2006 AMC 10A Problem 9

Attempt Problem 9 of the 2006 AMC 10A 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 10A solutions, or check the answer key.

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

9.

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
Small hint:
The sum of nn consecutive integers is nn times the middle value
Big hint:
Check n=2,3,4,5n = 2, 3, 4, 5
Solution:
The sum of nn consecutive integers equals nn times their median. For a sum of 1515: 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.
No set of 44 works (their sum is even), and 66 or more consecutive positive integers already exceed 15.15. There are 33 such sets.
Thus, the correct answer is C.
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