Difference between revisions of "2008 AMC 10B Problems/Problem 8"
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==Solution== | ==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 (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== | ||
+ | |||
+ | Let <math>x</math> and <math>y</math> be the number of roses and carnations bought. The equation should be <math>3x+2y = 50</math>. Since <math>50</math> is and odd number, the product of <math>3x</math> must be even and smaller than <math>50</math>. You can try all the even substitutes for <math>x</math> and you will end up with the numbers <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, <math>10</math>, <math>12</math>, <math>14</math>, and <math>16</math>. There are nine numbers in total, so the answers is <math>\boxed{9 \text{ (C)}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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, | ||
+ | <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, | ||
+ | <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> | ||
+ | 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}} |
Latest revision as of 14:38, 18 June 2020
Contents
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?
Solution
The cost of a rose is odd, hence we need an even number of roses. Let there be roses for some . Then we have dollars left. We can always reach the sum exactly by buying carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality , and as must be an integer, this solves to . Hence there are possible values of , and each gives us one solution.
Solution 2
Let and be the number of roses and carnations bought. The equation should be . Since is and odd number, the product of must be even and smaller than . You can try all the even substitutes for and you will end up with the numbers , , , , , , , , and . There are nine numbers in total, so the answers is .
Solution 3
Let represent the number of roses, and let represent the number of carnations. Then, we get the linear Diophantine equation, . Using the Euclidean algorithm, we get the initial solutions to be and , meaning the complete solution will be, , The solution range for which both and are positive is . There are possible values for .
2008 AMC 10B (Problems • Answer Key • Resources) | ||
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Followed by Problem 9 | |
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