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.
(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 |
simulation of manual trials
