Construct a positive decimal number of length $n$ or less that is divisible by $2^n-1$ for n in $Z^+_0$

Question

I got this question on a discrete mathematics test and I no idea on how to solve it:

Create a method for constructing a positive decimal number of length $n$ or less with digits from set $\{0,1,8,9\}$ that is divisible by $2^n-1$, where n is in the nonnegative integers.

I would appreciate any help. I spent so long thinking about it and nothing clicks.

Answer 1

We consider two points:

First:Note that the sum of given digits is $18$, that indicates the required number is divisible by $9$.

Second: Any number of form $(d_1d_2d_3...0d_1d_2d_3)$ is divisible by 11 iff number of digits $d_1, d_2, d_3 . . .$ is even, for example $23023=11\times2093$ or $472904729=11\times42991339$ etc.

Considering these points, we suppose $n=11$; with a four digit number like $abcd$; which is not divisible by 11 by given digits (i.e. 0,1,8 and 9), for $a=8$ , $b=0$ and $c=9$ we get:

$abcd=8091$

The required number must also be divisible by $11$, so the solution can be $809108091$ and we have:

$2^{11-1}-1=2^{10}-1=1023$

And:

$809108091=1023\times790917$

Numbers $8091080910$ and $80910809100$ can also be solutions. Other arrangements of digits does not give the solution.