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? | |
<math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20</math> | <math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20</math> | ||
Revision as of 23:45, 26 June 2020
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
