Proving finite bases for a Harshad number?

Question

I'm having some trouble with harshad numbers. How do I prove that the number 136 is Harshad only for the bases 2,3,4,5 and 9 and every multiple of 136 thereafter?

Answer 1

Long solution:

If $n \geq 12$ is any basis, then $$136=a_0+a_1n$$ and $$a_0+a_1| 136$$

Now for each $d |136$, meaning $d \in \{1, 2, 4, 8, 17, 34, 68, 136 \}$ you can simply solve the system of equations $$a_0+a_1n=136 \\ a_0+a_1=d$$ by observing that $$a_1(n-1)=136-d$$ leads to finitely many factorisations. You can eliminate many of them by observing that $$136=a_0+a_1n \geq a_1 \cdot 12 \Rightarrow a_1 \leq 11$$

The cases $n \leq 11$ can be studied by simply writing the number $136$ out in each of these basis.