2006 AMC 12A Problems/Problem 8
- The following problem is from both the 2006 AMC 12A #8 and 2008 AMC 10A #9, so both problems redirect to this page.
Problem
How many sets of two or more consecutive positive integers have a sum of ?
Solution
Notice that if the consecutive positive integers have a sum of , then their average (which could be a fraction) must be a divisor of . If the number of integers in the list is odd, then the average must be either or , and is clearly not possible. The other two possibilities both work:
If the number of integers in the list is even, then the average will have a . The only possibility is , from which we get:
Thus, the correct answer is
Question: (RealityWrites) Is it possible that the answer is , because should technically count, right?
Answer: (IMGROOT2) It isn't possible because the question asks for positive integers, and this means that negative integers or zero aren't allowed.
Note to readers: make sure to always read the problem VERY carefully before attempting; it could mean the difference of making the cutoff.
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.