Solve the equation $3x+7=6$ in $\mathbb{Z}_{13}$ The title is exercise 2.2 in The Fundamental Theorem of Algebra.  
The hint for the problem is: Find the value of $\frac{1}{3}$ in $\mathbb{Z}_{13}$ 

(please realize that my knowledge of the subject is what I read in Chapter 2)

I have gotten that $x = -\frac{1}{3}$.  
I know that $\mathbb{Z}_{13}$ is the integers modulo 13.  Thus $x \equiv n \pmod{13}$? for some integer n,  $0\le n < 13$.  
Thus for $x – n = km$ for some integer $k$, with $m \neq 0$ and an integer.  
How does m divide $(x-n)$, when $(x-n)$ is not an integer, if $x = -\frac{1}{3}$???? 
 A: The notation can indeed be confusing the first few times one encounters in. When one says $x=-1/3$ in the context of $\mathbb{Z}_{13}$, it is simply short-hand for saying that $x$ is some number from $0$ to $12$ such that $-3x \equiv 1 \pmod{13}.$ So remember that $x$ is not the usual real number $-1/3$, but we use this notation because it behaves like $-1/3$ would, in that $-3x=1$ in $\mathbb{Z}_{13}.$
A: Hint $\rm\ mod\,\ 3n\!+\!1\!:\ -1\equiv 3n\:\Rightarrow\:\dfrac{-1}3\,\equiv\, \dfrac{3n}3 \,\equiv\, n$
A: What you could have done also
$$ 3x + 7 = 6\pmod{13} \implies 3x = -1 \pmod{13} \implies 3x = 12 \pmod{13}$$
 At this point it should be pretty easy!
A: Note that $3x\equiv -1$ iff $3x\equiv 12$ modulo $13$. From there, it should be simple.
$x=-1/3$ only in the sense that it is the member of $\Bbb Z_{13}$ such that $3x\equiv -1$ modulo $13$. This is not the same as saying $x=-1/3$ in the sense of real numbers.
A: We look for an integer $x$ so that 3x=-1 mod 13. This is the same as $3x+1=0 \mod 13$ or $13 | 3x+1$. Now in this case we can just run through the integers $0\le x\le 12$ to find the answer $x=4 \mod 13$.
A: Hint:  what is $-1 \pmod {13}$ in the normal set of representatives $0 \le n \lt 13$?
