Sum of powers of $3$ modulo $8$

Question

What is $(3^1+3^2+\cdots+3^{2015})\bmod{8}?$

I'm really bad at these kinds of problems so a thorough explanation would be much appreciated.

Answer 1

As a hint, \begin{align} 3^1 &\equiv 3 \pmod 8\\ 3^2 &\equiv 1\pmod 8\\ 3^3 &\equiv 3\pmod 8 \\ 3^4 &\equiv 1\pmod 8\\ \end{align}

This gives a pattern $(3+1+3+1+\cdots) \bmod 8$ in your sum.

Answer 2

Hint: $3^1+3^2+\cdots+3^{2015}=\dfrac{3^{2016}-3}{3^1-1}=\dfrac{3^{2016}-3}{2}=\dfrac{3^{2(1008)}-3}{2}\equiv $ what? (mod $8$)

Answer 3

Hint:

As $3$ has order $2\bmod 8$, you have $$3^n\equiv\begin{cases}3&\text{if }n\text{ is odd},\\ 1&\text{if }n\text{ is even}. \end{cases}$$ Therefore $$\sum_{k=0}^{2015}3^k\equiv3\times(\text{number of odd integers}\le 2015)+(\text{number of even integers}\le2014)\mod 8.$$

Answer 4

Hint: $3^1+3^2+\cdots+3^{2015}=\dfrac{3^{2016}-3}{2}$

and $16|3^4-1$ so $16|(3^4)^{504}-1$ so $8|\dfrac{3^{2016}-1}{2}=\dfrac{3^{2016}-3}{2}+1$.

Answer 5

Here is a different take.

Let $x_n = 3^1+3^2+\cdots+3^{n}$. Then $x_{n+1}=3x_n+3$, with $x_0=0$.

Modulo $8$ we get a periodic sequence of period $4$: $$ 0,3,4,7,0,3,4,7\dots $$ Therefore, $x_{2015} = x_{2015 \bmod 4} = x_3 = 7$.