The remainder when $$7^{103} + 7^{101}$$ is divided by 9 is ____

$$7^{103} + 7^{101} = 7^{101}(7^2 + 1) = 7^{101} \cdot 50$$.

Mod 9: $$7 \equiv -2$$, so $$7^{101} \equiv (-2)^{101} = -2^{101}$$.

The powers of 2 mod 9 cycle every 6: $$2, 4, 8, 7, 5, 1$$. Since $$101 \bmod 6 = 5$$, we get $$2^{101} \equiv 5$$, hence $$7^{101} \equiv -5 \equiv 4 \pmod{9}$$.

Also $$50 \equiv 5 \pmod 9$$. So $$7^{101}\cdot 50 \equiv 4 \cdot 5 = 20 \equiv 2 \pmod 9$$.