2006 AMC 12A Problems/Problem 9
Contents
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 .
Solution 2
Since we know that the values of pencils and erasers are both whole numbers, and that there are erasers and pencils, we can choose numbers to be the value for the pencil, and if the remainder to is divisible by 3, we know our answer is correct. There are 7 numbers to check, although we can eliminate and 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, . 28 is not divisible by 3, so we know that this number is not correct. Moving on to 7, . 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 . so our answer is answer choice
~Shadow-18
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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All AMC 12 Problems and Solutions |
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