What Is the Remainder of a 3000-Layer 7 Power Tower Divided by 11?

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Crazy Power Tower Problem.

Consider an exponential tower of three thousand 7's.
What is the remainder when you divide the tower by 11?
Note that this notation means 7^(7^7) not (7^7)^7.
The final answer must be given as a single integer in the range 0-10.


Anyone who can help?
 
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Let x = 7 \uparrow \uparrow 2999\ (\mbox{mod }\phi (11)). Prove that 7 \uparrow \uparrow 3000 \equiv 7^x\ (\mbox{mod } 11). This is a starting point, hopefully you can see where to go with this.
 
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Hey, I'm not really familiar with that notation at all. How would you go about using that and what does the mod 11 mean??
 
mod 11 means the remainder on division by 11 and presumably the uparrow is hand notaiton for that repeated power you defined.
 
Without knowing what the arrows meant, you could have guessed, couldn't you? Anyways, see here (P.S. I changed the arrows in my post to double arrows in following with Knuth's up-arrow notation - see the link).
 
To expand a little, say you want to calculate 7^x (mod 11). If you can find some number n1 with 7^n1=1 (mod 11), then you only need to look at the exponent mod n1, because if a=b (mod n1), it's easy to show that 7^a=7^b (mod 11). As has been pointed out, you can take n1=\phi(11)=10, where \phi(n) is the Euler totient function.

So now if you want to calculate 7^(7^x) (mod 11), you only need to look at 7^x (mod 10). If you could find an n2 with 7^n2=1 (mod 10), you could use the same trick. Continuing this way, you can construct a sequence of positive integers n1,n2,n3,... , and it can be shown that this sequence is decreasing, so it must eventually get down to 1 (you should show this). What can you conclude from this? (note: there would be a problem if one of these nk were not relatively prime to 7. That won't happen here, but you should understand where the argument breaks down)
 
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Thanks for the help. I'm going to bring this into my group meeting and see what we can do with it.
 
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