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(2): 'Slide Rule' Logic by broll)

Yes, logarithms are needed for the terribly huge numbers involved, as well as extended precision arithmetic.

But it could be handled manually in a few steps given an available log function in extended precision calculator software.

Edited on January 2, 2013, 12:10 pm

Posted by Charlie on 2013-01-02 12:09:36