2016 AMC 10A Problem 14

A way to make $2016$ is fixed once you know how many threes are used; the twos then fill the rest. So count the possible numbers of threes.

1. Let $b$ be the number of threes. The threes add up to $3b$, and the twos must make up the rest, $2016 - 3b$.
2. The twos always add up to an even number, so $2016 - 3b$ must be even. Since $2016$ is even, $3b$ must be even, and that happens exactly when $b$ is even.
3. Look at the first few cases:

Each even $b$ works, as long as the threes do not go past $2016$.
4. The largest $b$ is when threes make the whole $2016$: $b = 2016 \div 3 = 672$, which is even, with $0$ twos.
5. So $b$ can be $0, 2, 4, \ldots, 672$. Dividing each by $2$ gives $0, 1, 2, \ldots, 336$, which is $337$ numbers (remember to count the $0$).

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

When one amount decides everything else, count the choices for that amount. An even/odd check told us which values of $b$ are allowed, and the allowed values go up in steps of $2$. The example in the problem, with $404$ threes, hints that $b$ is even. When counting a list like $0, 2, \ldots, 672$, watch for the off-by-one: include both ends.

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 $\boxed{\textbf{(C)}\ 337}$.