Find a number such that 0<x<111 and
(102^70 + 1)^35 = x(mod111)
Any help or tips are much appreciated.
Note for moderator: I originally posted this is the Number Theory forum by accident. I had multiple windows open ready a number of threads. And When I started a new thread, I posted in Number Theory instead of Abstract Algebra where I intended to.
I realized this before a mod messaged me. I was also told my message was moved, but after a few hours I can still see the message in the wrong forum, and now here in Abstract Algebra. If it is in this forum, I cannot see it and you can delete the double. Thank you.
This can also be considered to be a number theory problem.
It would be nice to use Fermat's Little Theorem / Euler's Theorem, but . In this case, you can use the Chinese Remainder Theorem - look at the number modulo 3 and 37 and then combine your results to get a common solution modulo 111.
Number theory it is.
Originally Posted by chrisc
thanks for the tip. I'll work on it after lunch and post an update :)
Hi again. So I am fairly new to the Chinese Remainder theorem. All my experiences with examples from text and class involve equations where the left side of the mod equation needs to be found.
n = amodb, find n.
Now I need to find a in this question.
I understand the idea of splitting the equation. That is simple. But what change do I need to make from my normal approach?
Would I still go about something like this:
3x + a = (102&70 + 1)^35
3x + a = amod37
Just doesn't seem right at this point.
Any further assistance is appreciated.