Difference between revisions of "2002 AMC 12B Problems/Problem 13"
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\qquad\mathrm{(E)}\ 441</math> | \qquad\mathrm{(E)}\ 441</math> | ||
== Solution == | == Solution == | ||
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| + | === Solution 1 === | ||
Let <math>a, a+1, \ldots, a + 17</math> be the consecutive positive integers. Their sum, <math>18a + \frac{17(18)}{2} = 9(2a+17)</math>, is a perfect square. Since <math>9</math> is a perfect square, it follows that <math>2a + 17</math> is a perfect square. The smallest possible such perfect square is <math>25</math> when <math>a = 4</math>, and the sum is <math>225 \Rightarrow \mathrm{(B)}</math>. | Let <math>a, a+1, \ldots, a + 17</math> be the consecutive positive integers. Their sum, <math>18a + \frac{17(18)}{2} = 9(2a+17)</math>, is a perfect square. Since <math>9</math> is a perfect square, it follows that <math>2a + 17</math> is a perfect square. The smallest possible such perfect square is <math>25</math> when <math>a = 4</math>, and the sum is <math>225 \Rightarrow \mathrm{(B)}</math>. | ||
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| + | === Solution 2 === | ||
== See also == | == See also == | ||
Revision as of 15:49, 2 July 2019
Problem
The sum of 
consecutive positive integers is a perfect square. The smallest possible value of this sum is
Solution
Solution 1
Let 
be the consecutive positive integers. Their sum, 
, is a perfect square. Since 
is a perfect square, it follows that 
is a perfect square. The smallest possible such perfect square is 
when 
, and the sum is 
.
Solution 2
See also
| 2002 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
