2023 AMC 10B 第 11 题

Question

Suzanne 去银行取了

\$800

。柜员用

\$20

、

\$50

和

\$100

纸币给她这笔钱，并且每种面额至少有一张。Suzanne 可能收到多少种不同的纸币组合？

Suzanne went to the bank and withdrew

\$800.

The teller gave her this amount using

\$20

bills,

\$50

bills, and

\$100

bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?

45

21

36

28

32

答案：B

知识点：丢番图方程奇偶性基本计数

难度评级：1500

小提示：

将

20a + 50b + 100c = 800

除以

10

；奇偶性会迫使

\$50

纸币张数为偶数

Divide

20a + 50b + 100c = 800

by

10;

parity then forces the number of

\$50

bills to be even

大提示：

写

b = 2t

，然后计数满足

t + c \le 7

的正整数对

(t, c)

。

Writing

b = 2t,

count positive integer pairs

(t, c)

with

t + c \le 7

解答：

设

a, b, c \ge 1

分别为

\$20, \$50, \$100

纸币张数。则

20a + 50b + 100c = 800

，化为

2a + 5b + 10c = 80

。因为

2a

和

10c

都是偶数，

5b

也必须为偶数，所以

b = 2t

。此时

a = 40 - 5t - 5c \ge 1

条件等价于

t + c \le 7

，其中

t, c \ge 1

。这样的正整数对数量为

1 + 2 + \cdots + 6 = 21

。所以正确答案是 B。

Let

a, b, c \ge 1

count the

\$20, \$50, \$100

bills. Then

20a + 50b + 100c = 800,

which divides down to

2a + 5b + 10c = 80.

Both

2a

and

10c

are even, so

5b

is too, forcing

b = 2t.

Now

a = 40 - 5t - 5c \ge 1

means

t + c \le 7.

With

t, c \ge 1,

the pairs number

1 + 2 + \cdots + 6 = 21.

Thus, B is the correct answer.

2023 AMC 10B 第 11 题

答案：B

难度评级：1500

小提示：

大提示：

解答：

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