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