SOLUTION: A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series.

Algebra ->  Sequences-and-series -> SOLUTION: A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series.      Log On


   



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) About Me  (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.
---------
If zero is considered to be positive, the answer is zero (0).
---------
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
--> 21%5E3%2A3%5E3 is the smallest sum.
= 63^3
= 250047