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

Revision as of 11:30, 7 June 2021

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 1

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 $25-3r \geq 0$ which solves to $r\leq 8 \frac13$ . $r$ must be an integer, so 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 an odd number, the product of $3x$ must be even and smaller than $50$ . You can try nonnegative even integers for $x$ and you will end up with the numbers $0$ , $2$ , $4$ , $6$ , $8$ , $10$ , $12$ , $14$ , and $16$ . There are $9$ numbers in total, so the answer 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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 17: / Line 17: @@
 ==Solution 1==
-The cost of a rose is odd, hence we need an even number of roses. Let there be <math>2r</math> roses for some <math>r\geq 0</math>. Then we have <math>50-3\cdot 2r = 50-6r</math> dollars left. We can always reach the sum exactly <math>50</math> by buying <math>(50-6r)/2 = 25-3r</math> carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality <math>25-3r \geq 0</math> which solves to <math>r\leq \frac{25}{3}</math>. <math>r</math> must be an integer, so there are <math>\boxed{9 \text{ (C)}}</math> possible values of <math>r</math>, and each gives us one solution.
+The cost of a rose is odd, hence we need an even number of roses. Let there be <math>2r</math> roses for some <math>r\geq 0</math>. Then we have <math>50-3\cdot 2r = 50-6r</math> dollars left. We can always reach the sum exactly <math>50</math> by buying <math>(50-6r)/2 = 25-3r</math> carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality <math>25-3r \geq 0</math> which solves to <math>r\leq 8 \frac13</math>. <math>r</math> must be an integer, so there are <math>\boxed{9 \text{ (C)}}</math> possible values of <math>r</math>, and each gives us one solution.
 ==Solution 2==

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