2012 AMC 12B 第 5 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

5.

两个整数的和为 2626。 当另外两个整数加到前两个整数上时,和为 4141。 最后再把另外两个整数加到前四个整数的和上时,和为 5757。 这 66 个整数中偶数的最少个数是多少?

Two integers have a sum of 26.26. When two more integers are added to the first two integers the sum is 41.41. Finally when two more integers are added to the sum of the previous four integers the sum is 57.57. What is the minimum number of even integers among the 66 integers?

11

22

33

44

55

答案:A
知识点:奇偶性
难度评级:1200
解答:

三个连续的整数对的和分别是 26264126=1541-26=15, 和 5741=1657-41=16

两个整数同奇偶时和为偶数,恰好一个是偶数时和为奇数。只有中间那一对的和为奇数,所以它必须至少含有一个偶数。

另外两对都可以全是奇数,所以最少可以只有 11 个偶数,例如 1,25,1,14,1,151,25,1,14,1,15

因此正确答案是 A

The three successive pairs have sums 26,26, 4126=15,41-26=15, and 5741=16.57-41=16.

A pair sums to an even number when its two integers share parity, and to an odd number when exactly one is even. Only the middle pair sums to an odd number, so it must contain at least one even integer.

The other two pairs can be all odd, so as few as 11 even integer is possible, for example 1,25,1,14,1,15.1,25,1,14,1,15.

Thus, the correct answer is A.

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