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.

February 22, 2026

$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.$

$ D = B \cdot k $ for some integer $ k \geq 1 $ (full utilization of batches, but $ k $ can vary)
But the key is: $ B $ must divide some multiple of 9 in the range [100, 199], i.e., $ B \mid D $ for some $ D \in [100,199] \cap 9\mathbb{Z} $
But since $ D $ must be divisible by 9, and $ B $ is a multiple of 11, we seek the smallest $ B \equiv 0 \pmod{11} $ such that $ B $ divides some multiple of 9 in [100, 199]
But to allow full batch deployment, $ B $ must *divide* some $ D $ divisible by 9 in [100,199].
So the condition is: $ B $ divides some $ D $ with $ D \in [100,199] $, $ 9 \mid D $
But since $ D $ must be divisible by 9, and $ B \mid D $, $ B $ must be such that it divides at least one multiple of 9 in the range.

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.

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