Difference between revisions of "2006 AMC 12A Problems/Problem 8"
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Thus, the correct answer is 3, answer choice <math>\mathrm{(C) \ }</math>. | Thus, the correct answer is 3, answer choice <math>\mathrm{(C) \ }</math>. | ||
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| + | Question: (RealityWrites) Is it possible that the answer is 4, because 0+1+2+3+4+5 should technically count, right? | ||
== See also == | == See also == | ||
Revision as of 10:06, 16 September 2021
- 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 15, then their average (which could be a fraction) must be a divisor of 15. If the number of integers in the list is odd, then the average must be either 1, 3, or 5, and 1 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 3, answer choice 
.
Question: (RealityWrites) Is it possible that the answer is 4, because 0+1+2+3+4+5 should technically count, right?
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. 
