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?
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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
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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.  So e >= 1 and p >= 2.
 
Considering the equation 13p + 3e = 100 modulo 3 (remainders when divided by 3) we have p + (0e) is congruent to 1 mod 3 or p is congruent to 1 mod 3.
 
Since p >= 2 possible values for p are  {4, 7, 11 ....}
 
Since 13 pencils cost less than 100 cents 13p < 100.
 
13 x 11 = 101 which is too high so p = 4 or 7.
 
  
If p = 4 then 13p = 52 and so 3e = 48 giving e = 16.
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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.
This contradicts the pencil being more expensive.
 
  
The only remaining value for p is 7 and the 13 pencils cost 7 x 13= 91 and so the 3 erasers cost 9 and each eraser cost 9/3 = 3.
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Since $p \geq 2$, possible values for $p$ are 4, 7, 10 ....
  
One pencil plus one eraser cost 7 + 3 = 10 cents.
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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.
  
Answer A (10)
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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 23:20, 30 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, $13p < 100$. $13 \times 10 = 130$ 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 $\mathrm{(A) \ }$.

See also