OHHHHHHHHHHHHHHHH ! I thought the 9 digits had to be a unique permutation of digits 1 to 9. Well there aren't any of those BUT!
There are 65 sets that match e(2) with e(3) and with e(5) (Danilins = 8 equations) some get pretty close to single digit answers to all 6 equations.
There are 5 of the closest = 9 not 8 in e2, e3 and e5.
First I created a file of the permutations of digits 1 to 9:
Then I checked all permutations with this code:
Nope! Nothing stopped and met Danilins 4, 8, 8, 9, 8, 6 answers so there is NOT 9 unique digits that solve Danilin's equations.
So how close can we get?
First I tried to just match e2, e3, and e5 and there I found 65 solutions!
So what combinations get us closest to e2=e3=e5 = 8?
How about e2 = e3 = e5 = 9 !
Yes only 5 of those:
Found with this modified code:
There are 65 sets that match e(2) with e(3) and with e(5) (Danilins = 8 equations) some get pretty close to single digit answers to all 6 equations.
There are 5 of the closest = 9 not 8 in e2, e3 and e5.
How I found these as alternate to Danilin's problem:
First I created a file of the permutations of digits 1 to 9:
Code: (Select All)
_Title "Recursive 9 Permutations of symbols to RA" 'mod B+ 2026-02-27
'ordered permutations.bas for SmallBASIC 0.12.6 [B+=MGA] 2016-07-21
'translated from JB v1.01 [B+=MGA] 2016-07-21
'recursive ordered permutations (non repeating)
'if 1st element < 2nd element < 3rd element... then list acsends
'if reverse case then list decsends
' 2026-02-27 this mod
Dim Shared ls, index
Dim Shared record As String * 9
s$ = "987654321"
ls = Len(s$)
index = 0
Dim Shared c$(ls) 'this holds each letter or digit or symbol
Open "9 permutations.RA" For Random As #1 Len = 9
perm s$ '<<<<<<<<<<<<< if s$ is ordered then the permutations will be so also!!!
Sub perm (r$)
'local lr, i, r1
If Len(r$) = 0 Then
index = index + 1
b$ = ""
For i = 1 To ls
b$ = b$ + c$(i)
Next
record = b$
Put #1, index, record
Else
lr = Len(r$)
For i = 1 To lr
c$(ls - lr + 1) = Mid$(r$, i, 1)
If i < Len(r$) And i > 1 Then
r1$ = Mid$(r$, 1, i - 1) + Mid$(r$, i + 1)
'ELSEIF i = 1 AND i = lr THEN
'r1$ = ""
ElseIf i = 1 Then
r1$ = Mid$(r$, i + 1)
Else
r1$ = Mid$(r$, 1, i - 1)
End If
perm r1$
Next
End If
End Sub
Then I checked all permutations with this code:
Code: (Select All)
_Title "Test Danlin's matrix with 9 permutations" 'b+ 2026-02-26
' 2026-02-27 rewrite for RA
Dim record As String * 9
Open "9 permutations.RA" For Random As #1 Len = 9
Dim c(1 To 9) As Integer, index As Long
While index < 362880
index = index + 1
Get #1, index, record
Print index; ":"; record
For i = 1 To 9
c(i) = Val(Mid$(record, i, 1))
Next
If c(1) * c(2) - c(3) = 4 Then
Print "Check First equation: "; record
'Sleep
If c(4) + c(5) + c(6) = 8 Then
Print "Check 2nd equation: "; record
'Sleep
If c(7) / c(8) + c(9) = 8 Then
Print "Check 3rd equation: "; record
Sleep
If c(1) / c(4) + c(7) = 9 Then
If c(2) + c(5) + c(8) = 8 Then
If c(3) + c(6) - c(9) = 6 Then
Print "!!!!!!!!!!!!!!!!!!!!Solution! "; c(1), c(2), c(3)
Print "!!!!!!!!!!!!!!!!!!!!Solution! "; c(4), c(5), c(6)
Print "!!!!!!!!!!!!!!!!!!!!Solution! "; c(7), c(8), c(9)
Sleep
End
End If
End If
End If
End If
End If
End If
Wend
Nope! Nothing stopped and met Danilins 4, 8, 8, 9, 8, 6 answers so there is NOT 9 unique digits that solve Danilin's equations.
So how close can we get?
