1. ## modulo question

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.

2. 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 $\displaystyle \gcd(102,111)=3$. 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.

3. Originally Posted by chrisc
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.
Number theory it is.

CB

4. thanks for the tip. I'll work on it after lunch and post an update

5. 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
so
3x + a = amod37
Just doesn't seem right at this point.

Any further assistance is appreciated.