Difference between revisions of "2008 AMC 10B Problems/Problem 8"
Line 19: | Line 19: | ||
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}</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}</math> possible values of <math>r</math>, and each gives us one solution. | ||
− | |||
{{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}} |
Revision as of 15:42, 6 February 2011
Problem
A class collects to buy flowers for a classmate who is in the hospital. Roses cost
each, and carnations cost
each. No other flowers are to be used. How many different bouquets could be purchased for exactly
?
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 |