2017 AMC 10A Problem 2

Compare the price per popsicle: singles cost $\$1$ each, a $3$-box costs $\tfrac{2}{3}$ dollar per popsicle, and a $5$-box costs $\tfrac{3}{5} = 0.60$ dollar per popsicle. The $5$-boxes are the best value, so start there.

Three $5$-boxes would cost $\$9$, too much. Two $5$-boxes cost $\$6$ and give $10$ popsicles, leaving $\$2$. That $\$2$ buys a $3$-box, for a total of $13$ popsicles for exactly $\$8$.

Can $14$ be reached? Every popsicle costs at least $\$0.60$ under any purchase plan, so $14$ popsicles cost at least $\$8.40 > \$8$. Hence $13$ is the maximum.

The answer is $\boxed{\textbf{(D)}\ 13}$.

With a fixed budget, maximize quantity by buying the cheapest-per-unit option first, then adjust the leftover money. The lower bound "each popsicle costs at least $\$0.60$" turns a guess into a proof: $\lfloor 8 / 0.6 \rfloor = 13$ is an absolute ceiling, and the explicit purchase reaches it.

Enumerate by the number of $5$-boxes: with two ($\$6$ left $\$2$): $10 + 3 = 13$; with one ($\$5$ left): two $3$-boxes plus one single gives $5 + 6 + 1 = 12$; with none: four $3$-boxes give $12$. The best is $13$.

• Buying only 5-boxes (two boxes, then two singles for 12 popsicles) instead of spending the last $2 on a 3-box. - Picking$15$because "three boxes of five" sounds right, without checking that they cost$\$9$.