Difference between revisions of "2017 AMC 12A Problems/Problem 1"
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Pablo buys popsicles for his friends. The store sells single popsicles for <math>\$1</math> each, 3-popsicle boxes for <math>\$2</math>, and 5-popsicle boxes for <math>\$3</math>. What is the greatest number of popsicles that Pablo can buy with <math>\$8</math>? | Pablo buys popsicles for his friends. The store sells single popsicles for <math>\$1</math> each, 3-popsicle boxes for <math>\$2</math>, and 5-popsicle boxes for <math>\$3</math>. What is the greatest number of popsicles that Pablo can buy with <math>\$8</math>? | ||
| − | <math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15</math> | + | |
| + | <math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15 </math> | ||
==Solution== | ==Solution== | ||
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>. | ||
Revision as of 15:01, 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. 
.
