AMC12 2006 A

AMC12 2006 A · Q9

AMC12 2006 A · Q9. It mainly tests Systems of equations, Money / coins.

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?

Oscar 用 $1.00$ 美元买了 $13$ 支铅笔和 $3$ 个橡皮。铅笔比橡皮贵，并且两种物品的价格都是整数美分。一支铅笔和一个橡皮的总价（美分）是多少？

(A) 10 10

(B) 12 12

(D) 18 18

(E) 20 20

Answer

Correct choice: (A)

正确答案：（A）

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 $\mathrm{(A) \ }$.

设一支铅笔的价格为 $p$，一个橡皮的价格为 $e$。则 $13p + 3e = 100$，且 $p > e > 0$。由于 $p$ 和 $e$ 为正整数，必有 $e \geq 1$ 且 $p \geq 2$。将方程 $13p + 3e = 100$ 对 3 取模（即比较两边除以 3 的余数），得 $p + 0e \equiv 1 \pmod 3$，所以 $p$ 除以 3 的余数为 1。由于 $p \geq 2$，$p$ 的可能值为 4, 7, 10 .... 又因为 13 支铅笔的总价小于 100 美分，$13p < 100$。$13 \times 10 = 130$ 太大，因此 $p$ 只能是 4 或 7。若 $p = 4$，则 $13p = 52$，从而 $3e = 48$，得 $e = 16$。这与铅笔更贵矛盾。故 $p$ 只能为 7；此时 13 支铅笔共 $7 \times 13= 91$ 美分，因此 3 个橡皮共 9 美分，每个橡皮 $\frac{9}{3} = 3$ 美分。因此一支铅笔加一个橡皮的总价为 $7 + 3 = 10$ 美分，对应选项 $\mathrm{(A) \ }$。

Topics