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 $b$").
On
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 $$b+1=12$$
If $N$ is a multiple of $b-1$ then by the rule of nines[1] the sum of the digits is a multiple of $b-1$ as well.
So $b-1| 6+ 7 + 8 + 9 = 30$. So $ b-1 = 1,2,3,5,6,10,15,30$
Bu $9$ is a digit so $b\ge 10$. And $b < 16$ so $b-1 < 15$.
... so $b-1 < 15$ and so $b-1 =10$ and $b =11$ and then successor of $b$ is $12$.
[1] Rule of nines: $\sum_{k=0}^n a_k b^k \equiv \sum_{k=0}^n a_k \pmod {(b-1)}$.
Pf: $\sum_{k=0}^n a_k b^k - \sum_{k=0}^n a_k = \sum_{k= 0}^n a_k (b^k-1) $
Now each $b^0 - 1 = 0$ and $b^1 - 1 =b-1$ and for $k > 1$ then $b^k -1 = (b-1)(b^{k-1} + b^{k-1} + ...... + 1)$ so $b-1|b^k-1$. So $b-1|\sum_{k= 0}^n a_k (b^k-1)$
And so $\sum_{k=0}^n a_k b^k - \sum_{k=0}^n a_k\equiv 0 \pmod {(b-1)}$.
In other words, $N =\sum_{k=0}^n a_k b^k$ and the sum of the digits of $N$ (in base $b$) have the same remainder when divided by $b-1$.
And therefor $b-1|N$ if and only if $b-1|$ the sum of the digits if $N$ (in base $b$)