What is the remainder of $3^{2020} / 7$

Question

I tried to solve this problem like:

Remainder of $3^2 / 7$ is $2$

Remainder of $3^3 / 7$ is $-1$

Then i square that, and I get

Remainder of $3^6 / 7$ is $1$

Then I can go all the way to:

Remainder of $3^{2016} / 7$ is $1$

Now how do I get to 2020? Or is there another way to solve this problem?

score 2 · Answer 1 · answered Jun 08 '19 at 07:22

2

$3^{2016}=(3^6)^{336}$. The remainder is $1^{336}=1$.

$3^{2020}=3^{2016}3^4$ ...

answered Jun 08 '19 at 07:22

CY Aries

23,393

Can you proceed now? – CY Aries Jun 08 '19 at 07:24
So the remainder then is 4 if I'm correct? – Gruja Jun 08 '19 at 07:25
Yes. Just equal to the remainder when $81$ is divided by $7$. – CY Aries Jun 08 '19 at 07:25

score 1 · Answer 2 · answered Jun 08 '19 at 09:57

Here is a slightly different way which does not require little Fermat in the form of $3^6 \equiv_7 1$:

\begin{eqnarray*}3^{2020} &\equiv_7 & 9^{1010} \\ &\equiv_7 & 2^{1010} \\ &\equiv_7 & 2^{3\cdot 336 + 2} \\ &\equiv_7 & \left(2^3\right)^{336}\cdot 2^2 \\ &\equiv_7 & 1^{336}\cdot 4 \\ &\equiv_7 & 4 \\ \end{eqnarray*}

What is the remainder of $3^{2020} / 7$

2 Answers2