I can't figure out what is going wrong?

I am currently trying to write a program that will accept a positive integer between 50 and 300 and count the number of factors the number has.  Output the numbers and the number of factors in a sentence.  Have the program decide whether the number is prime?

This is what I have so far, but it keep saying every number isn't prime, even if they are.
Also when I run the program if the number is under 50 or 300 it still shows the factors in the picture box. How can I fix this?
BostonpublicAsked:
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BostonpublicAuthor Commented:
oops forgot the post the code lol

Option Explicit
Dim Num As Integer, C As Integer, Counter As Integer, Response As Integer, itself As Integer

Private Sub cmdEnd_Click()
    End
End Sub

Private Sub Text1_Change()
   
End Sub

Private Sub cmdGo_Click()
    For C = 50 To 300
    Next C
    Num = Val(txtFactors.Text)
    Counter = 0
    For C = 1 To Num
        If Num Mod C = 0 Then
            Counter = Counter + 1
            PicAns.Print C
    End If
    Next C
    PicAns.Print "There were"; Counter; "Factors"
    If txtFactors.Text > 300 Then
    Response = MsgBox("You must enter a number smaller than 300", vbOKOnly + vbInformation, " Error")
    ElseIf txtFactors.Text < 50 Then
    Response = MsgBox("You must enter a number larger than 50", vbOKOnly + vbInformation, " Error")
    End If
    If Val(txtFactors.Text * 1 And itself) Then
    PicAns.Print "This is a prime number"
    Else: PicAns.Print "This isn't a prime number"
    End If
   
End Sub

Private Sub cmdReset_Click()
    txtFactors.Text = ""
    frmFactors.Cls
    PicAns.Cls
End Sub

Private Sub Picture1_Click()

End Sub

Private Sub Pic_Click()

End Sub

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BostonpublicAuthor Commented:
oops forgot the post the code lol

Option Explicit
Dim Num As Integer, C As Integer, Counter As Integer, Response As Integer, itself As Integer

Private Sub cmdEnd_Click()
    End
End Sub

Private Sub Text1_Change()
   
End Sub

Private Sub cmdGo_Click()
    For C = 50 To 300
    Next C
    Num = Val(txtFactors.Text)
    Counter = 0
    For C = 1 To Num
        If Num Mod C = 0 Then
            Counter = Counter + 1
            PicAns.Print C
    End If
    Next C
    PicAns.Print "There were"; Counter; "Factors"
    If txtFactors.Text > 300 Then
    Response = MsgBox("You must enter a number smaller than 300", vbOKOnly + vbInformation, " Error")
    ElseIf txtFactors.Text < 50 Then
    Response = MsgBox("You must enter a number larger than 50", vbOKOnly + vbInformation, " Error")
    End If
    If Val(txtFactors.Text * 1 And itself) Then
    PicAns.Print "This is a prime number"
    Else: PicAns.Print "This isn't a prime number"
    End If
   
End Sub

Private Sub cmdReset_Click()
    txtFactors.Text = ""
    frmFactors.Cls
    PicAns.Cls
End Sub

Private Sub Picture1_Click()

End Sub

Private Sub Pic_Click()

End Sub

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robg69Commented:
This guy has a prime number function:
http://www.geocities.com/SiliconValley/Network/4328/visual.htm

Another guy did a VB prog that will print all the prime numbers up to a number the user specifies, you may be able to use this:
http://www.codeguru.com/columns/VB/PK061802.html


Hope one of these can help ya.
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JacamarCommented:
Here's my code that will print out all the prime numbers between 50 and 300.  I hope this is of some use.  Let me know if you have any more problems with it.


Private Sub Command1_Click()
Dim inCounter(50 To 300) As Integer
For inI = 50 To 300
    For inJ = 1 To inI
        If inI Mod inJ = 0 Then inCounter(inI) = inCounter(inI) + 1
    Next inJ
    If inCounter(inI) < 3 Then Picture1.Print inI
Next inI
End Sub
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BostonpublicAuthor Commented:
it is only showing the fractions from 50 to whatever number I chose.
Dim inCounter(50 To 300) As Integer
For inI = 50 To 300
   For inJ = 1 To inI
       If inI Mod inJ = 0 Then inCounter(inI) = inCounter(inI) + 1
   Next inJ
   If inCounter(inI) < 50 Then PicAns.Print inI
Next inI
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JacamarCommented:
Here's a fixed version that will do what you want.



