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

(Solution)
(Solution)
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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>, and as <math>r</math> must be an integer, this solves to <math>r\leq 8</math>. Hence 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>, and as <math>r</math> must be an integer, this solves to <math>r\leq 8</math>. Hence 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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Let <math>r</math> represent the number of roses, and let <math>c</math> represent the number of carnations. Then, we get the linear Diophantine equation,
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<math>3r+2c=50</math>.
 +
Using the Euclidean algorithm, we get the initial solutions to be <math>r_0=50</math> and <math>c_0=-50</math>, meaning the complete solution will be,
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<math>r=50+\frac{2}{\gcd(2,3)}</math> <math>k=50+2k</math>, <math>c=-50-\frac{3}{\gcd(2,3)}k=-50-3k</math>
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The solution range for which both <math>r</math> and <math>c</math> are positive is <math>17</math> <math>\leq k</math> <math>\leq</math> <math>25</math>. There are <math>\boxed{9 \text{ (C)}}</math> possible values for <math>k</math>.
 
{{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}}
 
{{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:26, 27 December 2019

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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