Start as you wish had you prove that for any kdigit number M there exists a number n such that the string of first k digits of 2
^{n} equals M.
Find a power of 2 whose decimal expansion begins with the 12digit 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
'Slide Rule' Logic by broll)
Of course 2^1363 (integer power) begins "20131868734285".
It's unfortunate the original formulation mentioned a number n, but the idea is for an integer, otherwise we could use just 2^log2(201320132013) where log2(201320132013) ~= 37.5507004927368408372497476346649

Posted by Charlie
on 20130101 03:30:07 