2008 AMC 10B Problems/Problem 8

Revision as of 21:26, 27 December 2019 by Yermom (talk | contribs) (Solution)

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 $r$ 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 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

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