1. Start with the idea that this has already been checked up to 2^n, where n=4.6*10^7. (see http://mathworld.wolfram.com/Zero.html )
2. Based on something like 'old-fogey logic' in 'Start as you wish', assume the digits of 2^n are randomly distributed from 0-9. Actually the distribution don't even need to be completely random, as long as 0 has some moderate finite chance of occurring.
3. 2^n where n=4.6*10^7 has over 13 million digits. Assuming a probability of 9/10 for the first digit (not being zero)*9/10 for the second, etc. gives an outcome of about 10^(-595,000), which is really unlikely.
Edited on August 18, 2012, 1:31 pm
Posted by broll
on 2012-08-18 12:38:47