A probability problem by أحمد الحلاق

Working modulo $81$ in the ring $\mathbb{Z}[\sqrt{2}]$ , we have (successively cubing or squaring) $\begin{array}{rcl} (1 + \sqrt{2})^3 & \equiv & 7 + 5\sqrt{2} \\ (1 + \sqrt{2})^9 & \equiv & 16 + 13\sqrt{2} \\ (1 + \sqrt{2})^{27} & \equiv & 70 + 41\sqrt{2} \\ (1 + \sqrt{2})^{54} & \equiv & -11\sqrt{2} \\ (1 + \sqrt{2})^{108} & \equiv & -1 \\ (1 + \sqrt{2})^{216} & \equiv & 1 \end{array}$ and hence $\begin{array}{rcl} a + b\sqrt{2} & \equiv & (1 + \sqrt{2})^{2012} \; \equiv \; (1 + \sqrt{2})^{68} \; \equiv \; -11\sqrt{2}(1 + \sqrt{2})^{14} \\ & \equiv & -11\sqrt{2}(16 + 13\sqrt{2})(1 + \sqrt{2})^5 \; \equiv \; 17 + 42\sqrt{2} \end{array}$ Thus $b \equiv 42 \pmod{81}$ , and hence the highest common factor of $b$ and $81$ is the highest common factor of $42$ and $81$ , namely $\boxed{3}$ .