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?

1. Working in base ab throughout and ignoring n=0;   
2. Let b be the larger of a,b with a,b, relatively prime.   
3. First multiply b^n by a^n, less the first digit of a^n, x.   
4. Next multiply every digit of b^n except the first, again x, by x.   
5. The sum of 3 and 4 is C followed by a lot of zeros:   
   
C(ab)^(n-1) =(ab)^n-x^2(ab)^(n-1)   
6. Now we have x^2+C=10=ab   
7. The first digit of a^n is some power of (ab), say x(ab)^k; the first digit of b is similarly x(ab)^l   
8. Since a^n and b^n start with the same digit, we can also say that   
b^n/a^n is (roughly) some power of (ab), to be precise   
(ab)^l/(ab)^k or (ab)^(l-k); so x approximates to (ab)^.5)   
and C is less than x^2 since (x+1)^2>2x^2 only for x = {0,1,2}   
   
10. Let a = 2, b=7; we seek some x such that x^2+C = 10, giving {(1,D)(4,A)(9,5)}   
11. C is less than x^2 iff x= 3.   
12. The above doesn't guarantee solutions, it just states what form they will take if they exist.   
13. However, 2^17= 3c8,3a8, 7^17 = 3,839,d52,0c8,c06,1a7 is a possible solution.

A truly remarkable result - wholly counterintuitive.

Edited on November 20, 2010, 2:34 pm

Posted by broll on 2010-11-20 11:21:37