Challenge!!!

[vbud]

I have posted a similar thread once but no one seemed to have got it right. So here it goes again:

I have a series (Please find some random samples below) of numbers (left) that are generated by a system and their corresponding serials (to the right) which are plain autonumbers. These numbers are passed through the system for a sort of login and and are used only once.
When the number is enterred into the system, a validation is performed which checks if the numbers are correct, i.e if they fit the algo or not and if it has already been used before. The same pattern is used each time for validating. The challenge is to derive which pattern the system uses to generate those numbers and to be able to generate new numbers that will fit the algorithm and therefore allow new logins into the system. Anybody want to give it a try???
Note that the first four numbers are prefixes.

Numbers----------------------------Serials

1372 – 82309 – 63796 -------------- 17982309
1374 – 51171 – 42211 -------------- 18151171
1375 – 78274 – 42416 -------------- 18278274
1377 – 53203 – 41538 -------------- 18453203
1377 – 92076 – 80871 -------------- 18492076
1379 – 61492 – 26299 -------------- 18661492
1379 – 74160 – 99037 -------------- 18674160
1379 – 74167 – 31148 -------------- 18674167
1379 – 74195 – 48088 -------------- 18674195
1379 – 10314 – 10207 -------------- 18610314
1381 – 03451 – 80330 -------------- 18703451
1381 – 33852 – 46359 -------------- 18733852
1382 – 74085 – 35798 -------------- 18874085
1382 – 94057 – 52844 -------------- 18894057
1383 – 42560 – 52041 -------------- 18942560

[Spooner]

I haven't the foggiest idea what you're trying to say.

[opus]

Spooner, it's the same for me.
I don't care which sort of login you are going to break, but from the numbers you have given a serial can be made by:

You give 3 Numbers (Number1 [4 digits] Number2 [5digits] and Number3 [5digits])

To get the serial you only need Nimber1 and Number2, use following formula:

Serial=(Number1-1193)*100000 +Number2

The stuff about using the numbers only once is probably done by using checking the Number3, which should be used only once. So this number must be known by the checking system and the one who is logging in.

BTW: I like to take threads from this board along on my trainride home for the weekend, this one took about 10 minutes. The ride is MUCH longer!!!!

[vbud]

You got it wrong opus, the formula would not work for all numbers. Ok let give you some more info. The last five digits are not known be the system but when the numbers are enterred, the system performs some kind of validation to check if the numbers are correct or rather if they fit certain pattern that is still unknown to me. It is obvious that the last five digit of the serial are found in the second part of the key.
From what I have been able to gather from my research the last five numbers are validated if you increment it by 12100. So even if the other numbers are the same, you just need to add the above number to it so that he system would recognise it. This should have solved the problem but still the system would refuse this login and would say that it has already been used. So my point is that I can make the system validate the number but cant get a correct pattern to generate new numbers.

[opus]

OK, I missed this break in the patten.
But I don't get ypur thing with

the last five numbers are validated if you increment it by 12100

I don't see any help by adding 12100, and if the system is really checking if a number has been used, there must be a list of anny kind! If you can "make" login's, just by using a rule for the numbers, the system is not checking if the numbers have been used before!

[vbud]

ok maybe at some point the system does store the numbers somewhere, but consider this sample:

1383 – 42560 – 52041

Its serial (autonumber) is 18942560

Now if I were to add 12100 to the sample I shall get:

1383 – 42560 – 64141 which still has the serial 18942560 since the second part of the key (42560) has not been incremented. Now if I were to enter this key into the system, then it would say that this has already been used. You get my point now?

So at some point, maybe the system is still recognising the previous sample (1383 – 42560 – 52041) and it supposes that it is still dealing with that key. Or maybe that the newly generated key that used 12100 fits the algorithm that would validate the key in a first step during login. In anyway, it does not reject the number completely. So there should be a preliminary check going on that tells the system that this number was generated by itself.

[vbud]

Still no one on this???....