Private Sub Command1_Click()
Dim inCounter(50 To 300) As Integer
Dim inNum As Integer
Dim inFactor() As Integer
inNum = CInt(Text1.Text)

For inI = 1 To inNum
    If inNum Mod inI = 0 Then
        inCount = inCount + 1
        ReDim Preserve inFactor(1 To inCount)
        inFactor(inCount) = inI
    End If
Next inI

Dim stSentence As String
stSentence = "Factors are: "

For inI = 1 To inCount
    stSentence = stSentence & " " & inFactor(inI)
    stSentence = stSentence & ","
Next inI

stSentence = stSentence & " which is a total of " & inCount & " Factors."

Label1.Caption = stSentence
End Sub
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BostonpublicAuthor Commented:
thank you so much:)
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JacamarCommented:
Then, as you can see, if the inCount value is equal to 2, then the number is prime.  You can add statements as you wish to do whatever you want from there.
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BostonpublicAuthor Commented:
Oh quick question,
Why for when I do the factors of 71 it works but if i try it again it adds zeros to the forum.?
And how can I get the numbers to go vertically instead of horizontally?
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JacamarCommented:
It doesn't do that on mine (the zeros that is).  Here is code for the vertical thing.


Private Sub Command1_Click()
Label1.Alignment = 2
Dim inCounter(50 To 300) As Integer
Dim inNum As Integer
Dim inFactor() As Integer
inNum = CInt(Text1.Text)

For inI = 1 To inNum
   If inNum Mod inI = 0 Then
       inCount = inCount + 1
       ReDim Preserve inFactor(1 To inCount)
       inFactor(inCount) = inI
   End If
Next inI

Dim stSentence As String
stSentence = "Factors are: "

For inI = 1 To inCount
   stSentence = stSentence & vbLf & inFactor(inI)
Next inI

stSentence = stSentence & vbLf & " which is " & inCount & " Factors."

Label1.Caption = stSentence
End Sub
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BostonpublicAuthor Commented:
hmm maybe I have something wrong in my general statement?

this is what I have...
Option Explicit
Dim inI As Integer
Dim inCount As Integer
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JacamarCommented:
Dim inCount As Integer

Move this line in to the procedure
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JacamarCommented:
Sorry about that.....I guess I forgot to define that one in the procedure (I had no Option explicit, so it worked anyways).
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BostonpublicAuthor Commented:
thank you so much.

it works now.
when i error proof it like when they put a number greater than 300 or less than 50 i would just continue at the end of my code right?

Also how come my box where the factors are going jumped down?
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JacamarCommented:
Try this code.  It may be very suitable for you.  About the box jumping down, I don't know what that is.  Can you describe it a bit better?

Option Explicit
Dim inI As Integer
Private Sub Text1_Change()
Dim inCount As Integer
If IsNumeric(Text1.Text) Then
    If Text1.Text > 49 And Text1.Text < 301 Then
Label1.Alignment = 2
Dim inCounter(50 To 300) As Integer
Dim inNum As Integer
Dim inFactor() As Integer
inNum = CInt(Text1.Text)

For inI = 1 To inNum
  If inNum Mod inI = 0 Then
      inCount = inCount + 1
      ReDim Preserve inFactor(1 To inCount)
      inFactor(inCount) = inI
  End If
Next inI

Dim stSentence As String
stSentence = "Factors are: "

For inI = 1 To inCount
  stSentence = stSentence & vbLf & inFactor(inI)
Next inI

stSentence = stSentence & vbLf & " which is " & inCount & " Factors."

Label1.Caption = stSentence
End If
End If
End Sub
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BostonpublicAuthor Commented:
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supunrCommented:
Private Sub FindPrimes()
     Dim i as Long
     Dim j as Long
     Dim iRoot as Long
     Dim NoFactors as Long

     For i = 50 to 500
          iRoot = CLng(sqr(i)) ' SqureRoof of i
          NoFactors = 0
          For j = 2 to iRoot
               if (i MOD j = 0) then
                    NoFactors = NoFactors + 1
                    Exit For
               End if
          Next j
          if (NoFactors = 0) then
               Print i; " is a prime number."
          End if
     Next i

