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 Powers often in other bases. (Posted on 2013-04-07)
Create a table of powers of 10 in binary starting with 101 = 10102 then create a similar table in base 5 starting with 101 = 205.

If you look at the lengths of the numbers in the two tables combined, prove there is exactly one each of length 2, 3, 4...

 No Solution Yet Submitted by Jer No Rating

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 further hint | Comment 2 of 3 |
(In reply to Tables: no proof except for 2,3,4,...,67 by Charlie)

Let a = log2(10) ~=  3.321928094887

But log2(10) = log2(5) + 1

Similarly b = log5(10) = log5(2) + 1

But also, log2(5) = 1/log5(2).

So when we're taking logs of the powers of 10, using base-2 logs or base-5 logs, we're taking multiples of a + 1 or 1/a + 1, or looked at the other way b + 1 or 1/b + 1.

For curiosity's sake I tried using multiples of sqrt(2) + 1 and 1/sqrt(2) + 1 just to see if this is a property of all such reciprocal formulations. The results of the floor function being applied to multiples of these:

`2       14       37       59       612      814      1016      1119      1321      1524      1726      1828      2031      2233      2336      2538      2741      2943      3045      3248      3450      3553      3755      3957      4060      4262      4465      4667      4770      4972      5174      5277      5479      56`

5   point 10
10   X=sqrt(2)
20   Xr=1/X+1:X=X+1
30   for N=1 to 33
40      print int(N*X),int(N*Xr)
50   next

Indeed the puzzle proof seems based on the fact that log2(5) = 1/log5(2), as such reciprocal relations do not depend on the logarithmic nature of the reciprocals involved.

As the program listing shows, these are the integer part (floor function) of the muliples of these numbers.

Edited on April 7, 2013, 1:57 pm
 Posted by Charlie on 2013-04-07 12:46:19

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