\( n \equiv 2 \pmod{3}, n \equiv 1 \pmod{5} \): Let \( n = 5k + 1 \). Then \( 5k + 1 \equiv 2 \pmod{3} \Rightarrow 2k + 1 \equiv 2 \Rightarrow 2k \equiv 1 \Rightarrow k \equiv 2 \pmod{3} \Rightarrow k = 3m + 2 \Rightarrow n = 5(3m+2)+1 = 15m + 11 \). Smallest is \( n = 11 \).

["Solve the System of Congruences: ( n \equiv 2 \pmod{3} ) and ( n \equiv 1 \pmod{5} )", "Finding integers that satisfy multiple modular conditions is a foundational topic in number theory, and solving the system\n[\n\begin{cases}\nn \equiv 2 \pmod{3}, \\nn \equiv 1 \pmod{5}\n\end{cases}\n]\noffers an elegant illustration of the Chinese Remainder Theorem in action. In this article, we’ll walk through how to determine the general solution and identify the smallest positive integer satisfying both congruences.", "---", "### Understanding the Problem", "We seek all integers ( n ) such that:", "- When divided by 3, the remainder is 2,\n- When divided by 5, the remainder is 1.", "Rather than trial and error, we use algebraic reasoning based on modular arithmetic to derive the unique solution modulo 15—the least common multiple of 3 and 5.", "---", "### Step 1: Express ( n ) Using One Congruence", "Start with the second congruence:\n[\nn \equiv 1 \pmod{5}\n]\nThis means ( n ) can be written in the form\n[\nn = 5k + 1 \quad \ ext{for some integer } k.\n]", "We substitute this expression into the first congruence.", "---", "### Step 2: Substitute and Simplify Modulo 3", "Using ( n = 5k + 1 ), the first condition becomes:\n[\n5k + 1 \equiv 2 \pmod{3}\n]\nWe simplify modulo 3:\n- ( 5 \equiv 2 \pmod{3} ), so\n[\n2k + 1 \equiv 2 \pmod{3}\n]\nSubtract 1 from both sides:\n[\n2k \equiv 1 \pmod{3}\n]", "To solve ( 2k \equiv 1 \pmod{3} ), recall that 2 has a multiplicative inverse modulo 3 because ( \gcd(2,3) = 1 ). The inverse of 2 mod 3 is 2, since ( 2 \ imes 2 = 4 \equiv 1 \pmod{3} ). Multiplying both sides by 2:\n[\nk \equiv 2 \ imes 1 = 2 \pmod{3}\n]", "Thus,\n[\nk = 3m + 2 \quad \ ext{for some integer } m.\n]", "---", "### Step 3: Substitute Back to Find ( n )", "Recall ( n = 5k + 1 ), and substitute ( k = 3m + 2 ):\n[\nn = 5(3m + 2) + 1 = 15m + 10 + 1 = 15m + 11\n]\nTherefore, the general solution is\n[\nn \equiv 11 \pmod{15}\n]", "---", "### Step 4: Identify the Smallest Positive Solution", "The general solution ( n = 15m + 11 ) gives infinitely many solutions, but the smallest positive one occurs when ( m = 0 ):\n[\nn = 11\n]", "Verification:\n- ( 11 \div 3 = 3 ) remainder 2 → ( 11 \equiv 2 \pmod{3} ) ✓\n- ( 11 \div 5 = 2 ) remainder 1 → ( 11 \equiv 1 \pmod{5} ) ✓", "Thus, ( n = 11 ) is correct.", "---", "### Why This Matters", "This method showcases how the Chinese Remainder Theorem enables us to solve systems of linear congruences efficiently—no brute-forcing required. By expressing one variable in terms of a modulus and substituting, we reduce the problem to a simpler linear congruence, then climb backward to the full solution.", "---", "### Conclusion", "The system\n[\nn \equiv 2 \pmod{3}, \quad n \equiv 1 \pmod{5}\n]\nhas infinitely many solutions of the form\n[\nn = 15m + 11, \quad m \in \mathbb{Z}\n]\nwith the smallest positive solution being\n[\n\boxed{11}\n]", "Whether in modular arithmetic courses, coding challenges, or cryptography, mastering such techniques strengthens problem-solving skills essential for advanced mathematics and real-world applications.", "---", "Keywords: ( n \equiv 2 \pmod{3} ), ( n \equiv 1 \pmod{5} ), solving linear congruences, Chinese Remainder Theorem, modular arithmetic, smallest solution, math tutorial, step-by-step solution, integer congruences."]