Continuation "road to calvary".
looking for a 20 digit number, example 30568377312064202855 by 2 > 30 56 83 77 31 20 64 20 28 55 (10 pieces for mixing)
30568377312064202855 by 2 > 30 56 83 77 31 20 64 20 28 55
take the first 100 numbers power of 2 (4096-32768 or 4096-30000000)
these first 100 digits split by 2 (50 elements in list) and look at what positions our numbers (30 56 83 77 31 20 64 20 28 55) will be, then we sort them from smallest to largest,
some like this...
1 [0, 6, 7, 10, 16, 21, 24, 34, 35, 40]
2 [3, 4, 5, 11, 12, 23, 27, 40, 43, 48]
3 [8, 9, 10, 13, 17, 24, 27, 28, 35, 42]
4 [4, 7, 8, 18, 20, 22, 24, 27, 33, 36]
5 [4, 7, 8, 11, 13, 14, 21, 29, 38, 48]
6 [2, 5, 7, 12, 19, 23, 31, 33, 40, 44]
7 [0, 3, 7, 11, 13, 18, 20, 25, 35, 43]
8 [4, 13, 15, 17, 19, 25, 30, 33, 40, 46]
9 [1, 2, 4, 15, 18, 29, 35, 40, 43, 45]
10 [0, 3, 4, 5, 14, 17, 22, 32, 36, 38]
and see the difference between the numbers.
some like this...
95465 [0, 1, 10, 11, 14, 15, 22, 31, 40, 44, 47] 1 9 1 3 1 7 9 9 4 3
273880 [0, 2, 12, 15, 19, 21, 23, 31, 32, 37, 44] 2 10 3 4 2 2 8 1 5 7
310872 [0, 9, 10, 19, 23, 27, 28, 29, 41, 43, 47] 9 1 9 4 4 1 1 12 2 4
324831 [6, 10, 19, 20, 22, 23, 26, 33, 34, 35, 40] 4 9 1 2 1 3 7 1 1 5
327766 [1, 2, 5, 11, 14, 24, 27, 28, 45, 46, 49] 1 3 6 3 10 3 1 17 1 3
342364 [1, 5, 13, 15, 16, 20, 21, 26, 32, 38, 43] 4 8 2 1 4 1 5 6 6 5
484256 [3, 7, 10, 11, 15, 17, 18, 20, 41, 43, 49] 4 3 1 4 2 1 2 21 2 6
519343 [0, 2, 5, 6, 10, 24, 27, 28, 29, 40, 48] 2 3 1 4 14 3 1 1 11 8
540844 [2, 3, 6, 7, 8, 13, 29, 36, 37, 39, 45] 1 3 1 1 5 16 7 1 2 6
744223 [4, 8, 10, 13, 14, 16, 18, 26, 29, 44, 48] 4 2 3 1 2 2 8 3 15 4
773619 [2, 3, 11, 12, 13, 18, 20, 28, 31, 33, 49] 1 8 1 1 5 2 8 3 2 16
804181 [2, 6, 10, 12, 16, 33, 35, 37, 39, 45, 46] 4 4 2 4 17 2 2 2 6 1
816903 [3, 6, 11, 18, 20, 22, 23, 30, 31, 38, 45] 3 5 7 2 2 1 7 1 7 7
now we filter and look for the "formula")))
here we have several components in charge of search time, the size power of 2, first difference numbers to which to stick, the difference, and possible differences + ~10000000 for mixing 10 parts by 2. i.e. some like this
60466176 6x10 60466176×2^262144 = 15850845241344
9765625 5x10 9765625×2^262144 = 2560000000000
blablabla
besides need to do an offset (take the first 100 numbers power of 2) 0-100 num, 100-200 num, 200-300 num...
exemple 1
2^4096-2^32768
filt <=6 <=6 <=6 <=6 <=6 <=6 <=6 <=6 <=6
0 100
2^15047 [8, 11, 12, 15, 17, 19, 24, 24, 28, 34] 3 1 3 2 2 5 0 4 6
100 200
200 300
300 400
400 500
500 600
600 700
700 800
800 900
900 1000
2^21946 [12, 15, 15, 18, 22, 25, 31, 37, 39, 44] 3 0 3 4 3 6 6 2 5
1000 1100
exemple 2
2^4096-2^30000000
filt <=1 <=1 <=1 <=1 <=1 <=30 <=30 <=30 <=30
2^7788201 [1, 1, 2, 3, 4, 5, 19, 33, 35, 48] 0 1 1 1 1 14 14 2 13
2^20398384 [0, 0, 1, 2, 3, 4, 16, 17, 33, 41] 0 1 1 1 1 12 1 16 8
0-100
2^371466 [5, 6, 7, 7, 8, 9, 10, 11, 19, 20] 1 1 0 1 1 1 1 8 1
2^21846679 [2, 3, 4, 5, 5, 6, 21, 25, 27, 31] 1 1 1 0 1 15 4 2 4
100-200
200-300
2^23749755 [7, 8, 9, 10, 10, 11, 16, 18, 32, 36] 1 1 1 0 1 5 2 14 4
2^29029857 [5, 5, 6, 7, 8, 9, 14, 19, 35, 36] 0 1 1 1 1 5 5 16 1
300-400
2^1061949 [0, 0, 1, 2, 3, 4, 20, 38, 45, 47] 0 1 1 1 1 16 18 7 2
...
So this means only a computer genius that eats programs who can solve it? Lucky for you guys to crack it as it has huge rewards waiting to be claimed. I wonder if there were puzzles out there that does not require any computer skills just to crack it. Someone like me will surely cannot solve puzzles like this. 32BTC is a lot of money maybe I will have to find ways to participate and this is fun because the odds of me cracking it is so high maybe thousand times. 😅
Another puzzle that doesn't require any computer skills.
https://i.redd.it/n1x7g8ceaur51.pnghttps://www.blockchain.com/id/btc/address/1KfZGvwZxsvSmemoCmEV75uqcNzYBHjkHZ Designing and creating a puzzle seems more expensive than a reward.
If it works and you really found the keys to these addresses, then this should excite the crypto community.
-> you cannot import these key in to the wallet!
?
