2006 AMC 12A Problem 8

Attempt Problem 8 of the 2006 AMC 12A 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 2006 AMC 12A solutions, or check the answer key.

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8.

How many sets of two or more consecutive positive integers have a sum of 15?15?

11

22

33

44

55

Answer: C
Concepts:arithmetic sequencecasework
Difficulty rating: 1330
Solution:

The sum of nn consecutive integers equals nn times their median. For a sum of 15:15: n=2n = 2 gives 7+8,7 + 8, n=3n = 3 gives 4+5+6,4 + 5 + 6, and n=5n = 5 gives 1+2+3+4+5.1 + 2 + 3 + 4 + 5.

A run of four consecutive integers sums to an even number, and more than five terms already exceed 1+2+3+4+5=15.1 + 2 + 3 + 4 + 5 = 15. So there are 33 sets.

Thus, the correct answer is C.

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