2003 AMC 12B 第 7 题

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

小提示：

设镍币、角币、二十五美分硬币数分别为

n

、

d

、

q

，则有

n + d + q = 100

和

5n + 10d + 25q = 835

。

With

n

nickels,

d

dimes,

q

quarters:

n + d + q = 100

and

5n + 10d + 25q = 835

大提示：

消去

n

得

d + 4q = 67

，再利用

n,d,q \ge 0

且均为整数来限制

q

Eliminate

n

to get

d + 4q = 67,

then bound

q

so that

n,d,q \ge 0

are integers

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,