Difference between revisions of "2006 AMC 12A Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | How many | + | 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> | <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 | + | 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>1 + 2 + 3 + 4 + 5 = 15</math> | ||
| Line 15: | Line 15: | ||
*<math>15 = 7 + 8</math> | *<math>15 = 7 + 8</math> | ||
| − | Thus, the correct answer is C | + | Thus, the correct answer is 3, answer choice <math>\mathrm{(C) \ }</math>. |
== See also == | == See also == | ||
Revision as of 23:14, 30 January 2007
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 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:
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 3, answer choice 
.
See also
| {{{header}}} | ||
| Preceded by Problem 7 |
AMC 12A 2006 |
Followed by Problem 9 |
