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infinity bay resort west bay roatan - All Square Golf

📅 May 5, 2026 👤 scraface

May 05, 2026

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📌 But every positive integer divides some multiple of 9 (e.g., $ D = B \cdot 9 $ only if $ B \mid D $, so if we set $ D = B \cdot k $, we need $ 9 \mid Bk $. But we can choose $ k $).

📌 To allow $ B $ to divide a number $ D \in [100,199] \cap 9\mathbb{Z} $, it suffices that $ B $ divides at least one such $ D $.

📌 So we seek the smallest multiple of 11 that divides at least one multiple of 9 in [100,199].

📌 List multiples of 9 in range: 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, 198

📌 Now divide each by 11 and see if integer:

📌 Ã· 11 = 9.818 â not

📌 Ã· 11 â 10.636 â no

📌 So only 198 is divisible by 11.

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