So we are looking for the smallest three-digit number $x$ such that $x \equiv 1 \pmod{7}$ and $x \equiv 0 \pmod{60}$.
["Title: Find the Smallest Three-Digit Number Satisfying x ≡ 1 (mod 7) and x ≡ 0 (mod 60)", "When solving modular arithmetic problems in number theory, one common challenge is identifying the smallest three-digit number that meets two key conditions:\n- $ x \equiv 1 \pmod{7} $ (i.e., when divided by 7, the remainder is 1),\n- $ x \equiv 0 \pmod{60} $ (i.e., divisible by 60).", "This article explores how to find the smallest such three-digit number $ x $ satisfying both congruences using the Chinese Remainder Theorem (CRT) and modular arithmetic.", "---", "### Understanding the Problem", "We are given:\n1. $ x \equiv 0 \pmod{60} $ → So $ x = 60k $ for some integer $ k $.\n2. $ x \equiv 1 \pmod{7} $ → Substituting $ x = 60k $, this becomes $ 60k \equiv 1 \pmod{7} $.", "First, simplify $ 60 \mod 7 $:\n$$\n60 \div 7 = 8 \ ext{ remainder } 4 \quad \Rightarrow \quad 60 \equiv 4 \pmod{7}\n$$", "So the congruence becomes:\n$$\n4k \equiv 1 \pmod{7}\n$$", "We now seek the smallest positive integer $ k $ such that $ 4k \equiv 1 \pmod{7} $, meaning $ k $ is the multiplicative inverse of 4 modulo 7.", "---", "### Finding the Multiplicative Inverse of 4 modulo 7", "We test small values of $ k $ such that $ 4k \mod 7 = 1 $:", "- $ k = 1 $: $ 4 \cdot 1 = 4 \equiv 4 \pmod{7} $\n- $ k = 2 $: $ 4 \cdot 2 = 8 \equiv 1 \pmod{7} $ ← Success!", "Thus, $ k \equiv 2 \pmod{7} $, the smallest positive solution is $ k = 2 $.", "So the general solution is $ k = 7m + 2 $ for integer $ m $, and:\n$$\nx = 60k = 60(7m + 2) = 420m + 120\n$$", "Therefore, all solutions are of the form $ x = 420m + 120 $.", "---", "### Finding the Smallest Three-Digit Value of $ x $", "We now find the smallest $ m $ such that $ x \geq 100 $:\n$$\n420m + 120 \geq 100 \Rightarrow 420m \geq -20\n$$", "Since $ m $ is an integer and $ x $ must be a three-digit number, start with $ m = 0 $:\n- $ m = 0 $: $ x = 120 $ → This is a three-digit number.", "Check both conditions:\n- $ 120 \div 7 = 17 $ remainder $ 1 $ → $ 120 \equiv 1 \pmod{7} $ ✅\n- $ 120 \div 60 = 2 $ → $ 120 \equiv 0 \pmod{60} $ ✅", "Thus, $ x = 120 $ satisfies both congruences and is the smallest three-digit number meeting the criteria.", "---", "### Conclusion", "The smallest three-digit number $ x $ such that:\n- $ x \equiv 1 \pmod{7} $\n- $ x \equiv 0 \pmod{60} $", "is $ \boxed{120} $. This problem elegantly combines modular constraints with divisibility rules, and solving it reveals a clean solution rooted in the structure of residues modulo 7 and 60.", "---", "Key Takeaways for Future Solutions:\n- Use modular simplification to reduce complexity.\n- Solve linear congruences using modular inverses.\n- Combine constraints using least common multiples or parametrization.\n- Always verify results against the original conditions.", "This method applies broadly to similar problems in number theory and cryptography."]
