2005 AMC 10B 第 20 题

Question

使用数字

1

、

3

、

5

、

7

和

8

各一次能组成的所有

5

位数的平均数是多少？

What is the average (mean) of all

5

-digit numbers that can be formed by using each of the digits

1,

3,

5,

7,

and

8

exactly once?

48000

49999.5

53332.8

55555

56432.8

由对称性，每个数字在每个数位上出现的次数相同。

By symmetry, each digit appears equally often in each place

每个数位上的平均数字为

\dfrac{1+3+5+7+8}{5}

，再乘以

11111

。

The average digit in each place is

\dfrac{1+3+5+7+8}{5},

then multiply by

11111

由对称性，五个数字在每个数位上出现的次数相同，所以每个数位上的平均数字为

\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\text{。}

因此平均数为

\begin{gathered} \small 4.8(1 + 10 + 100 + 1000 + 10000) \\ = 4.8 \cdot 11111 = 53332.8 \end{gathered}\text{。}

所以正确答案是 C。

By symmetry, each of the five digits appears equally often in each place, so the average digit in every place is

\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8.

The average number is therefore

\begin{gathered} \small 4.8(1 + 10 + 100 + 1000 + 10000) \\ = 4.8 \cdot 11111 = 53332.8. \end{gathered}

Thus, C is the correct answer.

其他年份的第 20 题