2012 AMC 12B Problem 5

Attempt Problem 5 of the 2012 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12B solutions, or check the answer key.

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5.

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

Answer: A
Concepts:parity
Difficulty rating: 1200
Solution:

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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