AMC8 1996

AMC8 1996 · Q21

AMC8 1996 · Q21. It mainly tests Combinations, Parity (odd/even).

How many subsets containing three different numbers can be selected from the set $\{89, 95, 99, 132, 166, 173\}$ so that the sum of the three numbers is even?

从集合 \{89, 95, 99, 132, 166, 173\} 中能选出多少个包含三个不同数字的子集，使得三个数的和为偶数？

(A) 6 6

(B) 8 8

(D) 12 12

(E) 24 24

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The sum of three numbers is even if all three numbers are even, or if two numbers are odd and one is even. Since there are only two even numbers in the set, it follows that the three numbers must include two odd numbers and one even. The possibilities are: \[ \begin{array}{ccc} \{89,95,132\} & \{89,99,132\} & \{89,173,132\} \\ \{89,95,166\} & \{89,96,166\} & \{89,173,166\} \\ \\ \{95,99,132\} & \{95,173,132\} & \{99,173,132\} \\ \{95,99,166\} & \{95,173,166\} & \{99,173,166\} \end{array} \] Thus there are 12 possibilities.

答案（D）：三个数的和为偶数，当且仅当这三个数全为偶数，或者两个为奇数、一个为偶数。由于该集合中只有两个偶数，因此这三个数必须包含两个奇数和一个偶数。可能的组合为： \[ \begin{array}{ccc} \{89,95,132\} & \{89,99,132\} & \{89,173,132\} \\ \{89,95,166\} & \{89,96,166\} & \{89,173,166\} \\ \\ \{95,99,132\} & \{95,173,132\} & \{99,173,132\} \\ \{95,99,166\} & \{95,173,166\} & \{99,173,166\} \end{array} \] 因此共有 12 种可能。

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