Quote from: Andzhig on December 06, 2018, 05:08:14 AM

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Do you have a formula or anything close?

There is no formula, there is a random and our collective unconscious. Therefore, someone will write something, someone will think of something. Grin

You never know, I will write now and you will light up (light bulb lights up).

Take the last number found. 1011000110011101011011100001010010000110001001101000100001 take the first 3 bytes from it 10110001 10011101 01101110 here we have 10 zeros. Check how many combinations will be from the first byte 10110001, 8008 combinations which we need in place 583 (depending on the given data of course, table for counting in python, I'll leave it down) bin 101100011001110101101110 dec 11640174 multiply this number 34 times by 2 will get ~199976666598998016 practically what you need 199976667976342049. 199976667976342049-199976666598998016=1377344033 error on 1377344033 using a scanner "bitcrack" it's practically nothing. You need to guess just 3 bytes then "bitcrack" will do everything himself.

The most important question is how to guess these 3 bytes (and multiply by 2 35 times for 59 puzzle).

Now back to my writing.

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174176B015F4D
22BD43C2E9354
75070A1A009D4
EFAE164CB9E3C
180788E47E326C
236FB6D5AD1F44
6ABE1F9B67E114
9D18B63AC4FFDF
1EB25C90795D61C
2C675B852189A21
5 or 6 or 7 ...

Well, what's next, 288230376151711744-576460752303423488 ,if 6 it's 430000000000000000-500000000000000000 if 7 500000000000000000-570000000000000000

Depending on the initial number in hex, we get our initial bytes in bit.

hex ~610000000000000-6F0000000000000 ~436849163854938112-499899558638125056, 710000000000000-7F0000000000000 ~508906757892866048-571957152676052992.

6 or 7 implies 110 or 111

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("11000000"),("11000001"),("11000010"),("11000011"),("11000100"),("11000101"),("11000110"),("11000111"),
("11001000"),("11001001"),("11001010"),("11001011"),("11001100"),("11001101"),("11001110"),("11001111"),

("11010000"),("11010001"),("11010010"),("11010011"),("11010100"),("11010101"),("11010110"),("11010111"),
("11011000"),("11011001"),("11011010"),("11011011"),("11011100"),("11011101"),("11011110"),("11011111"),

("11100000"),("11100001"),("11100010"),("11100011"),("11100100"),("11100101"),("11100110"),("11100111"),
("11101000"),("11101001"),("11101010"),("11101011"),("11101100"),("11101101"),("11101110"),("11101111"),

("11110000"),("11110001"),("11110010"),("11110011"),("11110100"),("11110101"),("11110110"),("11110111"),
("11111000"),("11111001"),("11111010"),("11111011"),("11111100"),("11111101"),("11111110"),("11111111"),

As we judging by the previous numbers 6 we have mostly hex letters 6ABCDEF and seven with hex numbers 7123456789.

We narrow our space 6A0000000000000-6F0000000000000 477381560501272576-499899558638125056.

And we get 16 bytes.

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("11010000"),("11010001"),("11010010"),("11010011"),("11010100"),("11010101"),("11010110"),("11010111"),
("11011000"),("11011001"),("11011010"),("11011011"),("11011100"),("11011101"),("11011110"),("11011111"),

For each of which options are possible with 8-15 zeros.

But then you have to think further, or launch a "bitcrack" for everyone 16. It can run multiple copies at the same time. Or try to further filter for example in three bytes "111". On the example of our previous number 10110001 10011101 01101110 here are 2 of them "111" the number of options reduced from 8008 to 4802.

python script

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l1= [("10110001")]

#l1= [("11010000"),("11010001"),("11010010"),("11010011"),("11010100"),("11010101"),("11010110"),("11010111"),
# ("11011000"),("11011001"),("11011010"),("11011011"),("11011100"),("11011101"),("11011110"),("11011111")]

