How to compute $3^9 \pmod {10}$

Question

The solution given was:

$3^9 = (27)^3 = 7^3 = 49(7) = 9(7) = 63 = 3 $

I understand up to $\ 3^9 = (27)^3 $ But after that I am lost. Can someone explain how to solve this and what is going on here?

Answer 1

The $27$ was replaced with a $7$ because $27\pmod{10}=7$. The same concept arose with the $49$ later, as $49\pmod{10}=9$, and likewise $63\pmod{10}=3$. This is why the values greater than $10$ are switched.

Answer 2

$3^2 = 9 \equiv -1 \bmod 10 $.

Therefore $3^8 =(3^2)^4 \equiv (-1)^4 \bmod 10 \equiv 1 $.

Therefore $3^9 =3^{8+1} =3^8 3^1 \equiv 1\cdot 3 \bmod 10 \equiv 3 $.

Answer 3

When computing modulo $10$, all digits except units can be safely dropped.

For example, $7\cdot7\equiv\color{red}49$.

Then $3^3\equiv\color{red}27,3^9\equiv7^3\equiv7^2\cdot7\equiv\color{red}49\cdot7\equiv\color{red}63$.

Answer 4

You are asked to compute the unit digit of $3^9$.

So in every intermediate result, you can simply take the unit digit and continue without the rest of the number.

For example:

• Instead of $27$, you can simply continue with $7$
• Instead of $49$, you can simply continue with $9$