Solution: Let the number be $n$. Then $n \equiv 3 \pmod{7}$ and $n \equiv 5 \pmod{6}$. Solve the system:

["Title: Solve the System of Congruences: Finding the Smallest $ n $ Satisfying $ n \equiv 3 \pmod{7} $ and $ n \equiv 5 \pmod{6} $", "When faced with a system of congruences, solving for the smallest positive integer $ n $ can feel like uncovering a hidden pattern. This article walks you step-by-step through solving the system:", "$$\n\begin{align}\nn &\equiv 3 \pmod{7} \\nn &\equiv 5 \pmod{6}\n\end{align}\n$$", "Let $ n = \overline{m} $, and solve:\n$ n \equiv 3 \pmod{7} $ means $ n = 7k + 3 $ for some integer $ k $.", "Substitute into the second congruence:\n$$\n7k + 3 \equiv 5 \pmod{6}\n$$", "Now simplify:\n$$\n7k \equiv 2 \pmod{6}\n$$", "But $ 7 \equiv 1 \pmod{6} $, so:\n$$\n1 \cdot k \equiv 2 \pmod{6} \Rightarrow k \equiv 2 \pmod{6}\n$$", "Thus, $ k = 6m + 2 $ for some integer $ m $. Substitute back:\n$$\nn = 7k + 3 = 7(6m + 2) + 3 = 42m + 14 + 3 = 42m + 17\n$$", "The smallest positive solution occurs when $ m = 0 $:\n$$\nn = 42(0) + 17 = 17\n$$", "✅ Verification:\n- $ 17 \div 7 = 2 $ remainder $ 3 $ → $ 17 \equiv 3 \pmod{7} $ ✅\n- $ 17 \div 6 = 2 $ remainder $ 5 $ → $ 17 \equiv 5 \pmod{6} $ ✅", "This solution exemplifies the power of modular arithmetic and the Chinese Remainder Theorem in solving simultaneous congruences efficiently. Whether in cryptography, algorithm design, or number theory, such methods provide efficient ways to find solutions under modular constraints.", "Key Takeaways:\n- Use substitution and modular equivalence to reduce the system.\n- When one modulus divides the coefficient (like 7 ≡ 1 mod 6), simplify the equation cleanly.\n- Express $ k $ in terms of the new modulus, then substitute to find general form.\n- Plug back to find the smallest $ n $; always verify.", "Master these steps and tackle modular systems with confidence!"]