673$, or counting only $a$ values without noticing they must be multiples of $3$ apart.