First some observations:
A move doesn't change the sum of the triplet = 3n.
Any triplet will maintain the number of even/odd numbers.
The starting position has either one or two odds where the final position (0,0,3n) has either zero or one odds.
This means n is odd.
Also, a number can be factored out of a triplet without affecting whether it will work: (cx,cy,cz) = c(x,y,z)
You can work backwards from any position (x,y,z) to find the three positions that lead to this: (2x+y, x+2y, 3z)/3 or (2x+z, 3y, x+2z)/3 or (3x, 2y+z, y+2z)/3.
This leads to a lot of branches. I have found one line of branches that do work, but I don't know if there are others.
My solutions so far consist of n=3^m
(3^m-1, 3^m, 3^m+1) -> (3^m-3, 3^m, 3^m-3) = 3*(3^(m-1)-1,3^(m-1),3^(m-1)+1
continuing to keep the second number unchanged and factoring out the 3 eventually leads to 3^m*(0,1,2) -> 3^m*(0,0,3)
For example without factoring
(26,27,28) ->
(24,27,30) ->
(18,27,36) ->
(0,27,54) ->
(0,0,81)
Again, I have not shown there are no other solutions, but no other branches look promising.
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Posted by Jer
on 2021-03-09 10:15:09 |
Observations, some values
