The remainder when $$19^{200} + 23^{200}$$ is divided by 49, is _____.

Find the remainder when $$19^{200} + 23^{200}$$ is divided by 49.

Writing $$19 = 21 - 2$$ and $$23 = 21 + 2$$ where $$21 = 3 \times 7$$ allows us to work modulo 49 efficiently.

Using binomial expansion modulo 49, we obtain

$$19^{200} = (21-2)^{200} = \sum_{k=0}^{200} \binom{200}{k}21^k(-2)^{200-k}$$

$$23^{200} = (21+2)^{200} = \sum_{k=0}^{200} \binom{200}{k}21^k(2)^{200-k}$$

Adding these expressions shows that terms with odd $$k$$ cancel, and for even $$k$$ we get

$$19^{200} + 23^{200} = 2\sum_{j=0}^{100} \binom{200}{2j}21^{2j} \cdot 2^{200-2j}$$

Since $$21^2 = 441 = 9 \times 49$$, all terms with $$j \geq 1$$ are divisible by 49.

Hence, modulo 49 we have $$19^{200} + 23^{200} \equiv 2 \cdot 2^{200} = 2^{201} \pmod{49}$$.

Next, we compute $$2^{201} \pmod{49}$$.

By Euler's theorem, $$\phi(49) = 42$$ implies $$2^{42} \equiv 1 \pmod{49}$$.

Since $$201 = 42 \times 4 + 33$$, it follows that $$2^{201} \equiv 2^{33} \pmod{49}$$.

To compute $$2^{33} \pmod{49}$$ by successive squaring, we find:

$$2^1 = 2$$

$$2^2 = 4$$

$$2^4 = 16$$

$$2^8 = 256 = 5 \times 49 + 11 \equiv 11$$

$$2^{16} \equiv 11^2 = 121 = 2 \times 49 + 23 \equiv 23$$

$$2^{32} \equiv 23^2 = 529 = 10 \times 49 + 39 \equiv 39$$

$$2^{33} = 2^{32} \times 2 \equiv 39 \times 2 = 78 = 49 + 29 \equiv 29$$

The remainder is $$\boxed{29}$$.