Article 41878 of alt.security: Zetnor wrote in article <32F2D8A8.2D7D@ere.umontreal.ca>... > I would like if it's possible to find a program available on the Intenet > that can generate credit card numbers with the expiration date. > Found this on the net with AltaVista... it might help Re: CREDIT CARD NUMBER GENERATOR WANTED !!!!!!

Re: CREDIT CARD NUMBER GENERATOR WANTED !!!!!!

From           wa7twi@primenet.com (Richard
Kendrick)
Organization   Primenet
Date           6 Dec 1996 20:16:03 -0700
Newsgroups     alt.2600
Message-ID     <58anhj$au6@nnrp1.news.primenet.com>
References     1

reginald michel walsmit <reggie@reggie.demon.nl> wrote:

>Hi can anyone give me an easy to use, and working, credit card
number
>generator  that works for both the credit number as for the expiration
>date !!!!!!

>I got hundreds of passwords for sites, membership accounts included for
>some but want to have access to all the live shit on as many sites as
>possible and for that you need credit card number plus
expiration date.

>If you can give that program to me i will be verry gratefull !!!!!!!!!

This is all I'll give you. If you're really a hacker, write your own code.

Credit card checkers use a checksum algorithm. Every card
number
will conform to this checksum, but not every card number
that
passes this checksum is valid and can be used. It only means that
such a card number can be issued by the credit card
company.

The format rules for a card number depend on the type of
card
desired. All card numbers begin with a 3, 4, or 5. MasterCard
must be 16 digits. Visa numbers must be 13 or 16 digits, with
16 digits becoming more common. American Express numbers must
be 15 digits. The expiration dates are in a month/year (MM/YY)
format. The acutal dates you use are left to your imagination.

Here is how the algorithm works. First, check the format
of the
trial credit card number. It must have the correct number
of
digits and start with a valid number for the type of card used.
Then multiply the individual digits of the credit card
number
by a 2 or a 1 depending on the position of the digit. If the
product of an individual calculation is 10 or greater, subtract
9 from that product. Add the individual products to get a checksum.
If the checksum is divisible by 10, the card passes the test.

Here is the psuedo code:

	checksum = 0

	loop from first digit to last digit
		if first digit position or position is odd number
			then mulitiply by 2
		else if position is even number
			then multiply by 1
		if product is 10 or greater
			then subtract 9 from the product
		checksum = checksum + product
	end loop

	if checksum is divisible by 10
		then card passes checksum test
	else checksum is not divisible by 10
		then card fails checksum test

Here are two examples of a Visa card. One passes the test, the other
does not.

First card number:  4271-6201-6712-3487.

(first two groups of numbers)
   4     2     7     1       6     2     0     1
 x 2   x 1   x 2   x 1     x 2   x 1   x 2   x 1
----  ----  ----  ----    ----  ----  ----  ----
   8     2    14     1      12     2     0     1
             - 9           - 9
            ----          ----
               5             3

(second group of numbers)
   6     7     1     2       3     4     8     7
 x 2   x 1   x 2   x 1     x 2   x 1   x 2   x 1
----  ----  ----  ----    ----  ----  ----  ----
  12     7     2     2       6     4    16     7
 - 9                                   - 9
----                                  ----
   3                                     7

calculate the checksum from the individual numbers:

8 + 2 + 5 + 1 + 3 + 2 + 0 + 1 + 3 + 7 + 2 + 2 + 6 + 4 + 7 + 7 = 60

60 / 10 = 6     checksum passes test, card is valid

Second card number: 4271-5201-6712-3487

(first two groups of numbers)
   4     2     7     1       5     2     0     1
 x 2   x 1   x 2   x 1     x 2   x 1   x 2   x 1
----  ----  ----  ----    ----  ----  ----  ----
   8     2    14     1      10     2     0     1
             - 9           - 9
            ----          ----
               5             1

(second group of numbers)
   6     7     1     2       3     4     8     7
 x 2   x 1   x 2   x 1     x 2   x 1   x 2   x 1
----  ----  ----  ----    ----  ----  ----  ----
  12     7     2     2       6     4    16     7
 - 9                                   - 9
----                                  ----
   3                                     7

calculate the checksum from the individual numbers:

8 + 2 + 5 + 1 + 1 + 2 + 0 + 1 + 3 + 7 + 2 + 2 + 6 + 4 + 7 + 7 = 58

58 / 10 = 5.8     checksum does not pass test, card is not valid

NNNNNNN
end of file

>Maybe i could give you something back for it !!!!!!!!!!!!

I doubt it.

>Thanks anyway,

Your welcome.

Richard Kendrick CET N7NT (ex WA7TWI) |
Amateur Radio Extra Class             |  Do not adjust your mind, there
QRP #4129/G-QRP #8591/MI-QRP #M-1412  |  is a fault in reality.
Mesa, AZ  email: wa7twi@primenet.com  |