Is there a base for each positive integer where this number can be represented a string of n (base - 1) digit? How are called positive integer numbers that have the following property of being represented as:
$$
N = \sum_{k=0}^n{(B-1)B^k} = (B - 1)\sum_{k=0}^n{B^k}
$$
with $N$ a positive integer number, $B$ the base in which $N$ is represented, $B \lt N$
For example, base 10:
$$
99_{10} = (10-1)10^1 + (10-1)10^0
$$
$$
999_{10} = (10-1)10^2 + (10-1)10^1 + (10-1)10^0
$$
$$
...
$$
base 2:
$$
3_{10} = 11_2 = (2 - 1) 2^1 + (2 - 1) 2^0
$$
$$
7_{10} = 111_2 = (2 - 1) 2^2 + (2 - 1) 2^1 + (2-1)2^0
$$
$$
15_{10} = 1111_2 = (2-1)2^3 + (2-1) 2^2 + (2-1)2^1 + (2-1)2^0
$$
$$
...
$$
base 3:
$$
4_{10} = 22_{3} = (3-1)3^1 + (3 - 1)3^0
$$
$$
26_{10} = 222_{3} = (3-1)3^2 + (3-1)3^1 + (3-1)3^0
$$
$$
..
$$
base 16:
$$
255_{10} = ff_{16} = (16-1)16^1 + (16-1)16^0
$$
$$
4095_{10} = fff_{16} = (16-1)16^2+(16-1)16^1 + (16-1)16^0
$$
$$
...
$$
In this other hand, does a base $B$ exist for any positive integer $N$ where it can be written
$$
N = \sum_0^n{(B - 1)B^n}
$$
With the additional condition that $B \lt N$

When it is $1$'s repeated, it is called a repunit.  When it is $n$ repeated, it is called a repdigit.
We want $n=(b-1)\sum_{k=0}^{m-1}b^k=b^m-1 $
So whenever $n+1$ is an $m^{th}$ power, $n$ may be expressed as $m\ $ $b-1$'s
