Difference between revisions of "2008 AMC 10B Problems/Problem 8"
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==Solution 2== | ==Solution 2== | ||
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| + | Let x and y represent the number of roses bought. The equation should be <math>3x+2y = 50</math>. | ||
==Solution 3== | ==Solution 3== | ||
Revision as of 14:29, 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 x and y represent the number of roses bought. The equation should be 
.
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) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
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| All AMC 10 Problems and Solutions | ||
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