Determine the minimum value of a three digit octal (base 8) positive integer N that has its digits reversed when expressed in the duodecimal (base 12) system.
Determine the three digit duodecimal (base 12) positive integer N that has its digits reversed when expressed in the hexadecimal (base 16) system.
DefDbl A-Z
Dim crlf$
Function mform$(x, t$)
a$ = Format$(x, t$)
If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$
mform$ = a$
End Function
Private Sub Form_Load()
ChDir "C:\Program Files (x86)\DevStudio\VB\projects\flooble"
Text1.Text = ""
crlf$ = Chr(13) + Chr(10)
Form1.Visible = True
DoEvents
For v = 8 * 8 To 8 * 8 * 8 - 1
b8$ = base$(v, 8)
b12$ = base$(v, 12)
If Len(b12$) = 3 Then
good = 1
For i = 1 To 3
If Mid(b8$, i, 1) <> Mid(b12$, 4 - i, 1) Then good = 0: Exit For
Next
If good Then
Text1.Text = Text1.Text & v & " " & b8$ & " " & b12$ & crlf
DoEvents
End If
End If
Next
Text1.Text = Text1.Text & crlf
For v = 12 * 12 To 12 * 12 * 12 - 1
b12$ = base$(v, 12)
b16$ = base$(v, 16)
If Len(b16$) = 3 Then
good = 1
For i = 1 To 3
If Mid(b12$, i, 1) <> Mid(b16$, 4 - i, 1) Then good = 0: Exit For
Next
If good Then
Text1.Text = Text1.Text & v & " " & b12$ & " " & b16$ & crlf
DoEvents
End If
End If
Next
Text1.Text = Text1.Text & crlf
Text1.Text = Text1.Text & " done"
End Sub
Function base$(n, b)
v$ = ""
n2 = n
Do
d = n2 Mod b
n2 = n2 \ b
v$ = Mid("0123456789abcdefghijklmnopqrstuvwxyz", d + 1, 1) + v$
Loop Until n2 = 0
base$ = v$
End Function
finds
decimal 475 = octal 733 = duodecimal 337
decimal 1337 = duodecimal 935 = hexadecimal 539
Edited on December 29, 2014, 9:57 am
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Posted by Charlie
on 2014-12-29 09:56:16 |
computer solution
