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8.0

mod 11 vb code for check digit again URGENT

Asked by awolfe76 in Visual Basic Programming

Tags: check, code, digit

ok i posted this question before but without any really good information on it. i now have good information. so here is it.
i need check digit code, using mod 11, for an 11 digit number, the final(11th) digit is the check digit. the user will enter the 11 digit number into a text box, when they click the submit button the code should run on the number to validate it. it will tell whether or not check digit number is what it should be.

NCS MOD 11 CHECK DIGIT ROUTINE FOR 10 DIGIT NUMBER:              
                                                                 

A SERIES OF MULTIPLICATIONS AND ADDITIONS IS PERFORMED, WITH THE      

MULTIPLICATION RESULTS BEING ADDED TO A FINAL COUNTER.  ON            

COMPLETION OF ALL MULTIPLICATIONS, THE COUNTER ID DIVIDED BY 11,      

AND 11 IS SUBTRACTED FROM THE REMAINDER.  THE LAST DIGIT OF THIS      

SUBTRACTION IS THE CHECK DIGIT.                                      

                                                                 

EXAMPLE:  FINDER NUMBER - TOTAL OF 11 POSITIONS     RESULT            
                                                                     

          1438623406 C(CHECK DIGIT)                      0 (INITIAL VALUE)    

                                                                     

THE 10TH DIGIT IS MULTIPLIED BY 3    (6 X 3 = 18)   +   18            

                                                    ------            

                                                        18            

THE 9TH DIGIT IS MULTIPLIED BY 5     (0 X 5 = 0)    +    0            

                                                    ------            

                                                        18            

THE 8TH DIGIT IS MULTIPLIED BY 7     (4 X 7 = 28)   +   28            

                                                    ------      

                                                        46      

THE 7TH DIGIT IS MULTIPLIED BY 9     (3 X 9 = 27)   +   27      

                                                    ------      

                                                        73      

THE 6TH DIGIT IS MULTIPLIED BY 11    (2 X 11 = 22)  +   22      

                                                    ------      

                                                        95      

THE 5TH DIGIT IS MULTIPLIED BY 13    (6 X 13 = 78)  +   78      

                                                    ------      

                                                       173      

THE 4TH DIGIT IS MULTIPLIED BY 15    (8 X 15 = 120) +  120      

                                                    ------      

                                                       293      

THE 3RD DIGIT IS MULTIPLIED BY 17    (3 X 17 = 51)  +   51      

                                                    ------      

                                                       344      

THE 2ND DIGIT IS MULTIPLIED BY 19    (4 X 19 = 76)  +   76      

                                                    ------      

                                                       420      

THE 1ST DIGIT IS MULTIPLIED BY 21    (1 X 21 = 21)  +   21      

                                                    ------  

                                                       441  

                                                             

THIS RESULT IS DIVIDED BY 11:  441/11 = 40, WITH REMAINDER 1
                                                           

11 IS SUBTRACTED FROM THE REMAINDER: 1 - 11 = -10            
                                                             

THE LAST DIGIT OF THIS SUBTRACTION (0) IS THE CHECK DIGIT

i hope that this helps. again its urgent. and i appreciate all the help. thanks in advance.
[+][-]10/03/03 08:56 AM, ID: 9485668Accepted Solution

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About this solution

Zone: Visual Basic Programming
Tags: check, code, digit
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Solution Provided By: wuttrain
Participating Experts: 1
Solution Grade: A
 
[+][-]10/05/03 06:50 AM, ID: 9494350Author Comment

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