l2= [("10000000"),("10000001"),("10000010"),("10000011"),("10000100"),("10000101"),("10000110"),("10000111"),
   ("10001000"),("10001001"),("10001010"),("10001011"),("10001100"),("10001101"),("10001110"),("10001111"),
   ("10010000"),("10010001"),("10010010"),("10010011"),("10010100"),("10010101"),("10010110"),("10010111"),
   ("10011000"),("10011001"),("10011010"),("10011011"),("10011100"),("10011101"),("10011110"),("10011111"),
   ("10100000"),("10100001"),("10100010"),("10100011"),("10100100"),("10100101"),("10100110"),("10100111"),
   ("10101000"),("10101001"),("10101010"),("10101011"),("10101100"),("10101101"),("10101110"),("10101111"),
   ("10110000"),("10110001"),("10110010"),("10110011"),("10110100"),("10110101"),("10110110"),("10110111"),
   ("10111000"),("10111001"),("10111010"),("10111011"),("10111100"),("10111101"),("10111110"),("10111111"),
   ("11000000"),("11000001"),("11000010"),("11000011"),("11000100"),("11000101"),("11000110"),("11000111"),
   ("11001000"),("11001001"),("11001010"),("11001011"),("11001100"),("11001101"),("11001110"),("11001111"),
   ("11010000"),("11010001"),("11010010"),("11010011"),("11010100"),("11010101"),("11010110"),("11010111"),
   ("11011000"),("11011001"),("11011010"),("11011011"),("11011100"),("11011101"),("11011110"),("11011111"),
   ("11100000"),("11100001"),("11100010"),("11100011"),("11100100"),("11100101"),("11100110"),("11100111"),
   ("11101000"),("11101001"),("11101010"),("11101011"),("11101100"),("11101101"),("11101110"),("11101111"),
   ("11110000"),("11110001"),("11110010"),("11110011"),("11110100"),("11110101"),("11110110"),("11110111"),
   ("11111000"),("11111001"),("11111010"),("11111011"),("11111100"),("11111101"),("11111110"),("11111111"),
   ("00000000"),("00000001"),("00000010"),("00000011"),("00000100"),("00000101"),("00000110"),("00000111"),
   ("00001000"),("00001001"),("00001010"),("00001011"),("00001100"),("00001101"),("00001110"),("00001111"),
   ("00010000"),("00010001"),("00010010"),("00010011"),("00010100"),("00010101"),("00010110"),("00010111"),
   ("00011000"),("00011001"),("00011010"),("00011011"),("00011100"),("00011101"),("00011110"),("00011111"),
   ("00100000"),("00100001"),("00100010"),("00100011"),("00100100"),("00100101"),("00100110"),("00100111"),
   ("00101000"),("00101001"),("00101010"),("00101011"),("00101100"),("00101101"),("00101110"),("00101111"),
   ("00110000"),("00110001"),("00110010"),("00110011"),("00110100"),("00110101"),("00110110"),("00110111"),
   ("00111000"),("00111001"),("00111010"),("00111011"),("00111100"),("00111101"),("00111110"),("00111111"),
   ("01000000"),("01000001"),("01000010"),("01000011"),("01000100"),("01000101"),("01000110"),("01000111"),
   ("01001000"),("01001001"),("01001010"),("01001011"),("01001100"),("01001101"),("01001110"),("01001111"),
   ("01010000"),("01010001"),("01010010"),("01010011"),("01010100"),("01010101"),("01010110"),("01010111"),
   ("01011000"),("01011001"),("01011010"),("01011011"),("01011100"),("01011101"),("01011110"),("01011111"),
   ("01100000"),("01100001"),("01100010"),("01100011"),("01100100"),("01100101"),("01100110"),("01100111"),
   ("01101000"),("01101001"),("01101010"),("01101011"),("01101100"),("01101101"),("01101110"),("01101111"),
   ("01110000"),("01110001"),("01110010"),("01110011"),("01110100"),("01110101"),("01110110"),("01110111"),
   ("01111000"),("01111001"),("01111010"),("01111011"),("01111100"),("01111101"),("01111110"),("01111111")]

def funk():
   for element1 in (l1):
   for element2 in (l2):
   for element3 in (l2):
   bina = element1+element2+element3
   nuli = bina.count("0")
   if nuli == 10: #zeros count

   print(bina)
   pass
   else:
   pass
   return None

funk()

[/size]

when i take the only first 8 bits and start with 110 and the other bits change i have ths space from 60 to 6F

i search now with this setting with 5 cards with 245 Mkeys per card

cuBitCrack -d 0 -c -o found0.txt --keyspace 68*************:69************* -b 15 -t 512 -p 4120 -i adresses_50BTC.1.txt --continue test1.txt

but for every card another search space in this area

make this sense or not?