Solution: Let $x \equiv 1 \pmod{7}$ and $x \equiv 1 \pmod{11}$. By the Chinese Remainder Theorem, $x \equiv 1 \pmod{77}$. The two-digit numbers satisfying this are $1 + 77 = 78$. Thus, the answer is $\boxed{78}$.

["Chinese Remainder Theorem: Solving Simple Congruences Like $x \equiv 1 \pmod{7}$ and $x \equiv 1 \pmod{11}$", "When faced with a system of congruences, the powerful Chinese Remainder Theorem (CRT) simplifies finding solutions efficiently—even for two-digit integers. A classic example is solving:\n$$\nx \equiv 1 \pmod{7} \quad \ ext{and} \quad x \equiv 1 \pmod{11}\n$$\nAlthough seemingly complex, these conditions reveal a much elegant solution.", "Since both congruences share the same remainder—1—we can directly apply CRT intuition: if $x$ leaves a remainder of 1 when divided by 7 and also by 11, then $x - 1$ is divisible by both 7 and 11. Because 7 and 11 are coprime, their least common multiple is simply $7 \ imes 11 = 77$. Thus:\n$$\nx - 1 \equiv 0 \pmod{77} \quad \Rightarrow \quad x \equiv 1 \pmod{77}\n$$", "This means $x = 1 + 77k$ for integer values of $k$. We now seek two-digit numbers satisfying this form:\n$$\nx = 1 + 77k\n$$", "Testing small integer values of $k$:\n- For $k = 0$: $x = 1 + 77 \ imes 0 = 1$ (not two-digit)\n- For $k = 1$: $x = 1 + 77 \ imes 1 = 78$ (valid two-digit number)\n- For $k = 2$: $x = 1 + 154 = 155$ (too large)", "Therefore, the only two-digit solution is $x = 78$.", "This neat application of the Chinese Remainder Theorem demonstrates how modular arithmetic and prime moduli combine seamlessly to reduce multi-condition problems into a single, manageable equation. Whether in number theory, cryptography, or algorithm design, such streamlined solutions are invaluable.", "Answer: $\boxed{78}$"]