AMC10 2005 B

AMC10 2005 B · Q20

AMC10 2005 B · Q20. It mainly tests Averages (mean), Basic counting (rules of product/sum).

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?

使用数字 1、3、5、7、8 各一次可以形成的全部五位数的平均数（均值）是多少？

(A) 48000 48000

(B) 49999.5 49999.5

(D) 55555 55555

(E) 56432.8 56432.8

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) Each digit appears the same number of times in the 1’s place, the 10’s place, $\ldots$, and the 10,000’s place. The average of the digits in each place is $$\frac{1}{5}(1+3+5+7+8)=\frac{24}{5}=4.8.$$ Hence the average of all the numbers is $$4.8(1+10+100+1000+10000)=4.8(11111)=53332.8.$$

（C）每个数字在个位、十位、$\ldots$、万位上出现的次数相同。每一位上数字的平均值为 $$\frac{1}{5}(1+3+5+7+8)=\frac{24}{5}=4.8.$$ 因此所有这些数的平均值为 $$4.8(1+10+100+1000+10000)=4.8(11111)=53332.8.$$

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