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Smallest Perfect Cube Puzzle (Posted on 2015-07-15) Difficulty: 3 of 5
Determine the smallest positive perfect cube whose first three digits (reading from the left) are 201 (in this order) and the last digit is 5.

No Solution Yet Submitted by K Sengupta    
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Solution simulation of manual trials Comment 2 of 2 |
(In reply to computer assisted solution by Charlie)

ceil( 2010 ^(1/3))= 13 
the cube of the next odd multiple of 5 is 3375 

ceil( 20100 ^(1/3))= 28 
the cube of the next odd multiple of 5 is 42875 

ceil( 201000 ^(1/3))= 59 
the cube of the next odd multiple of 5 is 274625 

ceil( 2010000 ^(1/3))= 127 
the cube of the next odd multiple of 5 is 2460375 

ceil( 20100000 ^(1/3))= 272 
the cube of the next odd multiple of 5 is 20796875 

ceil( 201000000 ^(1/3))= 586 
the cube of the next odd multiple of 5 is 210644875 

ceil( 2010000000 ^(1/3))= 1263 
the cube of the next odd multiple of 5 is 2024284625 

ceil( 20100000000 ^(1/3))= 2719 
the cube of the next odd multiple of 5 is 20234828125 

ceil( 201000000000 ^(1/3))= 5858 
the cube of the next odd multiple of 5 is 201745589625 

Answer is 201745589625 


  Posted by Charlie on 2015-07-15 16:19:57
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