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Posted (edited)

So if there are 10 groups of numbers and 5 of those groups of numbers go into one code we need to try 100,000 codes?

 

If every here was active everyone here everyone could just try 14 codes and we would be done fast LOL.

Edited by T3A_guy
Posted (edited)

Okay. We need to make a system based on member numbers

 

Here is the formula: Start on combo number ((Member Number) * 14) - 14

 

and end with combo num(1718*14) =(1718*14) =ber (Member Number) * 14

 

So for example, my member number is 1718 so

 

(1718*14) - 14= 24038

and

 

(1718*14) = 24052

 

I would do combo's 24038 to 24052.

 

or

 

2, 4, 0, 3, 8

2, 4, 0, 3, 9

2, 4, 0, 4, 0

2, 4, 0, 4, 1

2, 4, 0, 4, 2

2, 4, 0, 4, 3

2, 4, 0, 4, 4

2, 4, 0, 4, 5

2, 4, 0, 4, 6

2, 4, 0, 4, 7

2, 4, 0, 4, 8

2, 4, 0, 4, 9

2, 4, 0, 5, 0

2, 4, 0, 5, 1

and

2, 4, 0, 5, 2

 

Combos = Number * Numbers = 36 comboss, lol not 10000
It's actually 50 different ways 0_o.

Okay so lets say the first combo of 5 letters is A the second is B all the way to the tenth being J.

 

I'll name more than 36 and more than 50 combos just here.

 

1) ABCDE

2) ABCDF

3) ABCDG

4) ABCDH

5) ABCDI

6) ABCDJ

7) ABCED

8 ) ABCEF

9) ABCEG

10) ABCEH

11) ABCEI

12) ABCEJ

13) ABCFD

14) ABCFE

15) ABCFG

16) ABCFH

17) ABCFI

18) ABCFJ

19) ABCGD

20) ABCGE

21) ABCGF

22) ABCGH

23) ABCGI

24) ABCGJ

25) ABCHD

26) ABCHE

27) ABCHF

28) ABCHI

29) ABCHJ

30) ABCID

31) ABCIE

32) ABCIF

33) ABCIG

34) ABCIH

35) ABCIJ

36) ABDCE

37) ABDCF

38) ABDCG

39) ABDCH

40) ABDCI

41) ABDCJ

42) ABDEC

43) ABDEF

44) ABDEG

45) ABDEH

46) ABDEI

47) ABDEJ

48) ABECD

49) ABEDC

50) ABEDE

51) ABEDF

 

51>50 and 51=/=50

 

I know in this post I might have confused you. LOL

A=1 and A or 1 are both a code of 5 numbers peaches found. For example if he found the code: 44G6H first then that would be code A or 1

same for the second code he found (B or 2)

Edited by T3A_guy

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