Difference between revisions of "2006 AMC 12A Problems/Problem 9"
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== Solution == | == Solution == | ||
| + | Let the price of a pencil be p and an eraser e.Then 13p + 3e = 100 | ||
| + | with p > e > 0. So e >= 1 and p >= 2. | ||
| + | Considering the equation 13p + 3e = 100 modulo 3 (remainders when divided by 3) we have p + (0e) = 1 or p = 1. | ||
| + | Since p >= 2 possible values for p are {4, 7, 11 ....} | ||
| + | Since 13 pencils cost less than 100 cents 13p < 100. | ||
| + | 13 x 11 = 101 which is too high so p = 4 or 7. | ||
| + | |||
| + | If p = 4 then 13p = 52 and so 3e = 48 giving e = 16. | ||
| + | This contradicts the pencil being more expensive. | ||
| + | |||
| + | The only remaining value for p is 7 and the 13 pencils cost 7 x 13= 91 and so the 3 erasers cost 9 and each eraser cost 9/3 = 3. | ||
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| + | One pencil plus one eraser cost 7 + 3 = 10 cents. | ||
| + | |||
| + | Answer A (10) | ||
== See also == | == See also == | ||
Revision as of 20:04, 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 p and an eraser e.Then 13p + 3e = 100 with p > e > 0. So e >= 1 and p >= 2. Considering the equation 13p + 3e = 100 modulo 3 (remainders when divided by 3) we have p + (0e) = 1 or p = 1. Since p >= 2 possible values for p are {4, 7, 11 ....} Since 13 pencils cost less than 100 cents 13p < 100. 13 x 11 = 101 which is too high so p = 4 or 7.
If p = 4 then 13p = 52 and so 3e = 48 giving e = 16. This contradicts the pencil being more expensive.
The only remaining value for p is 7 and the 13 pencils cost 7 x 13= 91 and so the 3 erasers cost 9 and each eraser cost 9/3 = 3.
One pencil plus one eraser cost 7 + 3 = 10 cents.
Answer A (10)
