| Brad
Rodgers |
If the sequence tk is defined by t0 = 3, tk+1 = 3^(tk ), how could one show that the last 10 digits of tk are all the same for k > = 10? I can prove that the last digit of tk is always 7 for k > 0, but not much more. Thanks, Brad |
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| Demetres
Christofides |
Why don't you try to prove more than what you are required in an inductive way? I don't know if this method would finally give a solution but here is how I would approach it: Try to prove that the last n digits of tk are the same for k ³ n. [Don't yet know if this is true or not but we are OK if it is true at least for n £ 10] So need 3^(3^3)=3^3 mod 10 3^[3^(3^3)]=3^(3^3) mod 100 etc. Now use Euler's theorem [a, n coprime then af(n)=1 mod n] and some sort of induction to see if you can get the result. Demetres |
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| Brad
Rodgers |
I think I can prove tk+1 = tk (mod 10k ) if I can prove 3^(4. 10k-1 ) = 1 (mod 4. 10k ) -- which I think I can do. But now that I can do this, I'm not sure I fully understand the problem; it could either be interpreted as a) For a given tk , the last k digits are the same (e.g. t5 = 19857677777 , or something like that, but much longer). or b) All tk have the same last n digits for k > = n. I initially interpreted the problem as asking for a), but you've interpreted the problem as b), which seems more reasonable; the proof is easier to see (since I still can't see how to prove a)), and I didn't really have this much trouble with the other problems. Again, I think I have the proof of b). Is a) even true? It clearly isn't for t2 ; MathCAD, Maple (and even C++...) can't check it for larger numbers. I'll give this some more thought when I get some rest. Brad |
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| Demetres
Christofides |
Well, proving b) shows that a) is not true. t2 = 87 mod 100, so from (b) tk = 87 mod 100 for all k > = 2. If you have any problems with the proof of (b) or you are unsure about anything then ask me. [I've checked and (b) is indeed a true statement] Demetres |