\( n \equiv 2 \pmod{3}, n \equiv 4 \pmod{5} \): \( n = 5k + 4 \). Then \( 5k + 4 \equiv 2 \pmod{3} \Rightarrow 2k + 1 \equiv 2 \Rightarrow 2k \equiv 1 \Rightarrow k \equiv 2 \pmod{3} \Rightarrow k = 3m + 2 \Rightarrow n = 15m + 14 \). Smallest \( n = 14 \).

["Solving the System of Congruences: Finding ( n \equiv 2 \pmod{3} ) and ( n \equiv 4 \pmod{5} )", "Solving systems of linear congruences is a fundamental problem in number theory, especially when dealing with modular arithmetic. A classic example is finding an integer ( n ) that satisfies:\n[\nn \equiv 2 \pmod{3} \quad \ ext{and} \quad n \equiv 4 \pmod{5}\n]", "Understanding how to combine these conditions can reveal the smallest such integer and provide insight into general methods in modular arithmetic.", "---", "Step 1: Express ( n ) from the second congruence", "Start with the second condition:\n[\nn \equiv 4 \pmod{5}\n]\nThis implies ( n ) can be written in the form\n[\nn = 5k + 4\n]\nfor some integer ( k ).", "---", "Step 2: Substitute into the first congruence", "Substitute ( n = 5k + 4 ) into the first condition:\n[\n5k + 4 \equiv 2 \pmod{3}\n]", "Simplify modulo 3:\n- ( 5 \equiv 2 \pmod{3} )\n- ( 4 \equiv 1 \pmod{3} )", "So the equation becomes:\n[\n2k + 1 \equiv 2 \pmod{3}\n]", "---", "Step 3: Solve the congruence", "Subtract 1 from both sides:\n[\n2k \equiv 1 \pmod{3}\n]", "We now find the multiplicative inverse of 2 modulo 3. Since ( 2 \ imes 2 = 4 \equiv 1 \pmod{3} ), the inverse is 2. Multiply both sides by 2:\n[\nk \equiv 2 \ imes 1 \equiv 2 \pmod{3}\n]", "Thus,\n[\nk = 3m + 2 \quad \ ext{for some integer } m\n]", "---", "Step 4: Substitute back to find ( n )", "Recall that ( n = 5k + 4 ), so substitute ( k = 3m + 2 ):\n[\nn = 5(3m + 2) + 4 = 15m + 10 + 4 = 15m + 14\n]", "This gives the general solution:\n[\nn \equiv 14 \pmod{15}\n]", "---", "Step 5: Find the smallest positive solution", "Setting ( m = 0 ) gives the smallest positive solution:\n[\nn = 15(0) + 14 = 14\n]", "---", "Verification", "- ( 14 \div 3 = 4 ) remainder ( 2 \Rightarrow 14 \equiv 2 \pmod{3} ) ✅\n- ( 14 \div 5 = 2 ) remainder ( 4 \Rightarrow 14 \equiv 4 \pmod{5} ) ✅", "Thus, ( n = 14 ) satisfies both congruences and is the smallest positive solution.", "---", "Conclusion", "By combining expressions from congruences and simplifying using modular arithmetic, we determined that the smallest solution to\n[\nn \equiv 2 \pmod{3} \quad \ ext{and} \quad n \equiv 4 \pmod{5}\n]\nis ( n = 14 ), following the logical steps: substitute ( n = 5k + 4 ), reduce modulo 3, solve the resulting equation, back-substitute, and find the minimal solution.", "This method exemplifies the powerful technique of solving systems of linear congruences using substitution and modular simplification.", "---", "Key takeaways:\n- Express one variable in terms of the modulus.\n- Substitute into the other congruence.\n- Simplify using modular arithmetic.\n- Solve the resulting simpler congruence.\n- Back-substitute to find the general solution.", "Understanding this process lays the foundation for more advanced topics like the Chinese Remainder Theorem."]