2006 AMC 12A Problem 9

Below is the professionally curated solution for Problem 9 of the 2006 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12A solutions, or check the answer key.

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Concepts:Diophantine Equation divisibility

Difficulty rating: 1430

Oscar buys $13$ pencils and $3$ erasers for $\$1.00.$ A pencil costs more than an eraser, and both items cost a whole number of cents. What is the total cost, in cents, of one pencil and one eraser?

$10$

$12$

$15$

$18$

$20$

Solution:

Let $p$ be a pencil's cost and $s$ the cost of one pencil plus one eraser, in cents. Then $13p + 3e = 3s + 10p = 100,$ so $3s$ is a multiple of $10$ less than $100.$ Hence $s \in \{10, 20, 30\},$ with $p = 7, 4, 1$ respectively.

Since a pencil costs more than an eraser, $p \gt \tfrac{s}{2},$ which holds only for $s = 10$ (pencil $7,$ eraser $3).$ So one pencil and one eraser cost $10$ cents.

Thus, the correct answer is A.

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