AMC10 2005 A

AMC10 2005 A · Q14

AMC10 2005 A · Q14. It mainly tests Basic counting (rules of product/sum), 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数字是首位和末位数字平均数的性质？

(A) 41 41

(B) 42 42

(D) 44 44

(E) 45 45

Answer

Correct choice: (E)

正确答案：（E）

Solution

The first and last digits must be both odd or both even for their average to be an integer. There are $5 \cdot 5 = 25$ odd-odd combinations for the first and last digits. There are $4 \cdot 5 = 20$ even-even combinations that do not use zero as the first digit. Hence the total is 45.

首位和末位数字必须同奇同偶，它们的平均数才为整数。首位和末位均为奇数的组合有 $5 \cdot 5 = 25$ 种。首位不为零的偶偶组合有 $4 \cdot 5 = 20$ 种。因此总数为 45。

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