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 Pandigital Sum (Posted on 2006-09-19)
Find a 2-digit integer, a 3-digit integer and a 4-digit integer, such that:
• among them contain all digits 1 - 9
• each integer is divisible by each of its digits
• the sum of the three integers is a 4-digit number that is also divisible by each of its digits, which are all different as well
•  See The Solution Submitted by Charlie Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): Solution | Comment 3 of 5 |
(In reply to re: Solution by bumble)

RE: "If you didn't use a program then that's an impressive find! I would like to know the reasoning."

I used an unimpressive program - hahaha. Reasoning it out would have been impressive.

But alas, I used brute force.

The prgram is pretty boring, not too sophisticated at all. It ran in about a second. Here it is:

Visual Basic Program:

Imports System
Imports System.IO
Imports System.Text
Imports System.Runtime.InteropServices
Imports System.Math
Module Module1
Dim digit0 As Integer
Dim digit1 As Integer
Dim digit2 As Integer
Dim digit3 As Integer
Dim digit4 As Integer
Dim digit5 As Integer
Dim digit6 As Integer
Dim digit7 As Integer
Dim digit8 As Integer
Sub Main()
Randomize()
For digit0 = 1 To 9
For digit1 = 1 To 9
If digit1 <> digit0 Then
For digit2 = 1 To 9
If digit2 <> digit0 And _
digit2 <> digit1 Then
For digit3 = 1 To 9
If digit3 <> digit0 And _
digit3 <> digit1 And _
digit3 <> digit2 Then
For digit4 = 1 To 9
If digit4 <> digit0 And _
digit4 <> digit1 And _
digit4 <> digit2 And _
digit4 <> digit3 Then
For digit5 = 1 To 9
If digit5 <> digit0 And _
digit5 <> digit1 And _
digit5 <> digit2 And _
digit5 <> digit3 And _
digit5 <> digit4 Then
For digit6 = 1 To 9
If digit6 <> digit0 And _
digit6 <> digit1 And _
digit6 <> digit2 And _
digit6 <> digit3 And _
digit6 <> digit4 And _
digit6 <> digit5 Then
For digit7 = 1 To 9
If digit7 <> digit0 And _
digit7 <> digit1 And _
digit7 <> digit2 And _
digit7 <> digit3 And _
digit7 <> digit4 And _
digit7 <> digit5 And _
digit7 <> digit6 Then
For digit8 = 1 To 9
If digit8 <> digit0 And _
digit8 <> digit1 And _
digit8 <> digit2 And _
digit8 <> digit3 And _
digit8 <> digit4 And _
digit8 <> digit5 And _
digit8 <> digit6 And _
digit8 <> digit7 Then
check_them_out()
End If
Next
End If
Next
End If
Next
End If
Next
End If
Next
End If
Next
End If
Next
End If
Next
Next
End Sub
Sub check_them_out()
Dim work2 As Double
Dim work3 As Double
Dim work4 As Double
Dim work5 As Double
Dim work6 As Double
Dim work7 As Double
Dim work8 As String
Dim work9(4) As Integer
work2 = _
(digit0 * (10 ^ 0)) + _
(digit1 * (10 ^ 1))
work3 = _
(digit2 * (10 ^ 0)) + _
(digit3 * (10 ^ 1)) + _
(digit4 * (10 ^ 2))
work4 = _
(digit5 * (10 ^ 0)) + _
(digit6 * (10 ^ 1)) + _
(digit7 * (10 ^ 2)) + _
(digit8 * (10 ^ 3))
'
work5 = work2 / digit0
work6 = Fix(work2 / digit0)
If work5 <> work6 Then
Exit Sub
End If
work5 = work2 / digit1
work6 = Fix(work2 / digit1)
If work5 <> work6 Then
Exit Sub
End If
work5 = work3 / digit2
work6 = Fix(work3 / digit2)
If work5 <> work6 Then
Exit Sub
End If
work5 = work3 / digit3
work6 = Fix(work3 / digit3)
If work5 <> work6 Then
Exit Sub
End If
work5 = work3 / digit4
work6 = Fix(work3 / digit4)
If work5 <> work6 Then
Exit Sub
End If
work5 = work4 / digit5
work6 = Fix(work4 / digit5)
If work5 <> work6 Then
Exit Sub
End If
work5 = work4 / digit6
work6 = Fix(work4 / digit6)
If work5 <> work6 Then
Exit Sub
End If
work5 = work4 / digit7
work6 = Fix(work4 / digit7)
If work5 <> work6 Then
Exit Sub
End If
work5 = work4 / digit8
work6 = Fix(work4 / digit8)
If work5 <> work6 Then
Exit Sub
End If
work7 = work2 + work3 + work4
If work7 < 1000 Or work7 > 9999 Then
Exit Sub
End If
work8 = Str(work7)
'
' Remove leading zero(s) from string:
work8 = CStr(CInt(work8))
'
For index1 As Integer = 1 To Len(work8)
work9(index1 - 1) = Int(Mid(work8, index1, 1))
If work9(index1 - 1) = 0 Then
Exit Sub
End If
Next
For index0 As Integer = 0 To 2
For index1 As Integer = index0 + 1 To 3
If work9(index0) = work9(index1) Then
Exit Sub
End If
Next
Next
For index0 As Integer = 0 To 3
work5 = work7 / work9(index0)
work6 = Fix(work7 / work9(index0))
If work5 <> work6 Then
Exit Sub
End If
Next
Console.WriteLine(" ")
Console.WriteLine("Solution:")
Console.WriteLine(work2 & " " & work3 & " " & work4)
Console.WriteLine("Sum = " & work7)
End Sub
End Module

Edited on September 19, 2006, 10:44 am
 Posted by Penny on 2006-09-19 10:15:26

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