If the tetradecimal (base 14) representation of 2^P and 7^P start with the same (non-zero) digit for a positive integer P, what is that digit? Can you prove that it must be so?

The digit is 3.  Which happens when p is an odd multiple of 7.

Perhaps it is easier to understand with regular numbers:
In plain old base 10 we have 2^p and 5^p beginning with 3 if they are equal.

2^p*5^p=10^p
Define {x} as the rational part of x in scientific notation.
(for example {2^15} = 3.2768)

{2^p}*{5^p}=10
{5^p}=10/{2^p}

If {2^p} is in the interval (1,2) then {5^p} is in the interval (1,5)
[note these are open intervals so they cannot both equal 1]
If {2^p} is in the interval (2,3) then {5^p} is in the interval (5,3.333)
If {2^p} is in the interval (3,4) then {5^p} is in the interval (3.333,2.5)
If {2^p} is in the interval (4,5) then {5^p} is in the interval (2.5,2)
If {2^p} is in the interval (5,10) then {5^p} is in the interval (2,1)

The intervals for {2^p} and {5^p} only overlap from 3 to 10/3.
So the first digit is 3.

For the base 14 problem in question we have the same idea {x} would mean the rational part in base 14 scientific notation.
The intervals then overlap from 3.5 to 4.
So the first digit is 3.

To generalize and frame this idea in base n.  We would need n to be composite so n=a*b.  Then a^P and b^P would be only be equal if they both have the first digit of sqrt(n).

Posted by Jer on 2010-11-19 17:26:33