Prove that $1982\cdot 1983|\underbrace{22 \ldots2}_{1980}$
Since $\underbrace{22 \ldots2}_{1980}=\frac{2(10^{1980}-1)}{10-1}$ then I need to show that $9\cdot 1982\cdot 1983| 2(10^{1980}-1)$, or $9\cdot 991\cdot 1983|(10^{1980}-1)$ since $(9,991,1983)=1$ if I show that $9|(10^{1980}-1)$, $991|(10^{1980}-1)$, $1983|(10^{1980}-1)$ then I finish the task.
Since $10\equiv 1 (mod{9})$ then $10^{1980}\equiv 1(mod{9})$ so $10^{1980}-1\equiv 0 (mod{9})$ so we show that $9|(10^{1980}-1)$.
In $991|(10^{1980}-1)$ we can use Euler's function since $(991,10)=1$ so $10^{\varphi (991)}\equiv 1 (mod 991)$ so $10^{990}\equiv 1 (mod{9})$ then if I square $10^{2\cdot 990}\equiv 1 (mod{991})$ so $10^{1980}-1\equiv 0 (mod{991})$.
Then $1983|(10^{1980}-1)$ $1983=3\cdot 661$ $3|10^{1980}-1$ it is trivial, and we can you Euler's theorem then $10^{660}\equiv 1(mod{661})$ and then $10^{3\cdot 660}\equiv 1 (mod{661})$ so $10^{1980}-1\equiv 0 (mod{661})$.
But I just not so sure when you show something like this and number $x,y,z \in \mathbb N$ (x,y,z)=1 and if say (y,x)=3 then that not change anything in task, if we want to show that x|n y|n and z|n? If you understand what I ask, here $(9,1983)=3$, but $(9,1982,1983)=1$
While $\gcd(9,1982,1983)=1$, you have $\gcd(9,1983)\ne 1$. You should rather factor $ 9\cdot 1982\cdot 1983$ as $27\cdot 661\cdot 991$.