AMC12 2005 A

AMC12 2005 A · Q11

AMC12 2005 A · Q11. It mainly tests Fractions, Parity (odd/even).

How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

有多少个三位数满足中间的数位是首位和末位数位的平均数？

(A) 41 41

(B) 42 42

(D) 44 44

(E) 45 45

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let the digits be $A, B, C$ so that $B = \frac {A + C}{2}$. In order for this to be an integer, $A$ and $C$ have to have the same parity. There are $9$ possibilities for $A$, and $5$ for $C$. $B$ depends on the value of both $A$ and $C$ and is unique for each $(A,C)$. Thus our answer is $9 \cdot 5 \cdot 1 = 45 \implies E$.

设三位数的数位为 $A, B, C$，则 $B = \frac {A + C}{2}$。要使其为整数，$A$ 与 $C$ 必须同奇同偶。$A$ 有 $9$ 种可能，$C$ 有 $5$ 种可能。$B$ 由 $A$ 与 $C$ 的取值唯一确定，因此答案为 $9 \cdot 5 \cdot 1 = 45 \implies E$。

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