AMC10 2012 B

AMC10 2012 B · Q9

AMC10 2012 B · Q9. It mainly tests Parity (odd/even).

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?

两个整数的和为 26。当再添加两个整数到前两个整数时，和为 41。最后当再添加两个整数到前四个整数的和时，和为 57。这 6 个整数中偶数的个数最小是多少？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (A)

正确答案：（A）

Solution

The sum of two integers is even if they are both even or both\nodd. The sum of two integers is odd if one is even and one is odd. Only the\nmiddle two integers have an odd sum, namely $41 - 26 = 15$. Hence at least one\ninteger must be even. A list satisfying the given conditions in which there is\nonly one even integer is 1, 25, 1, 14, 1, 15.

两个整数的和为偶数当且仅当它们都是偶数或都是奇数。两个整数的和为奇数当且仅当一个偶一个奇。只有中间两个整数的和为奇数，即 $41 - 26 = 15$。因此至少有一个整数必须是偶数。满足条件的列表中只有一个偶数的例子是 1、25、1、14、1、15。

Topics

NT.005 Parity (odd/even)