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 ==
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How many [[set]]s of two or more consecutive positive integers have a sum of <math>15</math>?
  
How many sets of two or more consecutive positive integers have a sum of <math>15</math>?
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<math> \textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) }  5</math>
 
 
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ }  5</math>
 
  
 
== Solution ==
 
== Solution ==
Notice that if the consecutive positive integers have a sum of 15, then their average (which could be a fraction) must be a factor 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:
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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>
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*<math>15 = 7 + 8</math>
 
*<math>15 = 7 + 8</math>
  
Thus, the correct answer is C (3).
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Thus, the correct answer is <math>\boxed{\textbf{(C) }3}.</math>
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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?
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Answer: (IMGROOT2) It isn't possible because the question asks for positive integers, and this means that negative integers or zero aren't allowed.
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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 ==
*[[2006 AMC 12A Problems]]
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{{AMC12 box|year=2006|ab=A|num-b=7|num-a=9}}
 
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{{AMC10 box|year=2006|ab=A|num-b=8|num-a=10}}
{{AMC box|year=2006|n=12A|num-b=7|num-a=9}}
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{{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 $15$?

$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) }  5$

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:

  • $1 + 2 + 3 + 4 + 5 = 15$
  • $4 + 5 + 6 = 15$

If the number of integers in the list is even, then the average will have a $\frac{1}{2}$. The only possibility is $\frac{15}{2}$, from which we get:

  • $15 = 7 + 8$

Thus, the correct answer is $\boxed{\textbf{(C) }3}.$


Question: (RealityWrites) Is it possible that the answer is $4$, because $0+1+2+3+4+5$ 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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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