Difference between revisions of "2006 AMC 12A Problems/Problem 9"
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{{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}} | ||
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[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] |
Revision as of 17:51, 3 July 2013
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 |
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