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

Revision as of 18:15, 2 February 2007

Problem

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?

$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18$

$\mathrm{(E) \ } 20$

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 integers, we must have $e \geq 1$ and $p \geq 2$ .

Considering the equation $13p + 3e = 100$ modulo 3 (that is, comparing the remainders 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 $p \geq 2$ , possible values for $p$ are 4, 7, 10 ....

Since 13 pencils cost less than 100 cents, $13p < 100$ . $13 \times 10 = 130$ is too high, so $p$ must be 4 or 7.

If $p = 4$ 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 $7 + 3 = 10$ 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
All AMC 12 Problems and Solutions

@@ Line 23: / Line 23: @@
 * [[2006 AMC 12A Problems]]
-{{AMC box|year=2006|n=12A|num-b=8|num-a=10}}
+{{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}}
 [[Category:Introductory Number Theory Problems]]

Page

Toolbox

Search

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

Revision as of 18:15, 2 February 2007

Problem

Solution

See also