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.

There is not doubt a more cunning way of solving this than the one I used which was to start with the 'Slide rule' logic. log 2^n = n log(2) for some n, where n on the bottom half of the slide rule will correspond to  rather more than log_10( 2.01320132013) i.e. 0.3038872065080277..(a) and rather less than log_10 (2.01320132014) i.e. 0.3038872065101849....(b).

Now n is some integer such that (a + C) < n Log_10(2) < (b + C): where C is a whole number representing the mantissa and n is the desired power.  But we can use natural logs for a more condensed approach:
(C log(10)+log(201320132013))/(log(2)) < n < (C log(10)+log(201320132014))/(log(2)).

For example, if C=1, then (C log(10)+log(201320132013))/(log(2)) = 40.872628587624206.., and 2^40.872628587624206... = 2.013201320130004... × 10^12. Of course this is neither a round power nor a whole number. Similarly, (log(10)+log(201320132014))/(log(2)) = 40.87262858763137... and 2 raised to this fractional power also is 2.01320132014000... × 10^12.

So far so good. What is needed is a round power somewhere near 37.55070049273684+3.32192809488736x.

I found x=399, 2^1363.0000103527900, which is pretty close, and indeed 2^1363 = 2.0131868734285... × 10^410.

Just a few digits short, but I felt the method was vindicated anyway.

Happy New Year, everyone!

 

Edited on January 2, 2013, 1:43 am

Posted by broll on 2012-12-31 17:16:15