2008 AMC 10B Problem 8

Once you know how many roses there are, the rest of the money must go to carnations. So count the possible numbers of roses.

1. Carnations cost $\$2$ each, so the money spent on carnations is always even.
2. The total is $\$50$, which is even. So the money spent on roses must also be even.
3. Roses cost $\$3$ each. The cost $3 \times (\text{roses})$ is even only when the number of roses is even.
4. The roses cannot cost more than $\$50$. Since $3 \times 16 = 48$ and $3 \times 18 = 54$, there are at most $16$ roses.
5. So the number of roses is one of $0, 2, 4, 6, 8, 10, 12, 14, 16$. That is $9$ choices.
6. Each one works: the leftover money is even, so it buys a whole number of carnations. For example, $0$ roses and $25$ carnations; $2$ roses ($\$6$) and $22$ carnations ($\$44$); up to $16$ roses ($\$48$) and $1$ carnation.

The answer is $\boxed{\textbf{(C)}\ 9}$.

With two unknown whole numbers and one equation, fix one of them and see what the other must be. Here the even/odd check tells you exactly which numbers of roses leave a whole number of carnations. Then count the list carefully, including both ends ($0$ roses and $16$ roses).

Let $r$ be the number of roses and $c$ the number of carnations, with $r, c \ge 0$ integers and
$$3r + 2c = 50.$$
Since $50$ and $2c$ are even, $3r$ must be even, so $r$ is even. Also $3r \le 50$ gives $r \le 16$. Thus $r \in \{0, 2, 4, \dots, 16\}$, and each such $r$ determines $c = (50 - 3r)/2$ as a nonnegative integer.

That list has $16/2 + 1 = 9$ values, so there are $9$ bouquets.

The answer is $\boxed{\textbf{(C)}\ 9}$.