Given that:
- A = 20232023
- B = sod(A)
- C = sod(B)
- D = sod(C)
Determine the value of
sod(D).
*** sod(x) denotes the sum of the digits of x.
For example: sod(38) = 3+8 = 11, and:sod(456) = 4+5+6 = 15
A = 2023^2023 is a 6689-digit number whose SOD is B = 30112.
That number's SOD is C = 7, which is already the digital root, so all further SODs remain 7, including D and SOD(D).
2023^2023 itself begins and ends this way:
1070707501339186485735906879443009775934515847505507436018532469684060484156622556457940618910046467
8304964606573388540129707028992875555668260045164201337947280471840711425484924104433016597154397436
2509619323608785152917779106088893982817212798718791957492336002333843267464768197005709952098633887
7776338279883387973701425369925459020411156250094194342515102162425104062929276329283880169188856404
0797531949249309696354735538708974766820381439310315732534339417331276457587304837036534719908204123
1475144327802290562525190329849102356561574498820213104727353018617519563164577093222874376050220765
. . .
6926617858337700321423446663548619796775903550730450291136525602901827729508470411962188045505294781
5561141941675315913977676493989085320754131372331816044886886805470037608384625712281888017417593895
6219125701092317566294104213614636298416918961851650427416335112019813272454543994057050522582438209
4110162514167600799908398986853750794197691447418194835402911822508534083176049177518043842613662817
4401924716636345245067611820413195683530161798856044103590452097729834905959718383942585076543511546
6752952804624316978136555655093707671592357331029387377373532624717724884848522267343889096826628906
5592590783310745734229998946030727992108471002229683507721491555270043497160122215953230856542098038
01090783071940824076100007700961621265485380856743026451816242152956789327206468799884567
clc
for p=2023
a=sym(2023)^p;
c=char(string(a));
length(c);
b=0;
for i=1:length(c)
b=b+str2num(c(i));
end
lc=length(c);
b;
c=sod(b);
d=sod(c);
disp([p lc b c d])
for i=1:100:6689
if i+99>6689
disp(extractAfter(char(string(a)),i-1))
else
disp(extractBetween(char(string(a)),i,i+99))
end
end
end
Early on a pattern may be found in 2023^k
length
k of A B C D SOD(D)
1 4 7 7 7 7
2 7 31 4 4 4
3 10 55 10 1 1
4 14 70 7 7 7
5 17 85 13 4 4
6 20 109 10 1 1
7 24 106 7 7 7
8 27 112 4 4 4
9 30 136 10 1 1
10 34 142 7 7 7
11 37 157 13 4 4
12 40 163 10 1 1
13 43 223 7 7 7
14 47 238 13 4 4
15 50 262 10 1 1
16 53 268 16 7 7
17 57 211 4 4 4
18 60 280 10 1 1
19 63 295 16 7 7
20 67 301 4 4 4
21 70 370 10 1 1
22 73 313 7 7 7
23 77 328 13 4 4
24 80 361 10 1 1
25 83 403 7 7 7
For 2023^k near k=2023
length
k of A B C D SOD(D)
2000 6612 29686 31 4 4
2001 6616 29089 28 10 1
2002 6619 29752 25 7 7
2003 6622 29470 22 4 4
2004 6626 29737 28 10 1
2005 6629 29887 34 7 7
2006 6632 29767 31 4 4
2007 6636 29584 28 10 1
2008 6639 29563 25 7 7
2009 6642 29731 22 4 4
2010 6646 30124 10 1 1
2011 6649 29860 25 7 7
2012 6652 29776 31 4 4
2013 6655 30187 19 10 1
2014 6659 29581 25 7 7
2015 6662 30208 13 4 4
2016 6665 30133 10 1 1
2017 6669 29734 25 7 7
2018 6672 30127 13 4 4
2019 6675 30268 19 10 1
2020 6679 29716 25 7 7
2021 6682 30181 13 4 4
2022 6685 29899 37 10 1
2023 6689 30112 7 7 7
2024 6692 30118 13 4 4
2025 6695 30394 19 10 1
2026 6698 30139 16 7 7
2027 6702 29992 31 4 4
2028 6705 30313 10 1 1
2029 6708 30544 16 7 7
2030 6712 29803 22 4 4
2031 6715 30052 10 1 1
2032 6718 30409 16 7 7
2033 6722 30082 13 4 4
2034 6725 30385 19 10 1
2035 6728 30157 16 7 7
In this neighborhood, apparently only the 10's have one more step to get to their digital roots of 1, which would be OK even if the number was in this area, as SOD(D) still follows the pattern as SOD(10)=1. But could we have predicted whether the level of SOD(D) would be sufficient to get there?
OEIS's A006050 indicates that in order to require 5 iterations of SOD (SOD(D) is only the fourth iteration (the number of such being called the persistence) ), the original number would need to have 23 digits--much more than the 6689 digits that 2023^2023 has.
|
Posted by Charlie
on 2022-02-22 14:11:08 |