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

(Solution)
(expand, add succession box)
Line 6: Line 6:
  
 
== Solution ==
 
== Solution ==
1+2+3+4+5 = 15;  4+5+6 = 15;  7+8 = 15
+
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:
  
The correct answer is C (3)
+
*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 <math>\frac{1}{2}</math>. The only possibility is <math>\frac{15}{2}</math>, from which we get:
 +
 
 +
*7+8 = 15
 +
 
 +
Thus, the correct answer is C (3).
  
 
== See also ==
 
== See also ==
* [[2006 AMC 12A Problems]]
+
*[[2006 AMC 12A Problems]]
*[[2006 AMC 12A Problems/Problem 7|Previous Problem]]
+
 
*[[2006 AMC 12A Problems/Problem 9|Next Problem]]
+
{{AMC box|year=2006|n=12A|num-b=7|num-a=9}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 21:08, 30 January 2007

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 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:

  • 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:

  • 7+8 = 15

Thus, the correct answer is C (3).

See also


{{{header}}}
Preceded by
Problem 7
AMC 12A
2006
Followed by
Problem 9