SOLUTION: A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series.
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Question 1105540: A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series.
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If zero is considered to be positive, the answer is zero (0).
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If not:
283 negative consecutive integers, then 0, then 283 positive consecutive integers has a sum of zero (0).
Moving the start, the smallest integer up 1 adds 567 to the sum.
567 = 21*27 = 21*3^3
--> is the smallest sum.
= 63^3
= 250047
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