Pick up at random a 3-digit number and decrease it by the sum of its digits.
Clearly,applying this procedure again and again you will hit zero no matter how big your initial number was .
How many iterations at most are needed and how many numbers cause it?
I would call it 80 steps for the largest ten 3-digit numbers.
clc
max=0;
for n0=100:999
n=n0;
for gen=1:1000
n=n-sod(n);
if n0==999
disp([gen n])
end
if n==0
break
end
end
disp([n0 gen])
if gen>max
max=gen;
end
end
disp(max)
requiring file sod.m:
function sd = sod(n)
s=0;
dgts=num2str(n) ;
l=length(dgts) ;
dgts=string(dgts);
for i=1:l
s=s+double(substr(dgts,i));
end
sd=s;
end
also substr.m:
function outstr = substr(x,r)
c=char(x);
c2=c(r);
outstr=string(c2);
end
produces, firstly a list of how many steps are needed for each 3-digit number:
100 11
101 11
102 11
103 11
104 11
105 11
106 11
107 11
108 11
109 11
110 12
111 12
112 12
113 12
114 12
115 12
116 12
117 12
118 12
119 12
120 13
121 13
122 13
123 13
124 13
...
983 79
984 79
985 79
986 79
987 79
988 79
989 79
990 80
991 80
992 80
993 80
994 80
995 80
996 80
997 80
998 80
999 80
and the 80 steps starting with 999:
after # remaining
steps
1 972
2 954
3 936
4 918
5 900
6 891
7 873
8 855
9 837
10 819
11 801
12 792
13 774
14 756
15 738
16 720
17 711
18 702
19 693
20 675
21 657
22 639
23 621
24 612
25 603
26 594
27 576
28 558
29 540
30 531
31 522
32 513
33 504
34 495
35 477
36 459
37 441
38 432
39 423
40 414
41 405
42 396
43 378
44 360
45 351
46 342
47 333
48 324
49 315
50 306
51 297
52 279
53 261
54 252
55 243
56 234
57 225
58 216
59 207
60 198
61 180
62 171
63 162
64 153
65 144
66 135
67 126
68 117
69 108
70 99
71 81
72 72
73 63
74 54
75 45
76 36
77 27
78 18
79 9
80 0
Edited on February 14, 2021, 9:03 am
Edited on February 14, 2021, 9:08 am
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Posted by Charlie
on 2021-02-14 08:59:24 |
computer solution
