# 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 $\boxed{9 \text{ (C)}}$ possible values of $r$, and each gives us one solution.

## Solution 2

Let $x$ and $y$ be the number of roses and carnations bought. The equation should be $3x+2y = 50$. Since $50$ is and odd number, the product of $3x$ must be even and smaller than $50$. You can try all the even substitutes for $x$ and you will end up with the numbers $0$, $2$, $4$, $6$, $8$, $10$, $12$, $14$, and $16$. There are nine numbers in total, so the answers is $\boxed{9 \text{ (C)}}$.

## Solution 3

Let $r$ represent the number of roses, and let $c$ represent the number of carnations. Then, we get the linear Diophantine equation, $3r+2c=50$. Using the Euclidean algorithm, we get the initial solutions to be $r_0=50$ and $c_0=-50$, meaning the complete solution will be, $r=50+\frac{2}{\gcd(2,3)}$ $k=50+2k$, $c=-50-\frac{3}{\gcd(2,3)}k=-50-3k$ The solution range for which both $r$ and $c$ are positive is $17$ $\leq k$ $\leq$ $25$. There are $\boxed{9 \text{ (C)}}$ possible values for $k$.

 2008 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions