Difference between revisions of "2006 AMC 12A Problems/Problem 8"

 
(See also)
(10 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 +
{{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 integer]]s have a sum of <math>15</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 [[divisor]] of 15. If the number of [[integer]]s 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:
 +
 +
*<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 3, answer choice <math>\mathrm{(C) \ }</math>.
  
 
== See also ==
 
== See also ==
* [[2006 AMC 12A Problems]]
+
{{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]]

Revision as of 17:51, 3 July 2013

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 $15$?

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

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 3, answer choice $\mathrm{(C) \ }$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png