Remainder of exponential expression divided by 13

By Fermat's Little Theorem, $10^{12} \equiv 1 \pmod{13}$.

$68 = 12 \times 5 + 8$, so: $10^{68} = (10^{12})^5 \cdot 10^8 \equiv 1^5 \cdot 10^8 \equiv 10^8 \pmod{13}$

Since $10 \equiv -3 \pmod{13}$: $10^8 \equiv (-3)^8 = 3^8 \pmod{13}$

Notice $3^3 = 27 \equiv 1 \pmod{13}$. Thus, $3^8 = (3^3)^2 \cdot 3^2 \equiv (1)^2 \cdot 9 = 9 \pmod{13}$.

So the remainder is 9.