Does B appear as the leftmost digit in the duodecimal (base 12) representation of any power of 2?
Does 9 appear as the leftmost digit in the duodecimal representation of any power of (37)12?
Is it possible to find a power of any counting number that has a given digit as its leftmost digit in the duodecimal system?
Bonus: What percentage of the powers of 2 in duodecimal system have 1 as their leftmost digit?
Note: In finding the powers of "any counting number," exclude powers of (10)12, whose leftmost digit is always 1.
pt1: 2^25 begins with B in base 12 as does 2^43.
pt2: 37b12 = 43 dec raised to the 31 (dec) power begins with 9.
pt3: From the second table below, any digit other than zero can be the leading digit in some power of 43 dec.
bonus: Examining the first 10000 powers of 2 confirms the suspicion that this percentage would be the base-12 log of 2, in accordance witn Benford's law as applied to base 12.
clearvars,clc
n=1;
for p=1:50
n=2*n;
fprintf('%3d %s
',p,dec2base(n,12))
end
disp(' ')
% pt 2
b=sym(43);
for p=2:50
n=b^p;
pos=[];
while n>0
pos(end+1)=mod(n,12);
n=floor(n/12);
end
pos=flip(pos);
disp([p pos(1:min(10,end))])
end
1 2
2 4
3 8
4 14
5 28
6 54
7 A8
8 194
9 368
10 714
11 1228
12 2454
13 48A8
14 9594
15 16B68
16 31B14
17 63A28
18 107854
19 2134A8
20 426994
21 851768
22 14A3314
23 2986628
24 5751054
25 B2A20A8
26 1A584194
27 38B48368
28 75A94714
29 12B969228
30 25B716454
31 4BB2308A8
32 9BA461594
33 17B8902B68
34 33B5605B14
35 67AB00BA28
36 1139A01B854
37 2277803B4A8
38 4533407A994
39 8A668139768
40 159114277314
41 2B6228532628
42 5B0454A65054
43 BA08A990A0A8
44 1B81597618194
45 3B42B73034368
46 7A85B26068714
47 1394BA50115228
48 2769B8A022A454
49 5317B5804588A8
50 A633AB408B5594
First 10 digits of 43(dec) to the n power:
n
2 1 0 10 1
3 3 10 0 1 7
4 1 1 8 10 5 8 1
5 4 1 2 9 6 3 11 7
6 1 2 8 5 0 1 8 2 6 1
7 4 4 8 1 11 6 0 4 11 9
8 1 3 8 9 3 0 2 7 5 10
9 4 8 4 5 1 9 9 4 9 11
10 1 4 9 11 10 5 6 0 8 3
11 5 0 3 9 6 5 8 8 5 9
12 1 6 0 1 7 2 2 6 2 4
13 5 4 6 5 8 8 11 0 2 7
14 1 7 3 3 2 6 3 11 5 9
15 5 9 0 8 6 0 8 2 1 8
16 1 8 7 5 6 5 8 5 3 8
17 6 1 10 8 10 2 5 3 0 1
18 1 10 0 9 5 8 6 8 9 9
19 6 7 0 9 11 5 8 1 7 1
20 1 11 7 3 11 8 1 4 1 8
21 7 0 7 3 2 10 0 9 10 1
22 2 1 3 2 0 7 2 0 11 3
23 7 6 6 4 4 1 8 5 4 4
24 2 3 0 4 9 6 10 1 3 2
25 8 0 10 5 2 3 6 2 6 6
26 2 4 11 1 4 7 2 7 3 1
27 8 7 7 9 11 5 10 4 0 2
28 2 6 11 5 0 8 2 0 0 4
29 9 2 10 11 1 5 3 2 1 4
30 2 9 1 5 1 10 1 10 4 7
31 9 10 8 1 5 7 4 8 2 5
32 2 11 5 3 1 3 1 5 9 4
33 10 6 11 10 1 6 2 3 8 7
34 3 1 11 0 5 3 5 2 3 4
35 11 3 10 6 6 11 3 7 1 11
36 3 4 6 10 9 6 10 5 10 7
37 1 0 1 4 8 8 3 7 7 1
38 3 7 4 11 11 1 9 0 2 5
39 1 0 11 6 10 8 11 3 3 8
40 3 10 5 5 8 6 0 4 10 4
41 1 1 10 5 7 5 5 7 5 5
42 4 1 8 6 1 8 7 1 8 5
43 1 2 10 1 6 0 1 9 7 1
44 4 5 2 3 4 6 6 5 4 5
45 1 3 10 7 2 1 3 5 1 3
46 4 8 10 11 8 6 7 3 3 5
47 1 4 11 11 3 11 7 8 0 9
48 5 0 10 9 7 2 8 5 10 9
49 1 6 2 2 8 4 10 8 5 1
50 5 5 1 11 8 1 6 4 2 5
>>
Edited on September 23, 2024, 9:20 am
|
|
Posted by Charlie
on 2024-09-23 09:18:30 |
solution
