2012 AMC 10B Problem 9

Question

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

Concepts:parity

Look at the parity of each added pair

The middle pair has odd sum, so it must contain one even and one odd integer

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.

Problem 9 in other years