All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Duplicate Digit Determination (Posted on 2006-12-03)
If 2^P and 5^P start with the same (non-zero) digit for positive integer P, what is that digit? Can you prove it must be so?

 See The Solution Submitted by Old Original Oskar! No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 1 of 6

Using common logs:

log(2)+log(5) = log(10) = 1, which is an integer.

so p*log(2) + p*log(5) will also be an integer, being log(10^p).

So the two mantissas add up to 1. If one of the mantissas is low and the other high, the two numbers will not begin with the same digit, so the mantissas have to be close enough to .5 to result in the same opening digit of the antilogs.

If both mantissas were .5, the opening digit would be 3, as the sequence of digits would be for sqrt(10).  As a potential mantissa for one of the numbers gets larger, the other gets smaller.  At some point (of varying the possible mantissa), one of them (first digit of antilog) changes to a digit other than 3.  Either at the same point or later, the other changes to a digit other than 3, but in the opposite direction. In either case the match is broken as soon as one or both do not begin with 3.

So 3 can be the only starting digit if the starting digit is shared by the two numbers.

 Posted by Charlie on 2006-12-03 11:05:20

 Search: Search body:
Forums (1)