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> \ | + | == Solution == |
| + | Let the price of a pencil be <math>p</math> and an eraser <math>e</math>. Then <math>13p + 3e = 100</math> with <math>p > e > 0</math>. Since <math>p</math> and <math>e</math> are [[positive integer]]s, we must have <math>e \geq 1</math> and <math>p \geq 2</math>. | ||
| + | |||
| + | Considering the [[equation]] <math>13p + 3e = 100</math> [[modulo]] 3 (that is, comparing the [[remainder]]s when both sides are divided by 3) we have <math>p + 0e \equiv 1 \pmod 3</math> so <math>p</math> leaves a remainder of 1 on division by 3. | ||
| + | |||
| + | Since <math>p \geq 2</math>, possible values for <math>p</math> are 4, 7, 10 .... | ||
| + | |||
| + | Since 13 pencils cost less than 100 cents, <math>13p < 100</math>. <math>13 \times 10 = 130</math> is too high, so <math>p</math> must be 4 or 7. | ||
| + | |||
| + | If <math>p = 4</math> then <math>13p = 52</math> and so <math>3e = 48</math> giving <math>e = 16</math>. This contradicts the pencil being more expensive. The only remaining value for <math>p</math> is 7; then the 13 pencils cost <math>7 \times 13= 91</math> cents and so the 3 erasers together cost 9 cents and each eraser costs <math>\frac{9}{3} = 3</math> cents. | ||
| + | |||
| + | Thus one pencil plus one eraser cost <math>7 + 3 = 10</math> cents, which is answer choice <math>\mathrm{(A) \ }</math>. | ||
| − | == Solution == | + | == Solution 2 == |
| + | Since we know that the values of pencils and erasers are both whole numbers, and that there are <math>3</math> erasers and <math>13</math> pencils, we can choose numbers to be the value for the pencil, and if the remainder to <math>1.00</math> is divisible by 3, we know our answer is correct. There are 7 numbers to check, although we can eliminate <math>1, 2, 3, 4,</math> and <math>5</math> 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, <math>13 \cdot 6 = 72</math>. 28 is not divisible by 3, so we know that this number is not correct. Moving on to 7, <math>13 \cdot 7 = 91</math>. 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 <math>\frac{9}{3} = 3</math>. <math>3 + 7 = 10</math> so our answer is answer choice <math>\mathrm{(A) \ }</math> | ||
| + | |||
| + | ~Shadow-18 | ||
== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}} | |
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Number Theory Problems]] | ||
Latest revision as of 17:14, 26 September 2020
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 |
| 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. 
