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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

15?

1

2

3

4

5

A run of

n

consecutive integers sums to

n

times its middle value

Test

n = 2, 3, 4, 5;

beyond

5

the smallest possible sum already exceeds

15

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.

Problem 8 in other years