First I tried to just match e2, e3, and e5 and there I found 65 solutions!
So what combinations get us closest to e2=e3=e5 = 8?
How about e2 = e3 = e5 = 9 !
Yes only 5 of those:
Code: (Select All)
index, record is 10546 968315427 e()'s match #1
9 * 6 - 8 = 46
3 + 1 + 5 = 9
4 / 2 + 7 = 9
9 / 3 + 4 = 7
6 + 1 + 2 = 9
8 + 5 - 7 = 6
index, record is 28070 935126847 e()'s match #2
9 * 3 - 5 = 22
1 + 2 + 6 = 9
8 / 4 + 7 = 9
9 / 1 + 8 = 17
3 + 2 + 4 = 9
5 + 6 - 7 = 4
index, record is 32390 926135847 e()'s match #3
9 * 2 - 6 = 12
1 + 3 + 5 = 9
8 / 4 + 7 = 9
9 / 1 + 8 = 17
2 + 3 + 4 = 9
6 + 5 - 7 = 4
index, record is 151910 629135847 e()'s match #4
6 * 2 - 9 = 3
1 + 3 + 5 = 9
8 / 4 + 7 = 9
6 / 1 + 8 = 14
2 + 3 + 4 = 9
9 + 5 - 7 = 7
index, record is 187190 539126847 e()'s match #5
5 * 3 - 9 = 6
1 + 2 + 6 = 9
8 / 4 + 7 = 9
5 / 1 + 8 = 13
3 + 2 + 4 = 9
9 + 6 - 7 = 8Found with this modified code:
Code: (Select All)
_Title "Alternate Danlin problem" 'b+ 2026-02-26
' 2026-02-27 mod for RA file
Dim record As String * 9
Dim c(1 To 9) As Integer
Dim eq(1 To 6) As Integer
Open "9 permutations.RA" For Random As #1 Len = 9
Open "Match e2 with e3 and e5 for unique 9 digitsets.dat" For Output As #2
index = 0
tryAgain:
index = index + 1
If index > 362880 Then Print: Print "Through Whole File All done!": End
Get #1, index, record
'Print index, record
For i = 1 To 9
c(i) = Val(Mid$(record, i, 1))
Next
If c(7) / c(8) = Int(c(7) / c(8)) Then
If c(1) / c(4) = Int(c(1) \ c(4)) Then
'Print "index, record is"; index, record
e(1) = c(1) * c(2) - c(3) '4
e(2) = c(4) + c(5) + c(6) '8 match e(2)
e(3) = c(7) / c(8) + c(9) '8 match with e(3)
e(4) = c(1) / c(4) + c(7) '9
e(5) = c(2) + c(5) + c(8) '8 match and with e(5)
e(6) = c(3) + c(6) - c(9) '6
If e(2) = 9 _AndAlso e(3) = 9 _AndAlso e(5) = 9 Then
'If e(2) = e(5) Then ' the middle one across = middle one down
' If 2 * e(1) = e(2) Then
' If e(4) = 1.1 / 8 * e(2) Then
' If e(6) = 3 / 4 * e(2) Then
count = count + 1
Print #2, "index, record is"; index, record, "e()'s match #"; _Trim$(Str$(count))
Print #2, c(1); " *"; c(2); " -"; c(3); " ="; c(1) * c(2) - c(3)
Print #2, c(4); " +"; c(5); " +"; c(6); " ="; c(4) + c(5) + c(6)
Print #2, c(7); " /"; c(8); " +"; c(9); " ="; c(7) / c(8) + c(9)
Print #2, c(1); " /"; c(4); " +"; c(7); " ="; c(1) / c(4) + c(7)
Print #2, c(2); " +"; c(5); " +"; c(8); " ="; c(2) + c(5) + c(8)
Print #2, c(3); " +"; c(6); " -"; c(9); " ="; c(3) + c(6) - c(9)
Print #2, ""
'Sleep
GoTo tryAgain:
'Else
' GoTo tryAgain:
'End If
'Else
' GoTo tryAgain:
'End If
' Else
' GoTo tryAgain:
' End If
' Else
' GoTo tryAgain:
' End If
Else
GoTo tryAgain:
End If
Else
GoTo tryAgain:
End If
Else
GoTo tryAgain:
End If
724 855 599 923 575 468 400 206 147 564 878 823 652 556 bxor cross forever
