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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| + | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
| + | https://youtu.be/VYo0SaDaMVs | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
Revision as of 14:43, 9 June 2023
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. 
.
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| 2017 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
