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 < | + | 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{(E) \ } 20</math> |
== Solution == | == Solution == | ||
− | Let the price of a pencil be < | + | 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]] < | + | 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 < | + | Since <math>p \geq 2</math>, possible values for <math>p</math> are 4, 7, 10 .... |
− | Since 13 pencils cost less than 100 cents, < | + | 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 < | + | 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 < | + | 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 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 and an eraser . Then with . Since and are positive integers, we must have and .
Considering the equation modulo 3 (that is, comparing the remainders when both sides are divided by 3) we have so leaves a remainder of 1 on division by 3.
Since , possible values for are 4, 7, 10 ....
Since 13 pencils cost less than 100 cents, . is too high, so must be 4 or 7.
If then and so giving . This contradicts the pencil being more expensive. The only remaining value for is 7; then the 13 pencils cost cents and so the 3 erasers together cost 9 cents and each eraser costs cents.
Thus one pencil plus one eraser cost cents, which is answer choice .
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 |