Question

两个整数的和为 $26$ 。再加入两个整数后，四个整数的和为 $41$ 。最后再加入两个整数后，六个整数的和为 $57$ 。这 $6$ 个整数中偶数个数的最小可能值是多少？

Two integers have a sum of $26.$ When two more integers are added to the first two integers the sum is $41.$ Finally when two more integers are added to the sum of the previous four integers the sum is $57.$ What is the minimum number of even integers among the $6$ integers?

$1$

$2$

$3$

$4$

$5$

答案：A

知识点：奇偶性

难度评级：960

解答：

前两个整数的和 $26$ 是偶数，所以它们可以都是奇数。接下来两个数之和为 $41-26=15$ 。

这个和是奇数，所以这一对必须有一个偶数和一个奇数。最后两个数之和为 $57-41=16$ ，所以它们可以都是奇数。因此至少有一个偶数，而且这个下界可以达到。

所以正确答案是 A。

The first two integers have even sum $26$ , so they can both be odd. The next two integers have sum $41-26=15$ , which is odd, so one of them must be even and one odd.

The last two integers have sum $57-41=16$ , so they can both be odd. Therefore at least one integer must be even, and one is attainable.

Thus, A is the correct answer.

2012 AMC 10B 第 9 题

答案：A

难度评级：960

解答：

其他年份的第 9 题