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> | ||
Revision as of 22:21, 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 
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, 
. 
which 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 
.
