$\displaystyle x\equiv1\ (mod\ 3)$

$\displaystyle x\equiv3\ (mod\ 5)$

$\displaystyle x\equiv5\ (mod\ 7)$

I did this exercise based on this post:

http://www.mathhelpforum.com/math-he...ongruence.html
And I just want to know if I did this properly and if this is the right approach.

$\displaystyle x\equiv1\ (mod\ 3)\ \Rightarrow\ 5x\ \equiv5\ (mod\ 15)\ \Rightarrow\ 10x\ \equiv10\ (mod\ 15)$

$\displaystyle x\equiv3\ (mod\ 5)\ \Rightarrow\ 3x\ \equiv9\ (mod\ 15)\ \Rightarrow\ -9x\ \equiv-27\ (mod\ 15)$

Adding gives $\displaystyle x\equiv-17 \ (mod\ 15)$

Comparing with the third equation:

$\displaystyle x\equiv-17\ (mod\ 15)\ \Rightarrow\ 7x\ \equiv-119\ (mod\ 105)\ \Rightarrow\ -14x\ \equiv238\ (mod\ 105)$

$\displaystyle x\equiv5\ (mod\ 7)\ \Rightarrow\ 15x\ \equiv75\ (mod\ 105)\ \Rightarrow\ 15x\ \equiv75\ (mod\ 105)$

Adding gives $\displaystyle x\equiv313\ (mod\ 15)$

So the solution is $\displaystyle \{313+15k:k\in\mathbb{Z}\}$

While I'm here, can anyone tell me what the easiest way to find the inverse of a congruence is? I seem to struggle with that.