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.
 A: This is a geometric series, which can be simplified to
$$12^1+12^2+\cdots+12^{100}=\frac{12}{11}(12^{100}-1)$$
Since $\gcd(12,7)=1$, Euler's Totient theorem applies. Can you take it from here?
A: 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 $\sum_{i=1}^{100}12^i\equiv 3\mod7$.
A: In the comments, you recognized that $12^1+12^2+12^3+\cdots+12^{100}$
$\equiv \underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\,5+4+6+2\pmod7.$
Note that the sum over each brace is a multiple of $7$, and you're almost done.
A: As $12\equiv-2\pmod7, 12^3\equiv(-2)^3\equiv-8\equiv-1,$ord$_712=6$
$\implies12^{6k+r}\equiv12^r\equiv(-2)^r\pmod7$
$$\sum_{r=1}^{100}12^r\equiv12^1+12^2+12^3+12^4+16\sum_{r=0}^512^r\pmod7$$
$$\equiv(-2)+(-2)^2+(-2)^3+(-2)^4+2\sum_{r=0}^5(-2)^r\pmod7$$
Finally $\displaystyle\sum_{r=0}^5(-2)^r\equiv\dfrac{(-2)^6-1}{-2-1}\equiv0\pmod7$ as $(-2-1,7)=1$
A: Fun Fact: $12= -2,2^{3k}= 1, 2^{3k+1}=2, 2^{3k+2}=4$ in $\mathbb Z/7\mathbb Z$. Thus, $$\sum_{i=1}^{100}12^i\equiv \sum_{i=1}^{100}(-2)^i\equiv -2^{0}+2^{99}+2^{100}+\sum_{a=0}^{32}(-2)^{3a}+(-2)^{3a+1}+(-2)^{3a+2}\pmod{7}$$
Now trivial right?
