Difference between revisions of "2017 AMC 12A Problems/Problem 1"
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By the greedy algorithm, we can take two 5-popsicle boxes and one 3-popsicle box with <math>\$8</math>. To prove that this is optimal, consider an upper bound as follows: at the rate of <math>\$3</math> per 5 popsicles, we can get <math>\frac{40}{3}</math> popsicles, which is less than 14. <math>\boxed{\textbf{D}}</math>. | By the greedy algorithm, we can take two 5-popsicle boxes and one 3-popsicle box with <math>\$8</math>. To prove that this is optimal, consider an upper bound as follows: at the rate of <math>\$3</math> per 5 popsicles, we can get <math>\frac{40}{3}</math> popsicles, which is less than 14. <math>\boxed{\textbf{D}}</math>. | ||
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+ | ==See Also== | ||
+ | {{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}} | ||
+ | {{AMC12 box|year=2016|ab=A|before=First Problem|num-a=2}} | ||
+ | {{MAA Notice}} |
Revision as of 14:03, 8 February 2017
Problem
Pablo buys popsicles for his friends. The store sells single popsicles for each, 3-popsicle boxes for , and 5-popsicle boxes for . What is the greatest number of popsicles that Pablo can buy with ?
Solution
By the greedy algorithm, we can take two 5-popsicle boxes and one 3-popsicle box with . To prove that this is optimal, consider an upper bound as follows: at the rate of per 5 popsicles, we can get popsicles, which is less than 14. .
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by First Problem |
Followed by Problem 2 |
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 12 Problems and Solutions |
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