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comment Solving a recurrence relation using back substitution.
Thank you! My only concern is regarding $\sum_{j=0}^{k-1}$, shouldn't it be $\sum_{j=0}^{k-2}$, since factoring $2^2$ back in would lead to the original last term $2^k$?
comment How to solve system of equations with mod?
Thank you, so much! I admit I'm still having trouble with it, though. From reading about the Euclidean Algorithms, am I right to assume that 15 is the greatest common divisor of 7 and 3? My only question regarding that aspect is how does it turn from 3 = 7 a (mod 26) to 19 = a (mod 26) by multiplying it by 15? I feel like it sounds like a pretty obvious question, but I can't figure it out... Also, if you would be so kind to provide me a fairly simple reading on the euclidean algorithm to be able to do what you did there, I would be quite grateful! (Wiki is hard ): )
comment How to solve system of equations with mod?
Awesome. I only have one last question to clarify. To obtain b, do I need to repeat the whole process, but this time substituting it the other way around or... Is there a faster way? I tried plugging the value of a in the second equation, but in the end I got a fraction... Unless I did it incorrectly, because just seeing mod there messes me up.