Create a table of powers of 10 in binary starting with 10
^{1} = 1010
_{2} then create a similar table in base 5 starting with 10
^{1} = 20
_{5}.
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...
(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 base2 logs or base5 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 1
4 3
7 5
9 6
12 8
14 10
16 11
19 13
21 15
24 17
26 18
28 20
31 22
33 23
36 25
38 27
41 29
43 30
45 32
48 34
50 35
53 37
55 39
57 40
60 42
62 44
65 46
67 47
70 49
72 51
74 52
77 54
79 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 20130407 12:46:19 