Remainder when divided by $7$

Question

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.

Answer 1

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.

Answer 2

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?

Answer 3

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?

Answer 4

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 $\sum_{i=1}^{100}12^i\equiv 3\mod7$.

Answer 5

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 $(-2-1,7)=1$