Prove that for any integer $n>0$, there exists a number consisting of 1's and 2's only, which is divisible by $2^n$.

Question

I really have no idea where to start, since I don't know what to apply here.

Answer 1

Hint: If you don't know where else to start, start with $n=1$. Can you find one? Then try $n=2$ and up to $n=5$. You should be able to find an $n$ digit one for $2^n$. You also should see a pattern in the numbers you find. Now prove the pattern works, perhaps by induction.

Answer 2

Suppose you have an $n$ digit number $d_n$ that is divisible by $2^n$. Since $10^n=2^n5^n$, if $d_n/2^n$ is even, then $d_n+2\cdot 10^n$ is divisible by $2^{n+1}$; if $d_n/2^n$ is odd, then $d_n+10^n$ is divisible by $2^{n+1}$, since $d_n/2^n$ and $10^n/2^n=5^n$ are both odd so their sum is even.

This, combined with the base $n=1$, proves it by induction.

Note that this works for any two digits one of which is even and the other is odd. In particular, any two consecutive digits will work.

Additional questions:

How many different digits do you have to choose from to make the sun always divisible by $5^n$?

What happens in bases other than 10?