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Limited-Time Offer: Scale-In One PS Game Hunt Only at PS Game Shop! - Crankk.io

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 But let's double-check by checking smaller common differences.

📌 Alternatively, suppose the condition is that $ x \equiv -3 \pmod{7,8,9} $, i.e., $ x + 3 \equiv 0 \pmod{504} $.

📌 Then $ x \equiv -3 \pmod{504} $, so $ x = 504k - 3 $.

📌 For $ x $ to be two-digit:

📌 \leq 504k - 3 \leq 99 \Rightarrow 13 \leq 504k \leq 102 \Rightarrow k = 0 \text{ or } 1

📌 Only $ k = 0 $ gives $ x = -3 $, $ k = 1 $ gives $ x = 501 $.

📌 No two-digit solution.

📌 But perhaps we made a misinterpretation. Let’s suppose the question means a number that is **three less than a common multiple** of $ 7, 8, 9 $, not necessarily the LCM.

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