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>\</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>
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<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18</math>
  
</math>\mathrm{(E) \ }  20<math>
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<math>\mathrm{(E) \ }  20</math>
  
 
== Solution ==
 
== Solution ==
Let the price of a pencil be </math>p<math> and an eraser </math>e<math>.  Then </math>13p + 3e = 100<math> with </math>p > e > 0<math>.  Since </math>p<math> and </math>e<math> are [[positive integer]]s, we must have </math>e \geq 1<math> and </math>p \geq 2<math>.
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Let the price of a pencil be <math>p</math> and an eraser <math>e</math>.  Then <math>13p + 3e = 100</math> with <math>p > e > 0</math>.  Since <math>p</math> and <math>e</math> are [[positive integer]]s, we must have <math>e \geq 1</math> and <math>p \geq 2</math>.
  
Considering the [[equation]] </math>13p + 3e = 100<math> [[modulo]] 3 (that is, comparing the [[remainder]]s when both sides are divided by 3) we have </math>p + 0e \equiv 1 \pmod 3<math> so </math>p<math> leaves a remainder of 1 on division by 3.
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Considering the [[equation]] <math>13p + 3e = 100</math> [[modulo]] 3 (that is, comparing the [[remainder]]s when both sides are divided by 3) we have <math>p + 0e \equiv 1 \pmod 3</math> so <math>p</math> leaves a remainder of 1 on division by 3.
  
Since </math>p \geq 2<math>, possible values for </math>p<math> are 4, 7, 10 ....
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Since <math>p \geq 2</math>, possible values for <math>p</math> are 4, 7, 10 ....
  
Since 13 pencils cost less than 100 cents, </math>13p < 100<math>.  </math>13 \times 10 = 130<math> is too high, so </math>p<math> must be 4 or 7.
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Since 13 pencils cost less than 100 cents, <math>13p < 100</math>.  <math>13 \times 10 = 130</math> is too high, so <math>p</math> must be 4 or 7.
  
If </math>p = 4<math> then </math>13p = 52<math> and so </math>3e = 48<math> giving </math>e = 16<math>.  This contradicts the pencil being more expensive.  The only remaining value for </math>p<math> is 7; then the 13 pencils cost </math>7 \times 13= 91<math> cents and so the 3 erasers together cost 9 cents and each eraser costs </math>\frac{9}{3} = 3<math> cents.
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If <math>p = 4</math> then <math>13p = 52</math> and so <math>3e = 48</math> giving <math>e = 16</math>.  This contradicts the pencil being more expensive.  The only remaining value for <math>p</math> is 7; then the 13 pencils cost <math>7 \times 13= 91</math> cents and so the 3 erasers together cost 9 cents and each eraser costs <math>\frac{9}{3} = 3</math> cents.
  
Thus one pencil plus one eraser cost </math>7 + 3 = 10<math> cents, which is answer choice </math>\mathrm{(A) \ }$.
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Thus one pencil plus one eraser cost <math>7 + 3 = 10</math> cents, which is answer choice <math>\mathrm{(A) \ }</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:24, 5 January 2008

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 $\mathrm{(A) \ }$.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AMC 12 Problems and Solutions
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