USSR puzzle digital

[bplus]

Ah-ha! Allowing for integer division instead of floating division I can:
Find a unique set of 9 digits that will produce = 8 in equations 2, 3 and 5
PLUS get the other equations all below 10 ie single digits!!!

There are 3 solutions like this (like in Danilin's problem) AND 1 is darn close to:
e(1) = 4
e(2) = 8
e(3) = 8
e(4) = 9
e(5) = 8
e(6) = 6

Code: (Select All)

index, record is 71038      829413657    e()'s match #1
8  * 2  - 9  = 7
4  + 1  + 3  = 8
6  \ 5  + 7  = 8
8  \ 4  + 6  = 8
2  + 1  + 5  = 8
9  + 3  - 7  = 5

index, record is 71158      829314657    e()'s match #2
8  * 2  - 9  = 7
3  + 1  + 4  = 8
6  \ 5  + 7  = 8
8  \ 3  + 6  = 8
2  + 1  + 5  = 8
9  + 4  - 7  = 6

index, record is 151676    629413857    e()'s match #3
6  * 2  - 9  = 3
4  + 1  + 3  = 8
8  \ 5  + 7  = 8
6  \ 4  + 8  = 9
2  + 1  + 5  = 8
9  + 3  - 7  = 5

I have solved a harder problem than Danilin's

Code: (Select All)

_Title "Alternate Danlin problem 2" 'b+ 2026-02-28

' 2026-02-27 mod for RA file



Dim record As String * 9

Dim c(1 To 9) As Integer

Dim e(1 To 6) As Integer

Open "9 permutations.RA" For Random As #1 Len = 9

Open "Match e2 with e3 and e5 = 8 for unique 9 digitsets.dat" For Output As #2

Do Until index = 362880

    index = index + 1

    Get #1, index, record

    For i = 1 To 9

        c(i) = Val(Mid$(record, i, 1))

    Next

    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) = 8 _AndAlso e(3) = 8 _AndAlso e(5) = 8 Then

        If e(1) < 10 _AndAlso e(4) < 10 _AndAlso e(6) < 10 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, ""

        End If

    End If

Loop

Oh look no Goto's b+LOL