Difference between revisions of "2002 AMC 12B Problems/Problem 13"
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Thus, the answer is <math>9*25 = \boxed{225}.</math> | Thus, the answer is <math>9*25 = \boxed{225}.</math> | ||
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== Solution 2 == | == Solution 2 == | ||
Revision as of 20:20, 31 August 2020
Problem
The sum of 
consecutive positive integers is a perfect square. The smallest possible value of this sum is
Solution
Solution 1
Let the first term be 
and the common difference be 
Thus, we know the sum of terms is 
Because 
is already a perfect square, we know that 
must also be a perfect square.
Since we want the smallest value of the sum, we know that 
and 
Thus, the answer is 
~coolmath2017
Solution 2
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
Notice that all five choices given are perfect squares.
Let 
be the smallest number, we have ![\[a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153\]](http://latex.artofproblemsolving.com/2/4/b/24ba9a08041a00d74638a06c236e5ee189a8420e.png)
Subtract 
from each of the choices and then check its divisibility by , we have 
as the smallest possible sum. 
~ Nafer
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. 
