Prove that for every positive integer m there exists a positive integer k, such that 2km is a digit anagram of km.
To get some insight I found the smallest satisfying k for m=1,60.
I list them below and also give the program that generated them.
I believe there must be some recursion used in the proof because the same sequences keep recurring: e.g 2517... 2148....
There are a lot of patterns below, even when digits begin to get used twice.
A key to the recursion I think is the similar answers to m=1,2,3
(One km product can be shown to satisfy many m's.) But that's all I got so far.
m k km 2km
----------------------------------------------------------------------
1 125874 125874 251748
2 62937 125874 251748
3 41958 125874 251748
4 269631 1078524 2157048
5 214857 1074285 2148570
6 20979 125874 251748
7 17982 125874 251748
8 157338 1258704 2517408
9 13986 125874 251748
10 125874 1258740 2517480
11 12987 142857 285714
12 89877 1078524 2157048
13 10989 142857 285714
14 8991 125874 251748
15 71619 1074285 2148570
16 78669 1258704 2517408
17 63972 1087524 2175048
18 6993 125874 251748
19 15039 285741 571482
20 62937 1258740 2517480
21 5994 125874 251748
22 12987 285714 571428
23 54243 1247589 2495178
24 52446 1258704 2517408
25 43497 1087425 2174850
26 10989 285714 571428
27 4662 125874 251748
28 44955 1258740 2517480
29 36306 1052874 2105748
30 41958 1258740 2517480
31 35082 1087542 2175084
32 46206 1478592 2957184
33 4329 142857 285714
34 31986 1087524 2175048
35 35451 1240785 2481570
36 29959 1078524 2157048
37 3402 125874 251748
38 28233 1072854 2145708
39 3663 142857 285714
40 269631 10785240 21570480
41 30429 1247589 2495178
42 2997 125874 251748
43 95409 4102587 8205174
44 33543 1475892 2951784
45 23873 1074285 2148570
46 27369 1258974 2517948
47 27342 1285074 2570148
48 26223 1258704 2517408
49 25461 1247589 2495178
50 214857 10742850 21485700
51 21324 1087524 2175048
52 49752 2587104 5174208
53 26469 1402857 2805714
54 2331 125874 251748
55 25587 1407285 2814570
56 22959 1285704 2571408
57 5013 285741 571482
58 18153 1052874 2105748
59 6993 412587 825174
60 20979 1258740 2517480
program anagram
implicit none
integer k,m,km,km2,d1,d2,flag
c note: I accidentally reversed the meaning of k and m w/o problem
do k=1,60
do m=1,1000000
km=k*m
km2=2*k*m
d1 = alog10(1.*km)
d2 = alog10(1.*km2)
flag=0
if(d1.eq.d2)call ana(km,km2,d1,flag)
if(flag.eq.1)then
print*,k,m,km,km2
go to 1
endif
enddo
1 enddo
end
subroutine ana(km,km2,d1,flag)
implicit none
integer i,km,km2,d1,flag,bin1(0:9),bin2(0:9),num1,num2,
1 dig1,dig2
flag=0
do i=0,9
bin1(i)=0
bin2(i)=0
enddo
num1=km
num2=km2
do i=d1,0,-1
dig1=num1/10**i
dig2=num2/10**i
bin1(dig1)=bin1(dig1)+1
bin2(dig2)=bin2(dig2)+1
num1=num1-dig1*10**i
num2=num2-dig2*10**i
enddo
do i=0,9
if(bin1(i).ne.bin2(i))go to 1
enddo
flag=1
1 return
end
Edited on January 28, 2021, 10:34 pm
seems true - here are examples, but no proof
