AMC12 2002 B

AMC12 2002 B · Q1

AMC12 2002 B · Q1. It mainly tests Fractions, Averages (mean).

The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$-digit number $M$, all of whose digits are distinct. The number $M$ doesn't contain the digit

集合 $\{9, 99, 999, 9999, \ldots, 999999999\}$ 中九个数的算术平均数是一个九位数 $M$，其所有数字均不相同。数 $M$ 不包含数字

(A) 0 0

(B) 2 2

(D) 6 6

(E) 8 8

Answer

Correct choice: (A)

正确答案：（A）

Solution

We wish to find $\frac{9+99+\cdots +999999999}{9}$, or $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$. This doesn't have the digit 0, so the answer is $\boxed{\mathrm{(A)}\ 0}$

我们要求 $\frac{9+99+\cdots +999999999}{9}$，即 $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$。它不含数字 0，所以答案是 $\boxed{\mathrm{(A)}\ 0}$

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