A number theory problem by A Former Brilliant Member

Here I will use this property:

$ak\equiv al\pmod{\! m}\iff k\equiv l\pmod{\!\frac{m}{(a,m)}}$ ,

where $(a,m)$ is $\gcd(a,m)$ , as usual.

$n\equiv 2\pmod{\! 6}\,\Rightarrow\, n=6k+2$

$n\equiv 6k+2\equiv -2k+2\equiv 4\pmod{\!8}\iff -2k\equiv 2\pmod{\! 8}$

\stackrel{:(-2)}\iff k\equiv -1\equiv 3\pmod{\! 4}\,\Rightarrow\, k=4m+3

$n=6(4m+3)+2=24m+20$

$n\equiv 24m+20\equiv -6m\equiv 6\pmod{\!10}$

\stackrel{:(-6)}\iff m\equiv -1\equiv 4\pmod{\! 5}\,\Rightarrow\, m=5h+4

$n=24(5h+4)+20=120h+\boxed{116}$