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...
This does not provide a general proof, but obviously proves for lengths 2,3,4,...,67.
base 10 base 2 base 5
10 1010 20
100 1100100 400
1000 1111101000 13000
10000 10011100010000 310000
100000 11000011010100000 11200000
1000000 11110100001001000000 224000000
10000000 100110001001011010000000 10030000000
100000000 101111101011110000100000000 201100000000
1000000000 111011100110101100101000000000 4022000000000
10000000000 1001010100000010111110010000000000 130440000000000
100000000000 1011101001000011101101110100000000000 3114300000000000
1000000000000 1110100011010100101001010001000000000000 112341000000000000
Lengths: length
representational length missed by
base 2 base 5 base 5
10 4 2 1
100 7 3
1000 10 5 4
10000 14 6
100000 17 8 7
1000000 20 9
10000000 24 11 10
100000000 27 12
1000000000 30 13
10000000000 34 15 14
100000000000 37 16
1000000000000 40 18 17
10000000000000 44 19
100000000000000 47 21 20
1000000000000000 50 22
10000000000000000 54 23
100000000000000000 57 25 24
1000000000000000000 60 26
10000000000000000000 64 28 27
100000000000000000000 67 29
1000000000000000000000 70 31 30
10000000000000000000000 74 32
100000000000000000000000 77 33
1000000000000000000000000 80 35 34
10000000000000000000000000 84 36
100000000000000000000000000 87 38 37
1000000000000000000000000000 90 39
10000000000000000000000000000 94 41 40
100000000000000000000000000000 97 42
1000000000000000000000000000000 100 43
10000000000000000000000000000000 103 45 44
100000000000000000000000000000000 107 46
1000000000000000000000000000000000 110 48 47
10000000000000000000000000000000000 113 49
100000000000000000000000000000000000 117 51 50
1000000000000000000000000000000000000 120 52
10000000000000000000000000000000000000 123 53
100000000000000000000000000000000000000 127 55 54
1000000000000000000000000000000000000000 130 56
10000000000000000000000000000000000000000 133 58 57
100000000000000000000000000000000000000000 137 59
1000000000000000000000000000000000000000000 140 61 60
10000000000000000000000000000000000000000000 143 62
100000000000000000000000000000000000000000000 147 63
1000000000000000000000000000000000000000000000 150 65 64
10000000000000000000000000000000000000000000000 153 66
100000000000000000000000000000000000000000000000 157 68 67
So up through 67, the lengths missed by the base-5 representations match those found in the base-2 representations.
10 cls
20
30 Pw=10
40 while Pw<1000000000000000000000000000000000000000000000000
45 Pl=L:L=len(fnCv$(Pw,5))
50 print Pw;tab(50);len(fnCv$(Pw,2));tab(60);L;
55 if L>Pl+2 then print "*";
56 if L>Pl+1 then print Pl+1:else print
60 Pw=Pw*10
70 wend
80 end
90 fnCv$(N,B)
100 R=N:V$=""
110 while R>0
120 Q=int(R/B):R=R-Q*B
130 V$=cutspc(str(R))+V$
140 R=Q
150 wend
160 return(V$)
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Posted by Charlie
on 2013-04-07 12:18:07 |
Tables: no proof except for 2,3,4,...,67
