2006 AMC 12A Problem 8

Below is the professionally curated solution for Problem 8 of the 2006 AMC 12A, 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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Concepts:arithmetic sequence casework

Difficulty rating: 1330

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

$1$

$2$

$3$

$4$

$5$

Solution:

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

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

Thus, the correct answer is C.

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