We are looking for a two-digit integer $ x $ such that $ x + 3 $ is divisible by $ 7 $, $ 8 $, and $ 9 $.

["Finding the Smallest Two-Digit Integer $ x $ Such That $ x + 3 $ Is Divisible by 7, 8, and 9", "Looking for a two-digit integer $ x $ that satisfies a special divisibility condition? Our latest mathematical search focuses on identifying the smallest two-digit number $ x $ for which $ x + 3 $ is divisible by 7, 8, and 9 simultaneously.", "---", "### What Does It Mean for $ x + 3 $ to Be Divisible by 7, 8, and 9?", "To solve this problem, we first recognize that if $ x + 3 $ is divisible by 7, 8, and 9, then $ x + 3 $ must be a common multiple of these three numbers. Since we are seeking the smallest two-digit $ x $, we focus on the least common multiple (LCM) of 7, 8, and 9.", "---", "### Step 1: Compute the Least Common Multiple (LCM)", "Find $ \ ext{LCM}(7, 8, 9) $:", "- Prime factorization:\n - $ 7 = 7 $\n - $ 8 = 2^3 $\n - $ 9 = 3^2 $", "The LCM takes the highest power of each prime factor:\n$$\n\ ext{LCM} = 2^3 \cdot 3^2 \cdot 7 = 8 \cdot 9 \cdot 7 = 504\n$$", "So, $ x + 3 $ must be a multiple of 504.", "---", "### Step 2: Find the Smallest $ x $ Such That $ x + 3 = 504k $", "We are looking for the smallest two-digit integer $ x $, so we test small integer values of $ k $:", "- For $ k = 1 $: $ x + 3 = 504 \Rightarrow x = 501 $ → Too large (three digits)\n- For $ k = 0 $: $ x + 3 = 0 \Rightarrow x = -3 $ → Not a positive two-digit integer", "Since 504 is already much larger than 99 (the largest two-digit number), 504k for any positive integer $ k $ will exceed two-digit values.", "Wait — this indicates a crucial insight.", "If $ x + 3 $ must be divisible by 7, 8, and 9, and the LCM is 504, then any such $ x + 3 $ must be 504, 1008, etc. But 504 > 105 (3 × 99), so no two-digit $ x $ satisfies $ x + 3 $ divisible by all three?", "But this contradicts the assumption of existence.", "Let’s double-check: Are there any two-digit $ x $ such that $ x + 3 $ is divisible by 7, 8, and 9?", "Since LCM = 504, the smallest such $ x + 3 $ is 504, which gives $ x = 501 $, a three-digit number.", "Hence, there is no two-digit integer $ x $ such that $ x + 3 $ is divisible by 7, 8, and 9.", "---", "### Clarifying the Challenge: Are We Solving a Contradiction?", "Yes — this prompts a thoughtful refinement: Perhaps the problem intends $ x + 3 $ to be divisible by at least one of the numbers, but the original says “divisible by 7, 8, and 9” — meaning all three simultaneously.", "Therefore, based on LCM calculations and digit constraints:", "> There is no two-digit integer $ x $ such that $ x + 3 $ is divisible by 7, 8, and 9.", "---", "### But What If We Relax or Re-Interpret?", "Suppose the problem seeks the smallest two-digit $ x $ where $ x + 3 $ is divisible by any subset — but the wording clearly uses “divisible by 7, 8, and 9”, i.e., the full set.", "Alternatively, perhaps the condition is that $ x + 3 $ is divisible by each of 7, 8, and 9 individually, which still requires divisibility by LCM(7,8,9) = 504.", "Still, 504 is beyond 105.", "---", "### Conclusion: No Solution Exists Among Two-Digit Integers", "Thus, the correct conclusion is:", "> There is no two-digit integer $ x $ such that $ x + 3 $ is divisible by 7, 8, and 9 simultaneously.", "This makes for a fascinating mathematical exercise — revealing constraints imposed by LCM and digit limits.", "---", "### Bonus Tip: When Does $ x + 3 $ Work for Two-Digit $ x $?", "We can reframe: Find smallest two-digit $ x $ such that $ x + 3 \leq 99 $. So:\n$$\nx \leq 96 \Rightarrow x + 3 \leq 99\n$$\nThus, $ x + 3 \leq 99 $, so $ x + 3 $ must be a multiple of 7, 8, and 9 and ≤ 99.", "But the smallest such multiple is 504 > 99 → impossible.", "So final answer:", "> \boxed{\ ext{No two-digit integer } x \ ext{ satisfies the condition.}}", "---", "### For Deeper Exploration", "Try similar problems:\n- Find $ x $ such that $ x + 2 $ divisible by 3 and 5 → $ x + 2 = 15k $, smallest $ x = 13 $\n- Or explore smallest $ x $ where $ x + 3 $ divisible by 7 and 8 — LCM(7,8)=56 → $ x + 3 = 56 \Rightarrow x = 53 $, a valid two-digit number.", "But for 7, 8, 9? No two-digit solution.", "Use this insight to refine wording in real-world problems — precision in divisibility claims avoids confusion.", "---", "#### Key Takeaway:\nWhile divisibility challenges are intellectually rewarding, not all modular conditions admit solutions within limited ranges — and 7, 8, 9’s LCM exceeds the two-digit threshold. Always verify bounds!", "---", "Tags: number theory, LCM, divisibility, math puzzle, two-digit numbers, problem-solving, integer x, competitive math", "Volume: Intermediate Math | LCM & GCD\nDifficulty: High School to Olympiad prep\nSearch Intent: Finding $ x $ satisfying multiple divisibility tests, LCM application, real-world math challenge"]