\( n \equiv 1 \pmod{3}, n \equiv 4 \pmod{5} \): Solve \( n = 5k + 4 \). Then \( 5k + 4 \equiv 1 \pmod{3} \Rightarrow 2k + 1 \equiv 1 \Rightarrow 2k \equiv 0 \Rightarrow k \equiv 0 \pmod{3} \Rightarrow k = 3m \Rightarrow n = 15m + 4 \). So smallest is \( n = 4 \).

["Solving a System of Congruences: ( n \equiv 1 \pmod{3} ) and ( n \equiv 4 \pmod{5} )", "Understanding how to solve systems of linear congruences is fundamental in number theory and has practical applications in cryptography, algorithm design, and scheduling problems. In this article, we explore the solution to the system:", "[\nn \equiv 1 \pmod{3}, \quad n \equiv 4 \pmod{5}\n]", "We assume ( n \equiv 4 \pmod{5} ), which allows us to express ( n ) in a simplified form for substitution.", "### Step 1: Substitution", "From the second congruence:\n[\nn = 5k + 4 \quad \ ext{for some integer } k\n]", "Substitute this expression into the first congruence:\n[\n5k + 4 \equiv 1 \pmod{3}\n]", "### Step 2: Simplify the Congruence", "Reduce modulo 3:\n- ( 5 \equiv 2 \pmod{3} )\n- ( 4 \equiv 1 \pmod{3} )", "So:\n[\n5k + 4 \equiv 2k + 1 \equiv 1 \pmod{3}\n]", "Subtract 1 from both sides:\n[\n2k \equiv 0 \pmod{3}\n]", "Since 2 and 3 are coprime, we can divide both sides by 2 modulo 3. The multiplicative inverse of 2 modulo 3 is 2 (because ( 2 \ imes 2 = 4 \equiv 1 \pmod{3} )), so:\n[\nk \equiv 0 \cdot 2 = 0 \pmod{3}\n]", "Thus, ( k \equiv 0 \pmod{3} ), which means:\n[\nk = 3m \quad \ ext{for some integer } m\n]", "### Step 3: Find ( n ) in Full Form", "Substitute ( k = 3m ) back into ( n = 5k + 4 ):\n[\nn = 5(3m) + 4 = 15m + 4\n]", "### Step 4: Smallest Positive Solution", "The general solution is ( n \equiv 4 \pmod{15} ). Therefore, the smallest positive integer solution occurs when ( m = 0 ):\n[\nn = 15(0) + 4 = 4\n]", "### Verification\nCheck:\n- ( 4 \div 3 = 1 ) remainder 1 → ( 4 \equiv 1 \pmod{3} ) ✅\n- ( 4 \div 5 = 0 ) remainder 4 → ( 4 \equiv 4 \pmod{5} ) ✅", "### Conclusion", "The system of congruences\n[\nn \equiv 1 \pmod{3}, \quad n \equiv 4 \pmod{5}\n]\nhas the general solution ( n = 15m + 4 ), and the smallest positive solution is:", "[\n\boxed{4}\n]", "This modular solution not only demonstrates algebraic reasoning but also highlights how combining modular constraints yields unique classes of integers—essential knowledge in computational mathematics and beyond."]