If $a \equiv b \pmod n \: \:$ and $\: \: a \equiv 0 \pmod n \: \:, \: b \equiv 0 \pmod n$
Is it implied that $a \equiv 0 \pmod b$ ? i.e if $b \leq a$
The specific context of this question is that of the multiples of 9. That is we know that for any multiple of 9, not only is the number divisible by 9, but also the sum of its digits. So is it true that the number is divisible by the sum of the digits?
Update Ok so I found a counterexample for the generalization of my question, but is the part about multiples of 9 true?
No, it is not true that $$a\equiv b\equiv 0\bmod n\implies a\equiv 0\bmod b$$ For example, we can look at $n=2$, $a=6$, and $b=4$.
For your other question: no, just because $$n\equiv 0\bmod 9\implies \operatorname{digit\_sum}(n)\equiv 0\bmod 9$$ does not mean that $$n\equiv 0\bmod 9\implies n\equiv 0\bmod \operatorname{digit\_sum}(n)$$ For example, consider $n=99$. Then $\operatorname{digit\_sum}(99)=18$ but $99\not\equiv 0\bmod 18$.