Prove: If x is a positive real number, then somewhere in the infinite sequence {x, 2x, 3x, ...} there is a number containing the digit 7.
If x is a positive real number, then somewhere in the finite sequence {x, 2x, 3x, ..., nx} there is a number containing the digit 7. Find the minimum value of n.
Note: Some numbers can be written in two ways (1.8=1.7999999...) only consider the form without all the 9's.
Source: Slightly adapted from a post by Victor Wang on Facebook.
mx=0;
for i=100000:999999
for n=1:200
ni=char(string(n*i));
ix=find(ni=='7');
if ~isempty(ix)
if n>mx
mx=n;
disp([n i n*i])
end
break
end
end
if n>=200
disp('raise limit on n')
break
end
end
n first 6 digits product
7 100000 700000
11 116000 1276000
12 116480 1397760
15 116500 1747500
33 166000 5478000
35 200000 7000000
finds the highest n that is necessary to obtain a 7 is 35, for any real number whose first six significant digits are anywhere from 100000 to 999999. In particular n=35 is necessary for real numbers starting with 2 and having all zeros after that: 2.00000... * 10^k.
Of course only the first six significant digits have been tested, so I have not proved that there isn't some set of digits after that that would cause a cascade of carries that would prevent a 7 from showing up in all multiples up to a higher n, but 35 is certainly necessary, and very likely sufficient (i.e., the minimum shown to be necessary).
Increasing the digit count to eight, finds nothing higher than 35 is needed:
mx=0;
for i=1:99999999
for n=2:35
ni=char(string(n*i));
ix=find(ni=='7');
if ~isempty(ix)
if n>mx
mx=n;
disp([n i])
end
break
end
end
if n>35
disp('raise limit on n')
break
end
end
That indeed 35 is necessary for 2, 20, 200..., or .00200..., etc. is shown by:
i=2;
for n=1:35
disp([n n*i])
end
showing
1 2
2 4
3 6
4 8
5 10
6 12
7 14
8 16
9 18
10 20
11 22
12 24
13 26
14 28
15 30
16 32
17 34
18 36
19 38
20 40
21 42
22 44
23 46
24 48
25 50
26 52
27 54
28 56
29 58
30 60
31 62
32 64
33 66
34 68
35 70 (the first 7 within a multiple of 2)
But another idea came to me. Subject every case with every possible carry into the last digit position:
clc
mx=0;
for i=100000:999999
for n=1:100
carryCt=0;
for carry=0:9
ni=char(string(n*i));
ix=find(ni=='7');
if ~isempty(ix)
carryCt=carryCt+1;
continue
end
end
if carryCt==10
if n>mx
mx=n;
disp([n i n*i])
end
break
end
end
if n>=100
disp('raise limit on n')
break
end
end
But the carries in some cases might be larger than single digit. It'll take longer, but we can speed things up by considering fewer digits now that we have the assurance that all carries will be accounted for.
clc
mx=0;
for i=1000:9999
for n=1:100
carryCt=0;
for carry=0:349
ni=char(string(n*i));
ix=find(ni=='7');
if ~isempty(ix)
carryCt=carryCt+1;
continue
end
end
if carryCt==350
if n>mx
mx=n;
disp([n i n*i])
end
break
end
end
if n>=100
disp('raise limit on n')
break
end
end
gets the same results, while accounting for carries through 349.
Edited on August 12, 2022, 3:04 pm
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Posted by Charlie
on 2022-08-12 15:03:15 |
solution
