The remainder, when $$7^{103}$$ is divided by $$17$$, is _____.

We need $$7^{103} \pmod{17}$$.

By Fermat's little theorem: $$7^{16} \equiv 1 \pmod{17}$$

$$103 = 16 \times 6 + 7$$

$$7^{103} = (7^{16})^6 \times 7^7 \equiv 1^6 \times 7^7 \equiv 7^7 \pmod{17}$$

Now compute $$7^7 \pmod{17}$$:

$$7^2 = 49 \equiv 49 - 34 = 15 \equiv -2 \pmod{17}$$

$$7^4 = (7^2)^2 \equiv (-2)^2 = 4 \pmod{17}$$

$$7^7 = 7^4 \times 7^2 \times 7 = 4 \times (-2) \times 7 = -56 \pmod{17}$$

$$-56 = -56 + 4 \times 17 = -56 + 68 = 12$$

$$7^{103} \equiv 12 \pmod{17}$$

The remainder is $$\mathbf{12}$$.