I’m expecting that people are currently looking at the current balance of 22, seeing that faul_sname has predicted [10, 30], and will upvote to try to get it out of that range. Which is a good thing for Grognor. But if you want me to pick a “reasonable” estimate, the same process will repeat itself, using whatever value I give. So I need to pick a value that’s high enough that I don’t think people will even try to reach it.
(EDIT: apparently it’s no longer possible to link to sections of Wikipedia articles using #. Above link is meant to point to the section of the article entitled “Rightmost decimal digits...”)
(EDIT: apparently it’s no longer possible to link to subsections of Wikipedia articles using #. Above link is meant to point to the section of the article entitled “Rightmost decimal digits...”)
I haven’t studied number theory, but I expect that someone who has would be able to answer this. Successive powers of three have final digits in the repeating pattern 1, 3, 9, 7, so if we can find N mod 4 for the N such that 3^N = 3^^^3, then we would have our answer.
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.]
How high is “reasonable”?
EDIT: The reason I ask is so that I can add it as a prediction statement on PredictionBook.
I’m expecting that people are currently looking at the current balance of 22, seeing that faul_sname has predicted [10, 30], and will upvote to try to get it out of that range. Which is a good thing for Grognor. But if you want me to pick a “reasonable” estimate, the same process will repeat itself, using whatever value I give. So I need to pick a value that’s high enough that I don’t think people will even try to reach it.
3^^^3 ;)
What if I predicted that the karma was going to end up even?
Edit: Or better, that it was going to end in a seven?
What’s the last digit (base 10) of 3^^^3, anyway?
7. See here
(EDIT: apparently it’s no longer possible to link to sections of Wikipedia articles using #. Above link is meant to point to the section of the article entitled “Rightmost decimal digits...”)
URL encode the apostrophe, and it works.
I haven’t studied number theory, but I expect that someone who has would be able to answer this. Successive powers of three have final digits in the repeating pattern 1, 3, 9, 7, so if we can find N mod 4 for the N such that 3^N = 3^^^3, then we would have our answer.
3^odd = 3 mod 4
so it ends in 7.
(but I repeat myself)
I think you’re mistaken. Counterexample: 3^9 = 19683.
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.
Nope
no, 7
(see other comment)
I would guess that Randaly meant to cheekily respond to
in place of my actual question.
Nope, I actually was wrong (thanks for the charitable interpretation, though!) (I cubed it an extra time)