Remainder when divided by $7$ What would be the remainder when
$12^1 +  12^2 + 12^3 +\cdots + 12^{100}$ is divided by $7$ ?
I tried cyclic approach (pattern method), but I couldn't solve this particular question.

We have
$$1+\sum_{i=1}^{100}12^i=\frac{12^{101}-1}{12-1}=\frac{12^{101}-1}{11}$$
Since $\gcd(7,11)=1$, we need only find the remainer of $12^{101}-1$ with $7$.
We have $12^{101}\equiv 5^{101}\bmod 7$ and $5^{101}=25^{50}\cdot5\equiv 5\mod7$
Thus $12^{101}-1\equiv 4\mod7$
and as we explained this gives $\frac{12^{101}-1}{11}\equiv4\mod7$
So $\sum_{i=1}^{100}12^i\equiv 3\mod7$.
