2006 AMC 12A Problems/Problem 9

Problem

Oscar buys $13$ pencils and $3$ erasers for $$ (Error compiling LaTeX. Unknown error_msg)1.00 $. A pencil costs more than an eraser, and both items cost a [[whole number]] of cents. What is the total cost, in cents, of one pencil and one eraser?$ \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18$$ (Error compiling LaTeX. Unknown error_msg)\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. Unknown error_msg)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. Unknown error_msg)p \geq 2 $, possible values for$ p$are 4, 7, 10 ....

Since 13 pencils cost less than 100 cents,$ (Error compiling LaTeX. Unknown error_msg)13p < 100 $.$ 13 \times 10 = 130 $is too high, so$ p$must be 4 or 7.

If$ (Error compiling LaTeX. Unknown error_msg)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. Unknown error_msg)7 + 3 = 10 $cents, which is answer choice$ \mathrm{(A) \ }$.

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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
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2006 AMC 12A Problems/Problem 9

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