Question

Find the least positive value of $x$ such that $126\equiv (x+4)\left(mod6\right)$.

Answer 1

$126\equiv (x+4)\left(mod6\right)$
$126-(x+4)=6n,\text{for some integer}n$.
$122-x=6n$
$122-x$ is a multiple of 6.
$\therefore$The least positive value of $x$ must be 2 as 122 - 2 = 120, which is the nearest multiple of 6 less than 122.