19683 = 3 mod 4
and 3^19683 = 150 … 859227, which ends in 7.
( The full number is 9392 digits long, which messes up the spacing in these comments. )
Oh, sorry; I agree that odd powers of three are 3 mod 4, but I had read VKS as claiming that odd powers of three had a final digit of seven; I probably misunderstood the argument. [EDIT: Yes, I was confused; I understand now.]
right, well, it’s just that 3^^^3 = 3^3^3^3^3...3^3^3 = 3^(3^3^3^3...3^3^3), for a certain number of threes. So, 3^^^3 is 3^(some odd power of three).
Yes, thanks; I apologize for having misunderstood you earlier.
That is entirely ok—I am badly in need of sleep and may have failed to optimise my messages for legibility.
19683 = 3 mod 4
and 3^19683 = 150 … 859227, which ends in 7.
( The full number is 9392 digits long, which messes up the spacing in these comments. )
Oh, sorry; I agree that odd powers of three are 3 mod 4, but I had read VKS as claiming that odd powers of three had a final digit of seven; I probably misunderstood the argument. [EDIT: Yes, I was confused; I understand now.]
right, well, it’s just that 3^^^3 = 3^3^3^3^3...3^3^3 = 3^(3^3^3^3...3^3^3), for a certain number of threes. So, 3^^^3 is 3^(some odd power of three).
Yes, thanks; I apologize for having misunderstood you earlier.
That is entirely ok—I am badly in need of sleep and may have failed to optimise my messages for legibility.