Modular Arithmetic

[Ychwack]

I have the congruence:

\(\displaystyle \begin{cases} x\, \equiv\, 1 & (\mbox{mod }\, 111)\\x\, \equiv\, 55 & (\mbox{mod }\, 63) \end{cases}\)

But I cannot seem to be able to solve it. I find that with the Chinese Remainder Theorem and so on I can usually solve them, but I'm stumped with this. I think it's because the modules are all divisible by 3, but I can't quite figure out how to "Convert" them to modules that are equivalent. Any idea?

 

[Deleted member 4993]

I have the congruence:

\(\displaystyle \begin{cases} x\, \equiv\, 1 & (\mbox{mod }\, 111)\\x\, \equiv\, 55 & (\mbox{mod }\, 63) \end{cases}\)

But I cannot seem to be able to solve it. I find that with the Chinese Remainder Theorem and so on I can usually solve them, but I'm stumped with this. I think it's because the modules are all divisible by 3, but I can't quite figure out how to "Convert" them to modules that are equivalent. Any idea?

111 * m + 1 = 63 * n + 55

37 * m = 21 * n + 18

'm' must be divisible by 3 → m = 3p

37 * p = 7 * n + 6

Now what .....