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I solved the last 8 digits of 3^^^3 (they're ...64,195,387). Take that ultrafinitists!
...62535796399618993967905496638003222348723967018485186439059104575627262464195387.
Boo-yah.
Edit: obviously this was not done by hand. I used Mathematica. Code:
Edit: this was all done to make up for my distress at only having an Erdos number of 3.
Impressive, I didn't think it could be automatized (and even if it could, that it could go so many digits before hitting a computational threshold for large exponentials). My only regret is that I have but 1 upvote to give.
It is not clear to me why the above code works. In particular, the 10^80 part.
A number mod 10^n yields the last n digits of the number.
Aaah. it was just saying how much to output. Phew. I was trying to figure out what 10^80 could have to do with 3^^^3 and failing, hard. So, it's a great relief that the answer is, 'nothing'.
Technically, I have not used the fact that the number in question is 3^^^3 -- I am treating it as 3^^X, where X is very large. So for huge numbers of digits, this will not give the correct answer, but I don't think it's practical to compute that many digits, in any case.
Should I explain how everything else works?
Nah, I just figured that the named functions were something relevant, and I'm satisfied with that level of knowledge at this point.
I.. just.. WHAT? The last digits are the easiest, of course, BUT STILL. What was your methodology? (because I can't be bothered to think of how to do it myself)
I started with some iterated powers of 3 and tried to find patterns. For instance, 3 to an odd (natural number) power is always 3 mod 4, and 3 to the power of (a natural number that's 3 mod 4) always has a 7 in the one's place.
http://en.wikipedia.org/wiki/Graham%27s_number#Rightmost_decimal_digits
That's awesome. Why did you do this? (ie. Did you get to publish it someplace...)
Partially just to prove it is a real number with real properties, but mostly because I wanted a challenge and wasn't getting it from my current math classes (I'm currently in college, majoring in math). As much as I'd like to say it was to outdo the AI at math (since calculators can't do anything with the number 3^^^3, even take its mod 2), I had to use a calculator for all but the last 3 digits.
You mean you did it manually? You didn't write code to do the grunt work?
In the interest of challenging my mental abilities, I used as few resources as possible (and I suck at writing code). It took fewer than 3^^^3 steps, thankfully.
In that case, you might find writing a program to solve it for you an even better challenge of your mental abilities.