Linear equation using modulo The question is:  If $3X \equiv 2 \pmod 7$, what is $x$?
I used this article to solve, but I can't really get it. It's quite complicated....
I need guidance please
 A: Hint: Use the Euclidean algorithm to find numbers $m$ and $n$ such that $3m+7n=1$. Then $3m\equiv 1\pmod 7$ (why?). The answer is then $X=2m$ (why?).
A: Here is a grubby but easy to understand way. Consider this table.
x: 0  1  2  3  4  5  6  
2x:0  2  4  6  1  3  5
3x:0  3  6  9  5  1  4

We see that 3*5 = 1 in this form of arithmetic, so 3 has multiplicative inverse 5.
You have $3x = 2$ so $5\cdot 3 x = 3$, hence $x = 3$.
A: Basically mod 7 means you are operating on numbers in the ring $\Bbb{Z}_7$. This is the set $\{0, 1, 2, 3, 4, 5, 6\}$, composed of seven numbers. It's called a ring because you go "around in circles", as shown below.
We are given a number times 3 equals two. Basically, if it results in a number that doesn't appear in the set $\Bbb{Z}_7$, for example $9$, then we have $$9 \mod 7 = 9 + 7i = 2 + 7 + 7i = 2 + 7(1 + i) = 2 + 0 = 2.$$ We then say that $9$ is contained in the congruence class of $[2]_7$.
Thus, $2$ would equal $9$ in $\Bbb{Z}_7$. The equation then becomes $$3X = 9 \mod 7.$$ Does this help?
