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Problem in IMEI with nokia switch

Viewing 7 posts - 1 through 7 (of 7 total)
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  • #40028 Reply

    hi all

    when you check subscriber IMEI number in nokia switch with command MVP, it is not showing the correct number? what could be the reason for that?

    The IMEI from switch differs from actual IMEI of particular mobile in last digit.

    waiting for reply

    #40029 Reply


    This is not a problem. In my Ericsson switch the situation is the same. The last digit is always 0. This is because the last digit (also called Luhn check digit) is not used to identify the mobile, it is used to verify if the IMEI is valid (correct) or not. You can calculate the last digit yourself, based on first 14 digits.


    #40030 Reply

    Thanks misa for your ans.

    can you tell me how we can find the last digit with first 14 digits? my mail id is

    thanks again

    #40031 Reply

    Let say the IMEI is presented as D14-D13-D12-…-D1-D0 (15 digits), where D0 is the check digit.

    The last digit D0 is calculated as follow:

    1. Double the values of the odd labelled digits D1, D3, D5 … D13.

    2. Add together the individual digits of all the seven numbers obtained in step 1, and then add this sum to the sum of all the even labelled digits D2, D4, D6 … D14.

    3. If the result obtained in step 2 ends with 0, then D0 = 0. If it does not end in 0, then D0 equals to the number subtracted from the next higher number which does end in 0.

    If you familiar with VB, I can give you a procedure written in VB for calculation of this last digit.

    #40032 Reply

    Thanks very very much Misa.

    I am familiar with VB. so if you send it, it will be very helpful for me.

    thanks very much again.


    #40033 Reply

    Here is the VB function:

    Function CheckDigit(sImei As String) As String
    Dim iDigit(0 To 14) As Integer, i As Integer, iCD As Integer, sCD As String
    For i = 1 To 14
    iDigit(i) = CInt(Mid(sImei, 15 – i, 1))
    If i Mod 2 = 1 Then
    iDigit(i) = 2 * iDigit(i)
    If iDigit(i) > 9 Then iDigit(i) = iDigit(i) – 9
    End If
    For i = 1 To 14
    iCD = iCD + iDigit(i)
    If iCD Mod 10 = 0 Then
    sCD = “0”
    sCD = CStr(((iCD \ 10) + 1) * 10 – iCD)
    End If
    CheckDigit = sCD
    End Function

    #40034 Reply

    Thanks Very Much Misa.

Viewing 7 posts - 1 through 7 (of 7 total)
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