MinibearRex comments on Rationality Quotes May 2012 - Less Wrong

6 Post author: OpenThreadGuy 01 May 2012 11:37PM

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Comment author: MinibearRex 02 May 2012 06:17:38PM -1 points [-]

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 ;)

Comment author: VKS comment score below threshold [+] (16 children)

Comment author: dlthomas 02 May 2012 09:24:22PM 1 point [-]

What's the last digit (base 10) of 3^^^3, anyway?

Comment author: komponisto 02 May 2012 10:05:40PM * 4 points [-]

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...")

Comment author: JGWeissman 02 May 2012 10:11:30PM 4 points [-]

(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...")

URL encode the apostrophe, and it works.

Comment author: Zack_M_Davis 02 May 2012 09:54:51PM 1 point [-]

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.

Comment author: VKS 02 May 2012 10:02:04PM * 4 points [-]

3^odd = 3 mod 4

so it ends in 7.

(but I repeat myself)

Comment author: Zack_M_Davis 02 May 2012 10:05:08PM * 0 points [-]

I think you're mistaken. Counterexample: 3^9 = 19683.

Comment author: JGWeissman 02 May 2012 10:09:33PM 0 points [-]

19683 = 3 mod 4

Comment author: VKS 02 May 2012 10:12:36PM * 1 point [-]

and 3^19683 = 150 ... 859227, which ends in 7.

( The full number is 9392 digits long, which messes up the spacing in these comments. )

Comment author: Zack_M_Davis 02 May 2012 10:18:09PM * 0 points [-]

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.]

Comment author: VKS 02 May 2012 10:24:57PM 2 points [-]

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).

Comment author: Zack_M_Davis 02 May 2012 10:28:21PM 1 point [-]

Yes, thanks; I apologize for having misunderstood you earlier.

continue this thread »

Comment author: Randaly 02 May 2012 09:35:28PM * -1 points [-]

Nope

Comment author: VKS 02 May 2012 09:47:51PM * 3 points [-]

no, 7

(see other comment)

Comment author: dlthomas 03 May 2012 12:20:04AM 1 point [-]

I would guess that Randaly meant to cheekily respond to

What is the last digit of the string "3^^^3"?

in place of my actual question.

Comment author: Randaly 03 May 2012 03:31:23AM * 1 point [-]

Nope, I actually was wrong (thanks for the charitable interpretation, though!) (I cubed it an extra time)

MinibearRex comments on Rationality Quotes May 2012 - Less Wrong

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