But more precisely: for a given $ B $, if there exists $ D \in [100, 199] $, $ D \equiv 0 \pmod{9} $, and $ 9 \mid D $, such that $ B \mid D $, then $ D = B \cdot k $, so $ B $ must be a divisor of some multiple of 9 in that interval.

Sun, 03 May 2026 05:27:38 +0000

Content under review