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

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== Problem ==
 
== Problem ==
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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?
  
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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?
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<math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20</math>
  
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<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18</math>
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== 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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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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Since <math>p \geq 2</math>, possible values for <math>p</math> are 4, 7, 10 ....
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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.
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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.
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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>.
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== Solution 2 ==
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Since we know that the values of pencils and erasers are both whole numbers, and that there are <math>3</math> erasers and <math>13</math> pencils, we can choose numbers to be the value for the pencil, and if the remainder to <math>1.00</math> is divisible by 3, we know our answer is correct. There are 7 numbers to check, although we can eliminate <math>1, 2, 3, 4,</math> and <math>5</math> mentally, since those numbers would probably have the cost of erasers higher than the cost of pencils (6 would too, but it is faster to just test 6). Trying 6, <math>13 \cdot 6 = 72</math>. 28 is not divisible by 3, so we know that this number is not correct. Moving on to 7, <math>13 \cdot 7 = 91</math>. We know that 9 is a multiple of 3, so this value works!
  
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<math>\mathrm{(E) \ } 20</math>
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Since the total cost of erasers is 9 cents, we know the individual eraser is <math>\frac{9}{3} = 3</math>. <math>3 + 7 = 10</math> so our answer is answer choice <math>\mathrm{(A) \ }</math>
  
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== Solution ==
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~Shadow-18
  
 
== See also ==
 
== See also ==
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* [[2006 AMC 12A Problems]]
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{{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}}
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*[[2006 AMC 12A Problems/Problem 8|Previous Problem]]
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{{MAA Notice}}
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*[[2006 AMC 12A Problems/Problem 10|Next Problem]]
 
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Latest revision as of 17:14, 26 September 2020

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?

$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20$

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 $\mathrm{(A) \ }$.

Solution 2

Since we know that the values of pencils and erasers are both whole numbers, and that there are $3$ erasers and $13$ pencils, we can choose numbers to be the value for the pencil, and if the remainder to $1.00$ is divisible by 3, we know our answer is correct. There are 7 numbers to check, although we can eliminate $1, 2, 3, 4,$ and $5$ mentally, since those numbers would probably have the cost of erasers higher than the cost of pencils (6 would too, but it is faster to just test 6). Trying 6, $13 \cdot 6 = 72$. 28 is not divisible by 3, so we know that this number is not correct. Moving on to 7, $13 \cdot 7 = 91$. We know that 9 is a multiple of 3, so this value works!

Since the total cost of erasers is 9 cents, we know the individual eraser is $\frac{9}{3} = 3$. $3 + 7 = 10$ so our answer is answer choice $\mathrm{(A) \ }$

~Shadow-18

See also

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

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