When Googol (10
100) is divided by 1,2,3,4,5,6... the remainders form the sequence 0,0,1,0,0,4...
When Googolplex (1010100) is divided by 1,2,3,4,5,6... the remainders form the sequence 0,0,1,0,0,4...
These sequences are not the same, however.
Find the smallest number that gives a different remainder upon dividing these two numbers.
Feel free to find more.
There's a powermod function in Matlab and other computer languages, such as MODPOW in UBASIC, that can be used even with the extended-precision numbers as provided by Matlab's sym type.
for i=1:50
p=powermod(10,100,i);
pp=powermod(sym(10),sym(10)^100,i);
disp([i p pp])
end
but extended precision could be avoided by substituting (((((((((10^(10^10)(^10^10))^(10^10))^(10^10))^(10^10))^(10^10))^(10^10))^(10^10))^(10^10))^(10^10):
ppp=10;
for j=1:10
ppp=powermod(ppp,10^10,i);
end
Either way it provides these sequences:
The first mismatch between the googol sequence and the googolplex sequence is in mod 17:
mod googol googolplex
1 0 0
2 0 0
3 1 1
4 0 0
5 0 0
6 4 4
7 4 4
8 0 0
9 1 1
10 0 0
11 1 1
12 4 4
13 3 3
14 4 4
15 10 10
16 0 0
17 4 1 mismatch
18 10 10
19 9 9
20 0 0
21 4 4
22 12 12
23 13 13
24 16 16
25 0 0
26 16 16
27 10 10
28 4 4
29 16 24 "
30 10 10
31 5 5
32 0 0
33 1 1
34 4 18 "
35 25 25
36 28 28
37 10 10
38 28 28
39 16 16
40 0 0
41 1 1
42 4 4
43 31 24 "
44 12 12
45 10 10
46 36 36
47 27 9 "
48 16 16
49 11 4 "
50 0 0
Are the discrepancies getting closer together?":
[51, 4, 1] "
[52, 16, 16]
[53, 28, 46] "
[54, 10, 10]
[55, 45, 45]
[56, 32, 32]
[57, 28, 28]
[58, 16, 24] "
[59, 16, 48] "
[60, 40, 40]
[61, 47, 47]
[62, 36, 36]
[63, 46, 46]
[64, 0, 0]
[65, 55, 55]
[66, 34, 34]
[67, 10, 10]
[68, 4, 52] "
[69, 13, 13]
[70, 60, 60]
[71, 20, 45] "
[72, 64, 64]
[73, 72, 1] "
[74, 10, 10]
[75, 25, 25]
[76, 28, 28]
[77, 67, 67]
[78, 16, 16]
[79, 67, 52] "
[80, 0, 0]
[81, 10, 10]
[82, 42, 42]
[83, 29, 10] "
[84, 4, 4]
[85, 55, 35] "
[86, 74, 24] "
[87, 16, 82] "
[88, 56, 56]
[89, 16, 16]
[90, 10, 10]
[91, 81, 81]
[92, 36, 36]
[93, 67, 67]
[94, 74, 56] "
[95, 85, 85]
[96, 64, 64]
[97, 9, 35] "
[98, 60, 4] "
[99, 1, 1]
[100, 0, 0]
... or has the frequency plateaued?
Inserting
if p~=pp
if floor(i/100)>floor(prevI/100)
disp([prevI i diffCt])
diffCt=0;
end
diffCt=diffCt+1;
prevI=i;
end
into the loop produces the number of occurrences between "century" changes:
previous curr. count
000 100 21
100 200 42
200 300 49
300 400 55
400 500 57
500 600 63
600 700 63
700 800 63
800 900 65
900 1000 66
1000 1100 69
1100 1200 74
1200 1300 67
1300 1400 73
1400 1500 69
1500 1600 75
1600 1700 72
1700 1800 76
1800 1900 72
1900 2000 74
2000 2100 77
2100 2200 75
2200 2300 77
2300 2400 75
2400 2500 75
2500 2600 77
2600 2700 80
2700 2800 77
2800 2900 76
2900 3000 79
3000 3100 80
3100 3200 79
3200 3300 81
3300 3400 77
3400 3500 82
3500 3600 79
3600 3700 79
3700 3800 79
3800 3900 83
3900 4000 79
4000 4100 77
4100 4200 85
4200 4300 79
4300 4400 80
4400 4500 85
4500 4600 81
4600 4700 82
4700 4800 82
4800 4900 80
4900 5000 80
5000 5100 82
5100 5200 82
5200 5300 82
so the number of mismatches per hundred continues to increase.
Looking up the first few numbers in OEIS:
A066298:
googol:
0, 0, 1, 0, 0, 4, 4, 0, 1, 0, 1, 4, 3, 4, 10, 0, 4, 10, 9, 0, 4, 12, 13, 16, 0, 16, 10, 4, 16, 10,
A067007:
googolplex:
0, 0, 1, 0, 0, 4, 4, 0, 1, 0, 1, 4, 3, 4, 10, 0, 1, 10, 9, 0, 4, 12, 13, 16, 0, 16, 10, 4, 24, 10,
Edited on May 12, 2022, 10:48 am
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Posted by Charlie
on 2022-05-12 10:46:20 |
computer solution
