The remainder on dividing $$1 + 3 + 3^2 + 3^3 + \ldots + 3^{2021}$$ by $$50$$ is ______.

We need to find the remainder when $$1 + 3 + 3^2 + 3^3 + \ldots + 3^{2021}$$ is divided by 50.

The sum is a geometric series: $$S = \frac{3^{2022} - 1}{3 - 1} = \frac{3^{2022} - 1}{2}$$

We need $$S \mod 50$$, which requires $$3^{2022} \mod 100$$ (since $$S = (3^{2022}-1)/2$$).

Finding $$3^{2022} \mod 100$$:

By Euler's theorem, $$\phi(100) = 40$$, so $$3^{40} \equiv 1 \pmod{100}$$.

$$2022 = 40 \times 50 + 22$$, so $$3^{2022} \equiv 3^{22} \pmod{100}$$.

Computing $$3^{22} \mod 100$$:

$$3^1 = 3, \quad 3^2 = 9, \quad 3^4 = 81, \quad 3^8 = 81^2 = 6561 \equiv 61 \pmod{100}$$

$$3^{16} \equiv 61^2 = 3721 \equiv 21 \pmod{100}$$

$$3^{22} = 3^{16} \cdot 3^4 \cdot 3^2 = 21 \times 81 \times 9 \pmod{100}$$

$$21 \times 81 = 1701 \equiv 1 \pmod{100}$$

$$1 \times 9 = 9 \pmod{100}$$

So $$3^{2022} \equiv 9 \pmod{100}$$.

Therefore: $$S = \frac{3^{2022} - 1}{2} \equiv \frac{9 - 1}{2} = \frac{8}{2} = 4 \pmod{50}$$

The remainder is $$4$$.