Question

两个整数的和为 $26$ 。当另外两个整数加到前两个整数上时，和为 $41$ 。最后再把另外两个整数加到前四个整数的和上时，和为 $57$ 。这 $6$ 个整数中偶数的最少个数是多少？

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：1200

小提示：

看每一对新加入的整数之和： $26$ ， $41-26$ ，和 $57-41$

Look at the sum of each new pair: $26,$ $41-26,$ and $57-41$

大提示：

两个整数的和为奇数时，恰好其中一个是偶数

A pair has an odd sum only when exactly one of its two integers is even

解答：

三个连续的整数对的和分别是 $26$ ， $41-26=15$ ，和 $57-41=16$ 。

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 $1$ 个偶数，例如 $1,25,1,14,1,15$ 。

因此正确答案是 A。

The three successive pairs have sums $26,$ $41-26=15,$ and $57-41=16.$

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 $1$ even integer is possible, for example $1,25,1,14,1,15.$

Thus, the correct answer is A.

2012 AMC 12B 第 5 题

答案：A

难度评级：1200

小提示：

大提示：

解答：

其他年份的第 5 题