Difference between revisions of "2008 AMC 10B Problems/Problem 8"
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==Problem== | ==Problem== | ||
| − | A class collects <math> | + | A class collects <math>$</math>50<math> to buy flowers for a classmate who is in the hospital. Roses cost </math>$<math>3</math> each, and carnations cost <math>$</math>2<math> each. No other flowers are to be used. How many different bouquets could be purchased for exactly </math>$<math>50</math>? |
<math> | <math> | ||
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{{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}} | ||
Revision as of 11:26, 4 July 2013
Problem
A class collects 
50
$
each, and carnations cost 2
$
?
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.
| 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 | ||
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