Difference between revisions of "2002 AMC 12B Problems/Problem 13"
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=== Solution 2 === | === Solution 2 === | ||
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+ | Notice that all five choices given are perfect squares. | ||
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+ | Let <math>a</math> be the smallest number, we have $a+(a+1)+(a+2)+...+(a+17)=17a+\sum^17 | ||
== See also == | == See also == |
Revision as of 15:52, 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
Notice that all five choices given are perfect squares.
Let be the smallest number, we have $a+(a+1)+(a+2)+...+(a+17)=17a+\sum^17
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 |
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