Question

Claudia 有 $12$ 枚硬币，每枚是 $5$ 美分或 $10$ 美分。用其中一枚或多枚硬币组合，恰好可以得到 $17$ 种不同金额。Claudia 有多少枚 $10$ 美分硬币？

Claudia has $12$ coins, each of which is a $5$ -cent coin or a $10$ -cent coin. There are exactly $17$ different values that can be obtained as combinations of one or more of her coins. How many $10$ -cent coins does Claudia have?

$3$

$4$

$5$

$6$

$7$

答案：C

知识点：钱币区间内整数计数

难度评级：1480

解答：

设 $5$ 美分硬币有 $x$ 枚，则 $10$ 美分硬币有 $12 - x$ 枚。

可以组成从 $5$ 美分到美分之间所有以 $5$ 美分为单位的金额。 $x=0$ ， $10,20,\ldots,120$ $12$ $x>0$ 。 $5$ $5x+10(12-x)=120-5x$

这样的 $5$ 的正倍数共有 $24 - x$ 种。题目给出这个数量为 $17$ ，所以 $x = 7$ 。

因此 $10$ 美分硬币有 $12 - 7 = 5$ 枚。

所以正确答案是 C。

Let the number of $5$ -cent coins be $x$ and the number of $10$ -cent coins be $12 - x.$

If $x=0,$ the only possible values are $10,20,\ldots,120$ cents, giving just $12$ values. Hence $x>0.$ Having at least one $5$ -cent coin then makes every multiple of $5$ from $5$ through the total value $5x+10(12-x)=120-5x$ obtainable.

There are $24 - x$ such multiples of $5,$ which means that $x = 7$ to get $17$ possible different values.

The number of $10$ -cent coins is therefore $12 - 7 = 5.$

Thus, C is the correct answer.

2015 AMC 10A 第 13 题

答案：C

难度评级：1480

解答：

其他年份的第 13 题