A Challenging Number Theory Proof – May or May Not Rely on Induction?
Prove that, for any positive integer N, there exists some positive integer X such that 2^N evenly divides X, where X is composed of N digits which are each either a 1 or a 2.
For example, if N = 2, X = 12, because 2^2 divides 12 and 12 is a 2-digit number composed of 1’s and 2’s.
If N = 3, X = 112, because 2^3 divides 112 and 112 is a 3-digit number composed of 1’s and 2’s.
It may be the case that this proof relies on an inductive argument, showing that it is possible to proceed from N = k to N = k+1 by adding either 10^(k+1) or 2*10^(k+1).
Alternatively, perhaps there is a way to convert an N-digit number composed of 1’s and 2’s into a binary representation by some systematic method that would make the proof more obvious.
Or perhaps we can combinatorially consider all possible values of such an N-digit number and systematically demonstrate that at least one must be the target value.