(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.

(In reply to re: exploration turned up something strange by Charlie)

So much for a trusty calculator. 

It turns out I encountered a flaw in the way it finds the intersection of two functions.  The first-two-digits function has discontinuities similar to a greatest integer function.   P-1 passes through most of these gaps.  Whatever variation of Newtons method it uses, sometimes things just go wrong in these cases.  Instead of reporting no solution it reported a solution that isnt even very close.

Yesterday, I though I was going to get solutions like 2^13.55075 = 12000.026738460294.  I will leave it at this.  This could make a nice extension of the problem.

Posted by Jer on 2011-03-17 13:58:24