2006 AMC 12A Problems/Problem 9

Problem

Oscar buys $13$ pencils and $3$ erasers for $1.00$ . 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?

$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20$

Solution

Let the price of a pencil be $p$ and an eraser $e$ . Then $13p + 3e = 100$ with $p > e > 0$ . Since $p$ and $e$ are positive integers, we must have $e \geq 1$ and $p \geq 2$ .

Considering the equation $13p + 3e = 100$ modulo 3 (that is, comparing the remainders when both sides are divided by 3) we have $p + 0e \equiv 1 \pmod 3$ so $p$ leaves a remainder of 1 on division by 3.

Since $p \geq 2$ , possible values for $p$ are 4, 7, 10 ....

Since 13 pencils cost less than 100 cents, $13p < 100$ . $13 \times 10 = 130$ is too high, so $p$ must be 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; then the 13 pencils cost $7 \times 13= 91$ cents and so the 3 erasers together cost 9 cents and each eraser costs $\frac{9}{3} = 3$ cents.

Thus one pencil plus one eraser cost $7 + 3 = 10$ cents, which is answer choice $\mathrm{(A) \ }$ .

Solution 2

Since we know that the values of pencils and erasers are both whole numbers, and that there are $3$ erasers and $13$ pencils, we can choose numbers to be the value for the pencil, and if the remainder to $1.00$ is divisible by 3, we know our answer is correct. There are 7 numbers to check, although we can eliminate $1, 2, 3, 4,$ and $5$ 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, $13 \cdot 6 = 72$ . 28 is not divisible by 3, so we know that this number is not correct. Moving on to 7, $13 \cdot 7 = 91$ . 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 $\frac{9}{3} = 3$ . $3 + 7 = 10$ so our answer is answer choice $\mathrm{(A) \ }$

~Shadow-18

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2006 AMC 12A Problems/Problem 9

Contents

Problem

Solution

Solution 2

See also