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. | ||
− | </math> | + | 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>. | |
− | Since 13 pencils | + | == 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}} | {{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] |
Latest revision as of 18:14, 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 |
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