Start as you wish had you prove that for any k-digit number M there exists a number n such that the string of first k digits of 2ⁿ equals M.

Find a power of 2 whose decimal expansion begins with the 12-digit string "201320132013". It need not be the smallest such number.

Bonus: Find the smallest such number and prove it to be the smallest.

(In reply to re: 'Slide Rule' Logic by Charlie)

I agree.

Actually I did not repeat in my suggested approach that part of the analysis which explained why I did not seek a closer answer, and perhaps I should have done.

As I mentioned, the log equivalents of the two numbers are rather close. For example, 2^40.87262858762420... and 2^40.87262858763137... differ only 10 digits after the decimal point. Ideally one would look for a value where 37.55070049273684... plus a multiple of 3.32192809488736..., and 37.55070049274400... plus a multiple of 3.32192809488736..., reconverted into conventional numbers,  straddled an integral power of 2.

Equivalently, one is looking for multiples of 3.32192809488736... whose fractional value (numbers after the decimal point) approximates to some integer, N, plus somewhere between 0.4492995072506..., and 0.4492995072631... The problem here is that WolframAlpha wasn't really interested in dealing with such inequalities as I was able to frame along these lines, while Excel didn't have the required accuracy.

No doubt a custom program would supply an answer, but all rules and no mercy isn't really my cup of tea these days.

 

 

 

Edited on January 2, 2013, 2:44 am

Posted by broll on 2013-01-02 01:48:36