The remainder when $$3^{2022}$$ is divided by $$5$$ is

$$3^{2022} = (3^2)^{1011}$$

$$(3^2)^{1011} \equiv (-1)^{1011} \pmod{5}$$

$$(-1)^{1011} = -1$$

Now we know that a remainder of $$-1$$ is not the standard way. So to convert a negative remainder back to a positive one, we can just add the divisor ($$5$$):

Hence, $$3^{2022} \equiv 4 \pmod{5}$$

Alternative Method 1:

Using Fermat's Little Theorem: $$a^{p-1} \equiv 1 \pmod{p}$$

So, we can say, $$3^{5-1} \equiv 3^4 \equiv 1 \pmod{5}$$

Now, $$2022 = 4 \times 505 + 2$$

Hence, we can say, $$3^{2022} = 3^{4 \times 505 + 2} = (3^4)^{505} \times 3^2$$

$$3^{2022} \equiv (1)^{505} \times 3^2 \pmod{5}$$

$$3^{2022} \equiv 1 \times 9 \pmod{5}$$

$$3^{2022} \equiv 4 \pmod{5}$$

Alternative Method 2:

We can also find the pattern. 

$$3^1 \equiv 3 \pmod{5}$$

$$3^2 = 9 \equiv 4 \pmod{5}$$

$$3^3 = 27 \equiv 2 \pmod{5}$$

$$3^4 = 81 \equiv 1 \pmod{5}$$

$$3^5 = 243 \equiv 3 \pmod{5}$$

So the cycle repeats: 3, 4, 2, 1

Exponent is 2022. 

Hence, $$2022 \div 4 = 505 \text{ with a remainder of } 2$$

Since the remainder is $$2$$, the result would be same as the second position in our cycle ($$3^2 \equiv 4 \pmod{5}$$).