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

We need to find the remainder when $$19^{200} + 23^{200}$$ is divided by 49.

Write $$19 = 21 - 2$$ and $$23 = 21 + 2$$. Then:

$$19^{200} + 23^{200} = (21-2)^{200} + (21+2)^{200}$$

Expanding using the binomial theorem, odd powers of 21 cancel, leaving:

$$= 2\sum_{k=0}^{100}\binom{200}{2k}(21)^{2k} \cdot 2^{200-2k}$$

Since $$21^2 = 441 = 9 \times 49 \equiv 0 \pmod{49}$$, all terms with $$k \geq 1$$ vanish modulo 49. Only the $$k = 0$$ term survives:

$$19^{200} + 23^{200} \equiv 2 \cdot 2^{200} = 2^{201} \pmod{49}$$

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

$$201 = 42 \times 4 + 33$$, so $$2^{201} \equiv 2^{33} \pmod{49}$$.

Computing $$2^{33} \pmod{49}$$ step by step:

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

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

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

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

The remainder is 29.