End Sub

Good Luck!
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BostonpublicAuthor Commented:
thank you
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BostonpublicAuthor Commented:
oh darnit i dont think that is what the teacher wanted. I think she wanted the user to be able to enter a number between 50 and 300 and the program is to find the factors of it and say whether or not the number is prime. And somehow error proof it in case the user enters a number less than 50 or greater than 300.
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JacamarCommented:
Well then add a statement such as the following at the end of the code.

if inCount < 3 then
Label1.Caption = Label1.Caption & vblf & "This number is prime"
end if
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BostonpublicAuthor Commented:
oh ok thanks
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BostonpublicAuthor Commented:
Dim i As Long
    Dim j As Long
    Dim iRoot As Long
    Dim NoFactors As Long
For i = 50 To 300
         iRoot = CLng(Sqr(i)) ' SqureRoof of i
         NoFactors = 0
         For j = 2 To iRoot
              If (i Mod j = 0) Then
                   NoFactors = NoFactors + 1
                   Exit For
              End If
         Next j
         If (NoFactors = 0) Then
              Print i; " is a prime number."
         End If
    Next i
If inCount < 3 Then
PicAns.Print PicAns & vbLf & "This number is prime"
End If

does that look right?
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JacamarCommented:
No, cause that does what I did for you in the first place.  It goes through and finds all the prime numbers.  Use this.


Option Explicit
Dim inI As Integer
Private Sub Text1_Change()
Dim inCount As Integer
If IsNumeric(Text1.Text) Then
   If Text1.Text > 49 And Text1.Text < 301 Then
Label1.Alignment = 2
Dim inCounter(50 To 300) As Integer
Dim inNum As Integer
Dim inFactor() As Integer
inNum = CInt(Text1.Text)

For inI = 1 To inNum
 If inNum Mod inI = 0 Then
     inCount = inCount + 1
     ReDim Preserve inFactor(1 To inCount)
     inFactor(inCount) = inI
 End If
Next inI

Dim stSentence As String
stSentence = "Factors are: "

For inI = 1 To inCount
 stSentence = stSentence & vbLf & inFactor(inI)
Next inI

stSentence = stSentence & vbLf & " which is " & inCount & " Factors."

Label1.Caption = stSentence
Label1.caption = label1.caption & vblf & "This Number is Prime"
End If
End If
End Sub
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BostonpublicAuthor Commented:
do you have any idea why the box is still moving down on me though?

but other than that it worked, thank you so much.

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BostonpublicAuthor Commented:
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JacamarCommented:
Sure, see if this works.  I think i know what was wrong



Option Explicit
Dim inI As Integer
Private Sub Text1_Change()
Dim inCount As Integer
If IsNumeric(Text1.Text) Then
  If Text1.Text > 49 And Text1.Text < 301 Then
Label1.Alignment = 2
Dim inCounter(50 To 300) As Integer
Dim inNum As Integer
Dim inFactor() As Integer
inNum = CInt(Text1.Text)
inCount = 0
For inI = 1 To inNum
If inNum Mod inI = 0 Then
    inCount = inCount + 1
    ReDim Preserve inFactor(1 To inCount)
    inFactor(inCount) = inI
End If
Next inI

Dim stSentence As String
stSentence = "Factors are: "

For inI = 1 To inCount
stSentence = stSentence & vbLf & inFactor(inI)
Next inI

stSentence = stSentence & vbLf & " which is " & inCount & " Factors."

Label1.Caption = stSentence
Label1.Caption = Label1.Caption & vbLf & "This Number is Prime"
End If
End If
End Sub
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JacamarCommented:
What is your email.....i'll send you what I have here that i've been tinkering with.
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BostonpublicAuthor Commented:
it's elemenopee3@hotmail.com :)
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BostonpublicAuthor Commented:
it is still saying every number is prime even if they aren't.
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BostonpublicAuthor Commented:
it is still saying every number is prime even if they aren't.
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JacamarCommented:
ok, check your email......subject is "Factors"
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BostonpublicAuthor Commented:
i am downloading it now
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BostonpublicAuthor Commented:
it's taking forever do you use msn? maybe it'll send faster?
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BostonpublicAuthor Commented:
actually nm let me look at it =)
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JacamarCommented:
yeah I have msn
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BostonpublicAuthor Commented:
it worked thank you so so much.
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