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Powers often in other bases. (Posted on 2013-04-07) Difficulty: 3 of 5
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    
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Hints/Tips 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       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 2013-04-07 12:46:19

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