Find a positive integer N such that the digits of N combined with the digits of N^2 and the digits of N^3 are exactly the digits of N^6.
Extra credit: Find additional solutions, other than multiples of 10.
Sorry, I did not include the actual solutions I found.
And part about the sixth root of 0.1 being 0.68129206905 is that this is a starting point for searching so the numbers of digits come out right.
Take 4 digit numbers, for example. My claim is that from 1000 to 6812, the numbers of digits won't be right, but from 6813 to 9999 they will. I did not intend to imply that 6812920691 was a solution.
So here are the ones I found.
n = 9305649
n^2 = 86595103311201
n^3 = 805823636532774274449
n^6 = 649351733194904702291358580642658374253601
n = 71032113
n^2 = 5045561077244769
n^3 = 358396864587252160266897
n^6 = 128448312545973161578627675360153279566274008609
n = 97560675
n^2 = 9518085306455625
n^3 = 928590827205392632546875
n^6 = 862280924369995358075005678535785615549072265625
There are probably more.
And multiplying any of these n values by a power of 10 will yield more solutions.
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Posted by Larry
on 2023-10-11 14:21:54 |
D'oh; I forgot to post my solutions, here they are
