<s>[INST] Integer $N$, in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b<16$, then what is the successor of $b$? 
An integer $N$, expressed in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b$ is less than $16$, then what is the successor of $b$?

I couldn't develop anything more than 
$$N_b = \left(\;n\cdot (b-1)\;\right)_b = 6789_b$$ 
(with subscripts indicating "in base $b$").
 [/INST] It seems there are multiple answers.
Since $N$ can be written as $6789$ in base $b$ we know $$6b^3+7b^2+8b+9=N$$
If we divide this polynomial by $b-1$ we get $$6b^2+13b+21+\frac{30}{b-1}$$ but we know that $b-1$ divides $N$ so $\frac{30}{b-1}$ is an integer. $$b-1=1,2,3,5,6,10,15,30$$ $$b=2,3,4,5,7,11,16,31$$ but $b < 16$ and $b>9$ as noted by John Omielan hence $$b+1=12$$
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