Solving the equation for $x$ in $Z_n$ How do you solve for x in the the $Z_n$ specified? For example, for the equation:
1) $3\odot x\oplus8\equiv1(\rm{mod} 10)$ or
2) $342\odot x\oplus 448\equiv73(\rm{mod}1003)$ 
How would you solve for x?
 A: You could just use a computer program. To do it by hand, see below.
In general, $ax+b \equiv c \pmod{n}$ is solvable for $x \in \mathbb{Z}_n$ if and only if $\gcd(a, n) \mid c-b$. When there is a solution, it is unique $\pmod{\frac{n}{\gcd(a,n)}}$. To get a solution, use the extended Euclidean algorithm to determine $\gcd(a,n)$ and to find integers $s, t$ such that $as+nt=\gcd(a,n)$. Then $a(s\frac{c-b}{\gcd(a,n)})=(c-b)-nt\frac{c-b}{\gcd(a,n)}$, so solutions are given by $x \equiv s(\frac{c-b}{\gcd(a,n)})\pmod{\frac{n}{\gcd(a,n)}}$. 
Things become a lot simpler when $\gcd(a,n)=1$ (which is the case for the 2 examples you provided). In that case, there is always a unique solution for $x \in \mathbb{Z}_n$.
To demonstrate, I will apply the method described above to your examples.
For $1)$, we have $$10=3(3)+1$$ $$1=10(1)+3(-3)$$ $$3=1(3)+0$$
Therefore $x \equiv (1-8)(-3) \equiv 1 \pmod{10}$.
For $2)$, we shall speed up the process by using a variant of the Euclidean algorithm which allows for negative remainders: $$1003=342(3)+(-23)$$ $$-23=1003+342(-3)$$ $$342=(-23)(-15)+(-3)$$ $$-3=342+(1003+342(-3))(15)=1003(15)+342(-44)$$ $$-23=(-3)8+1$$ $$1=(1003+342(-3))+(1003(15)+342(-44))(-8)=1003(-119)+342(349)$$ $$-3=1(-3)+0$$
Therefore $x\equiv (73-448)(349) \equiv 518 \pmod{1003}$.
