Let's rather prove, by induction, that for every n there exists a n-digit number multiple of 2^n, written only with sixes and sevens.Let's call such a number M(n). Obviously, M(1)=6. [You could also start with M(0)=0, but it's less clear.] We'll prove that if M(n)/2^n is even, then M(n+1) equals M(n) with a 6 appended at the left, and when M(n)/2^n is odd, then M(n+1) equals M(n) with a 7 at the left.
First case: M(n)/2^n=2K. M(n+1)= 6x10^n+M(n)= 6x10^n+2Kx2^n, which is a multiple of 2^(n+1).
Second case: M(n)/2^n=2K+1. M(n+1)= 7x10^n+M(n)= 7x10^n+2Kx2^n+2^n= 7x5^nx2^n+Kx2^(n+1)+2^n= (7x5^n+1)x2^n+Kx2^(n+1), which also is a multiple of 2^(n+1); note that the first term in parenthesis comes out to be even.
Thus, we've proved that M(n) exists for all n>0. It's also easy to prove that M(n) is unique, realizing that if you take out the leftmost digit, you are left with a possible M(n-1), and so all the way down to M(1), which is certainly unique. The following Python program does the needed calculations:
m = m + 6 * 10L ** i
m = m + 7 * 10L ** i
i = i + 1
This program produces: