Solve 2^27841 mod 34 by Hand: Discrete Math Problem Solution

raross
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could someone show me how u would solve 2^27841 mod 34 by hand? I know what theorm to use, I am just having trouble using it? Thanks
 
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What theorem would you use? Anyways, if it helps, 27841 = 11x2531 and 34 = 2x17. I found that 11 was a factor of 27841 by trial and error, and then by a lot more trial and error, found that 2531 is prime. Hopefully I didn't make a mistake in the calculations.
 
Is there any other way to do this without changing the base?
 
Euler's totient theorem?
Is a bit tricky because 34 and 2 are not co-prime.
but
2^{17} \equiv 2 \mod 34
Then we can use that
27841 \equiv 1 \mod 16
to get
2^{27841} \equiv 2^{1} \equiv 2 \mod 34
 
hrm yeah that works. How would you solve it with modular exponentiation?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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