Difference between revisions of "2006 AMC 12A Problems/Problem 8"
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+ | {{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #8]] and [[2006 AMC 10A Problems/Problem 9|2008 AMC 10A #9]]}} | ||
== Problem == | == Problem == | ||
+ | How many [[set]]s of two or more consecutive positive integers have a sum of <math>15</math>? | ||
− | + | <math> \textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math> | |
− | <math> | + | == Solution == |
+ | 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>4 + 5 + 6 = 15</math> | ||
+ | |||
+ | If the number of integers in the list is even, then the average will have a <math>\frac{1}{2}</math>. The only possibility is <math>\frac{15}{2}</math>, from which we get: | ||
+ | |||
+ | *<math>15 = 7 + 8</math> | ||
+ | |||
+ | Thus, the correct answer is <math>\boxed{\textbf{(C) }3}.</math> | ||
+ | |||
+ | |||
+ | 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. | |
== See also == | == See also == | ||
− | + | {{AMC12 box|year=2006|ab=A|num-b=7|num-a=9}} | |
− | + | {{AMC10 box|year=2006|ab=A|num-b=8|num-a=10}} | |
− | + | {{MAA Notice}} | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 23:15, 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.