Question

你有九枚硬币：包括美分、五分镍币、一角硬币和二十五分硬币，总价值为 $\$1.02$ ，且每种硬币至少有一枚。你必须有多少枚一角硬币？

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $\$1.02,$ with at least one coin of each type. How many dimes must you have?

$1$

$2$

$3$

$4$

$5$

答案：A

知识点：钱币模运算分类讨论

难度评级：1560

解答：

美分硬币的数量必须与 $102$ 模 $5$ 的余数相同，所以可能有 $2$ 枚或 $7$ 枚美分硬币。如果有七枚，只剩两枚硬币给五分镍币、一角硬币和二十五分硬币，不可能每种至少一枚。

所以有 $2$ 枚美分硬币。剩下 $7$ 枚硬币价值 $100$ 分。设五分镍币、一角硬币和二十五分硬币数量分别为 $n,d,q$ ，则 $n+d+q=7$ ，且 $5n+10d+25q=100$ 。

将价值方程除以 $5$ ，再减去硬币数量方程，得到 $d+4q=13$ 。唯一正整数解是 $q=3$ ， $d=1$ ， $n=3$ 。

因此必须有 $1$ 枚一角硬币。

所以正确答案是 A。

The number of pennies must have the same remainder as $102$ modulo $5$ , so there are either $2$ or $7$ pennies. Seven pennies would leave only two coins for nickels, dimes, and quarters, impossible because at least one of each type is needed.

So there are $2$ pennies. The remaining $7$ coins are worth $100$ cents. If $n,d,q$ are the numbers of nickels, dimes, and quarters, then $n+d+q=7$ and $5n+10d+25q=100$ .

Dividing the value equation by $5$ and subtracting the coin-count equation gives $d+4q=13$ . The only positive solution is $q=3$ , $d=1$ , and $n=3$ .

Thus there must be $1$ dime.

Thus, A is the correct answer.

2000 AMC 8 第 20 题

答案：A

难度评级：1560

解答：

其他年份的第 20 题