I need to solve a linear congruence:
$17x \equiv 3 \pmod {29}$
The multiplicative inverse is 12. So, multiplying both sides by 12 I get,
$12\cdot 17x \equiv 12 \cdot 3 \pmod{29}$
The next line says :
$x \equiv 36 \pmod{29}$
What is $1 \pmod{29}$ The $12 \cdot 17$ disappears as it is like:
$12 \cdot 17 \equiv 1 \pmod{29}$
How $12\cdot 17$ from the Left Hand Side could be ignored here ?What is the property if $1 \pmod{29}$ that could do this?
In fact we're working with congruence classes modulo $29$. A congruence class of $x$ modulo $m$ is defined as the set of all numbers that are congruent to $x$ modulo $m$. So for example $[1]_{29}= \{1,30,59,...\}$.
So eventually we have that $[1][x] \equiv [36] \pmod{29}$, as $[12]\cdot [17] = [1]$. Obviously we have that $[1][x] = [x]$, hence we have that $[x] \equiv [36] \pmod {29}$. But as this is an equivalence relation we can say that $x \in [36]_{29}$.
Usually as congruence class representatives are used the numbers from $0$ to $m-1$, so therefore we say that $x$ is congruent to $7$ modulo $29$, but there's nothing wrong saying that $x$ is congruent to $36$ modulo $29$