Prove that $22\ldots 22$ ($1980$ twos) $\equiv 0 \pmod{1982}$

Question

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.

Answer 1

$\underbrace{22\ldots 22}_{1980} = \frac {2}{9}\sum_\limits {i=0}^{1979} 10^i = \frac {2}{9} (10^{1980} -1)$

$10^{990}\equiv 1 \pmod {991}$ 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 $\frac {2}{9}(10^{1980} - 1)$

Answer 2

$991$ is prime so $10^{990} \equiv 1 \mod 991$

So $10^{1980} = (10^{990})^2 \equiv 1 \mod 991$

So $991| 10^{1980}-1 = 99999...... = 9*11111.....$ and $\gcd (9,991) = 1$ so $991 | 111111.....$ so $1982| 222222.......= \frac {10^{1980} - 1}9 * 2$.