$$50^{th}$$ root of a number $$x$$ is $$12$$ and $$50^{th}$$ root of another number $$y$$ is $$18$$. Then the remainder obtained on dividing $$(x + y)$$ by $$25$$ is ______.

We are given that the $$50^{th}$$ root of $$x$$ is $$12$$ and the $$50^{th}$$ root of $$y$$ is $$18$$.

So $$x = 12^{50}$$ and $$y = 18^{50}$$.

By Euler's theorem, since $$\gcd(12, 25) = 1$$ and $$\phi(25) = 20$$, we have $$12^{20} \equiv 1 \pmod{25}$$.

$$12^{50} = 12^{20 \cdot 2} \cdot 12^{10} = (12^{20})^2 \cdot 12^{10} \equiv 12^{10} \pmod{25}$$

Now compute $$12^{10} \pmod{25}$$:

$$12^2 = 144 \equiv 19 \pmod{25}$$

$$12^4 = 19^2 = 361 \equiv 11 \pmod{25}$$

$$12^8 = 11^2 = 121 \equiv 21 \pmod{25}$$

$$12^{10} = 12^8 \cdot 12^2 = 21 \times 19 = 399 \equiv 24 \pmod{25}$$

$$18^2 = 324 \equiv 24 \equiv -1 \pmod{25}$$

$$18^{50} = (18^2)^{25} = (-1)^{25} = -1 \equiv 24 \pmod{25}$$

$$(x + y) \pmod{25} = (24 + 24) \pmod{25} = 48 \pmod{25} = 23$$

The remainder is $$\boxed{23}$$.