For any positive integer $n$ there exists an $n$-digit number divisible by $2^n$ and containing only the digits $2$ and $3$

Question

Prove that for any positive integer $n$ there exists an $n$-digit number divisible by $2^n$ and containing only the digits $2$ and $3$

Need help solving by induction

Answer 1

For a number $a$ to be divisible by $2^n$, you need the last $n$ digits to form a number divisible by $2^n$. So inductively assume you have a number $a_n$, of $n$ digits, all equal to $2$ or $3$, divisible by $2^n$. Then take $a_{n+1}$ to be either $2\times 10^n+a_n$ or $3\times 10^n+a_n$. Both of these are divisible by $2^n$. All that remains for you to prove is that one is divisible by $2^{n+1}$.