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

Revision as of 18:00, 31 January 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 $\mathrm{(A) \ }$ .

@@ Line 22: / Line 22: @@
 == See also ==
 * [[2006 AMC 12A Problems]]
-*[[2006 AMC 12A Problems/Problem 8|Previous Problem]]
-*[[2006 AMC 12A Problems/Problem 10|Next Problem]]
+{{AMC box|year=2006|n=12A|num-b=8|num-a=10}}
 [[Category:Introductory Number Theory Problems]]

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Preceded by Problem 8	AMC 12A 2006	Followed by Problem 10

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

Revision as of 18:00, 31 January 2007

Problem

Solution

See also