Difference between revisions of "2006 AMC 12A Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | Notice that if the consecutive positive integers have a sum of 15, then their | + | Notice that if the consecutive positive integers have a sum of <math>15</math>, then their average (which could be a fraction) must be a divisor of <math>15</math>. If the number of integers in the list is odd, then the average must be either <math>1, 3, </math> or <math>5</math>, and <math>1</math> is clearly not possible. The other two possibilities both work: |
*<math>1 + 2 + 3 + 4 + 5 = 15</math> | *<math>1 + 2 + 3 + 4 + 5 = 15</math> | ||
| Line 15: | Line 15: | ||
*<math>15 = 7 + 8</math> | *<math>15 = 7 + 8</math> | ||
| − | Thus, the correct answer is | + | Thus, the correct answer is <math>\boxed{\textbf{(C) }3}.</math> |
| − | Question: (RealityWrites) Is it possible that the answer is 4, because 0+1+2+3+4+5 should technically count, right? | + | Question: (RealityWrites) Is it possible that the answer is <math>4</math>, because <math>0+1+2+3+4+5</math> 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. | + | 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 == | == See also == | ||
Revision as of 22:15, 16 December 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 , 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. 
