Question

Penniless Pete 的存钱罐里没有便士，但有 $100$ 枚硬币，全部是镍币、角币和二十五美分硬币，总价值为 $\$8.35$ 。存钱罐不一定三种硬币都有。角币数量可能的最大值与最小值之差是多少？

Penniless Pete's piggy bank has no pennies in it, but it has $100$ coins, all nickels, dimes, and quarters, whose total value is $\$8.35.$ It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?

$0$

$13$

$37$

$64$

$83$

答案：D

知识点：方程组丢番图方程极限情形界定

难度评级：1430

解答：

设镍币、角币、二十五美分硬币数分别为 $n$ ， $d$ ， $q$ 则 $n + d + q = 100$ ，且把总价值方程除以 $5$ 得 $n + 2d + 5q = 167$ 。

两式相减得 $d + 4q = 67$ ，所以 $d = 67 - 4q$ 。

当 $q = 0$ 时 $d$ 最大，得到 $d = 67$ （此时 $n = 33$ ）。当 $q = 16$ 时，角币数最小，得到 $d = 3$ （此时 $n = 81$ ）。差为 $67 - 3 = 64$ 。

因此，正确答案是 D。

Let $n,$ $d,$ $q$ be the numbers of nickels, dimes, quarters. Then $n + d + q = 100$ and $n + 2d + 5q = 167$ (dividing the value equation by $5$ ).

Subtracting gives $d + 4q = 67,$ so $d = 67 - 4q.$

The largest $d$ is at $q = 0,$ giving $d = 67$ (with $n = 33$ ). The smallest occurs at $q = 16,$ giving $d = 3$ (with $n = 81$ ). The difference is $67 - 3 = 64.$

Thus, the correct answer is D.

2003 AMC 12B 第 7 题

答案：D

难度评级：1430

解答：

其他年份的第 7 题