# Difference between revisions of "2008 AMC 10B Problems/Problem 8"

## Problem

A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?

$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$

## Solution

The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $r\geq 0$. Then we have $50-3\cdot 2r = 50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2 = 25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality $25-3r \geq 0$, and as $r$ must be an integer, this solves to $r\leq 8$. Hence there are $\boxed{9->C}$ possible values of $r$, and each gives us one solution.

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