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

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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:
 
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>7 + 8 = 15</math>
+
*<math>15 = 7 + 8</math>
  
 
Thus, the correct answer is C (3).
 
Thus, the correct answer is C (3).

Revision as of 20:13, 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:

  • $15 = 7 + 8$

Thus, the correct answer is C (3).

See also


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