# Difference between revisions of "2006 AMC 12A Problems/Problem 9"

## Problem

Oscar buys $13$ pencils and $3$ erasers for $$(Error compiling LaTeX. ! Missing  inserted.)1.00$. A pencil costs more than an eraser, and both items cost a [[whole number]] of cents. What is the total cost, in cents, of one pencil and one eraser?$ \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18$$ (Error compiling LaTeX. ! Missing $inserted.)\mathrm{(E) \ } 20$== Solution == Let the price of a pencil be$p$and an eraser$e$. Then$13p + 3e = 100$with$p > e > 0$. Since$p$and$e$are [[positive integer]]s, we must have$e \geq 1$and$p \geq 2$.

Considering the [[equation]]$(Error compiling LaTeX. ! Missing$ inserted.)13p + 3e = 100$[[modulo]] 3 (that is, comparing the [[remainder]]s when both sides are divided by 3) we have$p + 0e \equiv 1 \pmod 3$so$p$leaves a remainder of 1 on division by 3. Since$ (Error compiling LaTeX. ! Missing $inserted.)p \geq 2$, possible values for$p$are 4, 7, 10 ....

Since 13 pencils cost less than 100 cents,$(Error compiling LaTeX. ! Missing$ inserted.)13p < 100$.$13 \times 10 = 130$is too high, so$p$must be 4 or 7. If$ (Error compiling LaTeX. ! Missing $inserted.)p = 4$then$13p = 52$and so$3e = 48$giving$e = 16$. This contradicts the pencil being more expensive. The only remaining value for$p$is 7; then the 13 pencils cost$7 \times 13= 91$cents and so the 3 erasers together cost 9 cents and each eraser costs$\frac{9}{3} = 3$cents.

Thus one pencil plus one eraser cost$(Error compiling LaTeX. ! Missing$ inserted.)7 + 3 = 10$cents, which is answer choice$\mathrm{(A) \ }\$.