(A) For a base ten positive integer P drawn at random between 10 and 99 inclusively, determine the probability that the first two digits (reading left to right) in the base ten expansion of 2^P is equal to P-1.

(B) For a base ten positive integer P drawn at random between 10 and 99 inclusively, determine the probability that the first two digits (reading left to right) in the base ten expansion of 6^P is equal to P-1.

This is an interesting problem.  Its too bad I have to idea how to solve it analytically.

I just made a table on my trusty graphing calculator:
Y1=X-1
Y2=int(10*(2^X)/10^(int(log(2^x))))
and found the same solutions.

It does make an interesting graph that implies 25 solutions total.
(of course only 3 of them are integers: 21, 35, 76)

The smallest I find at about about 19.93156857
The strange thing is 2^19.93156857 = 1000000

I am not sure what is going on here.  I will explore some more.

Posted by Jer on 2011-03-16 14:14:58