<s>[INST] Prove that $22\ldots 22$ ($1980$ twos) $\equiv 0 \pmod{1982}$ 
Prove that $$\underbrace{22\ldots 22}_{1980} \equiv  0 \pmod{1982}$$

I tried and found that we need to prove that $991$ divides  $\underbrace{11\ldots 11}_{1980}$. I am stuck after this.
 [/INST] $\underbrace{22\ldots 22}_{1980} = \frac {2}{9}\sum_\limits {i=0}^{1979} 10^i = \frac {2}{9} (10^{1980} -1)$
$10^{990}\equiv 1 \pmod {991}$ by Fermat's little theorem.
$10^{1980} = (10^{990})^2\equiv 1^2 \pmod {991}$
$10^{1980} - 1 \equiv 0 \pmod {991}\\
(10^{1980} - 1) \equiv 0 \pmod {991}$
$\frac 19(10^{1980} - 1)$ is whole number.
$991$ is prime and $991|9\cdot \frac 19(10^{1980} - 1)$ and  implies $991|9$ or $991| \frac 19(10^{1980} - 1)$
If $991$ divides $\frac {1}{9}(10^{1980} - 1), 1982$ divides $\frac {2}{9}(10^{1980} - 1)$
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