Difference between revisions of "2006 AMC 12A Problems/Problem 9"
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== Problem == | == Problem == | ||
| − | Oscar buys <math>13</math> pencils and <math>3</math> erasers for <math>$1.00</math>. 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? | + | Oscar buys <math>13</math> pencils and <math>3</math> erasers for <math>$1.00</math>. 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? |
<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18</math> | <math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18</math> | ||
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== Solution == | == Solution == | ||
| − | Let the price of a pencil be p and an eraser e.Then 13p + 3e = 100 | + | 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 $p \geq 2$. |
| − | with p > e > 0. | ||
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| − | + | Considering the [[equation]] $13p + 3e = 100$ [[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. | |
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| − | + | Since $p \geq 2$, possible values for $p$ are 4, 7, 10 .... | |
| − | + | Since 13 pencils cost less than 100 cents, $13p < 100$. $13 \times 10 = 130$ which 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. | |
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| + | Thus one pencil plus one eraser cost $7 + 3 = 10$ cents, which is answer choice $\mathrm{(A) \ }$. | ||
== See also == | == See also == | ||
Revision as of 22:20, 30 January 2007
Problem
Oscar buys 
pencils and 
erasers for 
. 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?
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, $13p < 100$. $13 \times 10 = 130$ which 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 $\mathrm{(A) \ }$.
