Every integer $n > 1$ has a multiple less than $n^4$ whose decimal expansion has at most four distinct digits

Question

Prove that every integer $n > 1$ has a multiple less than $n^4$ whose decimal expansion has at most four distinct digits.

Proof：Choose $k$ such that $ 2^ {k-1} $$ \leqslant $ n < $ 2^ {k} $ .The result is easy to check when $k \leqslant 5 $. So assume that $k \geqslant 6$. There are $ 2^ {k} $ > n nonnegative numbers less than $10^k$ and having only digits $0$ and $1$. Two of them must give the same remainder when divided by $n$, hence their difference is a number with digits $0,1,8$ or $9$, which is less than $10^k < 16^{k-1} \leqslant n^4 $. (the inequality $10^k < 16^{k-1}$ is equivalent to $1.6^k>16$ and holds since $1.6^6>16$ and $k \geqslant 6$)

From Number Theory: Concepts And Problems by Titu Andreescu exercise 2.93

My question: How $10^k$ comes and how the inequality $10^k < 16^{k-1}$ is equivalent to $1.6^k>16$ for $k \geqslant 6$? Can anyone explain this proof further?

Thanks in advance.

Answer 1

$$10^k< 16^{k-1} $$ is equal to $$10^{k} < \frac{16^{k}}{16}$$ Dividing by $10^{k}$ and multiplying by 16 gives $$16 < \frac{16^{k}}{10^{k}}$$ or $$16 < 1.6^{k}$$

This is only true for $k>6$ (just put in the numbers). I hope that helps :)