Solution: Let the integer be $ n $. We have $ n \equiv 0 \pmod{16} $ and $ n \equiv 1 \pmod{5} $. Let $ n = 16k $. Substituting into the second congruence:

["Title: Solve the System of Congruences: $ n \equiv 0 \pmod{16} $ and $ n \equiv 1 \pmod{5} $", "Meta Description: Learn how to solve the system of congruences $ n \equiv 0 \pmod{16} $ and $ n \equiv 1 \pmod{5} $ using substitution and the Chinese Remainder Theorem.", "---", "## Understanding the Problem", "We are given two modular congruences involving an integer $ n $:", "1. $ n \equiv 0 \pmod{16} $\n2. $ n \equiv 1 \pmod{5} $", "We are instructed to set $ n = 16k $, since $ n \equiv 0 \pmod{16} $ means $ n $ is a multiple of 16. Substituting this expression into the second congruence allows us to solve for $ k $, and hence find $ n $.", "---", "## Substituting $ n = 16k $ into the Second Congruence", "Replace $ n $ with $ 16k $ in $ n \equiv 1 \pmod{5} $:", "$$\n16k \equiv 1 \pmod{5}\n$$", "Now reduce $ 16 \mod 5 $:", "$$\n16 \div 5 = 3 \ ext{ remainder } 1 \quad \Rightarrow \quad 16 \equiv 1 \pmod{5}\n$$", "So the congruence simplifies to:", "$$\n1k \equiv 1 \pmod{5} \quad \Rightarrow \quad k \equiv 1 \pmod{5}\n$$", "---", "## Solving for $ k $", "The solution $ k \equiv 1 \pmod{5} $ means $ k $ can be expressed as:", "$$\nk = 5m + 1 \quad \ ext{for some integer } m\n$$", "Now substitute back to find $ n $:", "$$\nn = 16k = 16(5m + 1) = 80m + 16\n$$", "---", "## Final Answer", "Thus, the general solution is:", "$$\nn \equiv 16 \pmod{80}\n$$", "All integers $ n $ satisfying both original congruences are of the form:", "$$\nn = 80m + 16, \quad \ ext{where } m \in \mathbb{Z}\n$$", "The smallest positive solution is $ n = 16 $, which satisfies:", "- $ 16 \div 16 = 1 $, so $ 16 \equiv 0 \pmod{16} $\n- $ 16 \div 5 = 3 $ remainder $ 1 $, so $ 16 \equiv 1 \pmod{5} $", "---", "## Why This Works: The Chinese Remainder Theorem", "Although the moduli 16 and 5 are not coprime (they share no common factor other than 1), the system still always has a unique solution modulo $ \mathrm{lcm}(16, 5) = 80 $. This proves that a solution exists and is unique modulo 80.", "---", "Conclusion:\nBy letting $ n = 16k $ and substituting into the second congruence, we reduced the system to a simple linear congruence in $ k $. Solving step-by-step leads directly to the smallest solution $ n = 16 $, and all solutions follow the pattern $ n = 80m + 16 $. This method is a powerful illustration of modular arithmetic in solving simultaneous congruences.", "For faster scanning and search visibility: use keywords like modular congruence solution, Chinese Remainder Theorem, solve $ n \equiv 0 \mod 16 $, $ n \equiv 1 \mod 5 $, and integer system of congruences."]