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Power Tower problem HELP!

  1. Sep 7, 2006 #1
    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???
    Last edited: Sep 7, 2006
  2. jcsd
  3. Sep 7, 2006 #2


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    Let [itex]x = 7 \uparrow \uparrow 2999\ (\mbox{mod }\phi (11))[/itex]. Prove that [itex]7 \uparrow \uparrow 3000 \equiv 7^x\ (\mbox{mod } 11)[/itex]. This is a starting point, hopefully you can see where to go with this.
    Last edited: Sep 8, 2006
  4. Sep 8, 2006 #3
    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??
  5. Sep 8, 2006 #4

    matt grime

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    mod 11 means the remainder on division by 11 and presumably the uparrow is hand notaiton for that repeated power you defined.
  6. Sep 8, 2006 #5


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    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).
  7. Sep 8, 2006 #6


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    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=[itex]\phi(11)[/itex]=10, where [itex]\phi(n)[/itex] 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)
    Last edited: Sep 8, 2006
  8. Sep 8, 2006 #7
    Thanks for the help. I'm gonna bring this into my group meeting and see what we can do with it.
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