Is there any point in this option (geometric).
take a circle ,top right 000,001,002,003 ... left 999,998,997 ...
000
999 001
998 002
997 003
... ....
21 num length
970 436 974 005 023 690 481
by age
005 023 436 481 690 970 974
and connect the points in a circle.
angles are obtained, not less than 90 degrees (or a jump from 481 690 970 gives an acute angle? check who has the software at hand for this)
90 (working space degrees)
|
|
___________|___________ 180
and that 90×90×90×90×90×90×90= 47829690000000 iterate step by step (for mixing).
Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 282. (Read 242901 times)
-snip-
32 BTC is a lot of money, maybe you can give us some info about you so that we can trust you.
32 BTC is a lot of money, maybe you can give us some info about you so that we can trust you.
In fact, 32BTC was the old total bounty reward. The current amount is approximately 100BTC
Thank you so much for this offer, but we don't know how we will trust you!
There are many scammers on the internet, and they always have attractive offers.
32 BTC is a lot of money, maybe you can give us some info about you so that we can trust you.
There are many scammers on the internet, and they always have attractive offers.
32 BTC is a lot of money, maybe you can give us some info about you so that we can trust you.
All the info you need is on the first post, but you will need to read it.
Thank you so much for this offer, but we don't know how we will trust you!
There are many scammers on the internet, and they always have attractive offers.
32 BTC is a lot of money, maybe you can give us some info about you so that we can trust you.
There are many scammers on the internet, and they always have attractive offers.
32 BTC is a lot of money, maybe you can give us some info about you so that we can trust you.
BTCBTCI think the best existing tool currently is yours or otherwise private. I would think the best way would be to modify (ocl)vanitygen according to BurtW's suggestion. You would need to limit the random number generator to a certain amount of bits and keep the rest.
can you please share a python program to find private key from public key using baby giant step method
https[Suspicious link removed]
Thanks Bigvito. Its helpful.
Can you please share the same program in python
Oh I forgot about this one is in python, this was last updated last year.
https://github.com/Telariust/pollard-kangaroo
can you please share a python program to find private key from public key using baby giant step method
https[Suspicious link removed]
Thanks Bigvito. Its helpful.
Can you please share the same program in python
It's so amazing how long this puzzle has stood the test of time compared to that puzzle that was just published by MrBeast the other week that got solved in all of 9 hours for less money and took like $250k to develop, apparently.
If someone eventually managed to brute force Puzzle 64 (and beyond) can't someone who's monitoring the keys just use Pollard Kangaroo once the public key is broadcasted and also sweep it with a higher transaction fee since double spending is still an issue with Bitcoin?
If thats the case how can you prevent someone from sweeping the same key if you both became aware of the private key?
If thats the case how can you prevent someone from sweeping the same key if you both became aware of the private key?
can you please share a python program to find private key from public key using baby giant step method
https://github.com/JeanLucPons/BSGS
https://github.com/JeanLucPons/BSGS/releases/download/1.1/BSGS.exe
can you please share a python program to find private key from public key using baby giant step method
another search option, but not fast 
each of the 3 according to their position...
i.e 2^... trimming and structuring from smallest to largest, so that all sets are completely filled from 1 to 1000, from 1 to 1000000 etc.
2 numbers
970
436
974
005
6 05 0 0 1
51 50 1 1 0
023
690
481
3 numbers
970
80 079 2 1 0
98 097 1 2 0
710 709 2 0 1
791 790 1 0 2
908 907 0 2 1
971 970 0 1 2
436
347 346 1 0 2
365 364 2 0 1
437 436 0 1 2
464 463 0 2 1
635 634 2 1 0
644 643 1 2 0
974
480 479 2 1 0
498 497 1 2 0
750 749 2 0 1
795 794 1 0 2
948 947 0 2 1
975 974 0 1 2
005
6 005 0 0 2
16 015 0 0 2
26 025 0 0 2
36 035 0 0 2
46 045 0 0 2
51 050 0 0 1
52 051 0 0 1
53 052 0 0 1
54 053 0 0 1
55 054 0 0 1
56 055 0 0 1
57 056 0 0 1
58 057 0 0 1
59 058 0 0 1
60 059 0 0 1
66 065 0 0 2
76 075 0 0 2
86 085 0 0 2
96 095 0 0 2
106 105 1 1 2
151 150 2 2 1
206 205 1 1 2
251 250 2 2 1
306 305 1 1 2
351 350 2 2 1
406 405 1 1 2
451 450 2 2 1
501 500 1 1 0
502 501 1 1 0
503 502 1 1 0
504 503 1 1 0
505 504 1 1 0
506 505 1 1 0
507 506 1 1 0
508 507 1 1 0
509 508 1 1 0
510 509 1 1 0
511 510 2 2 0
521 520 2 2 0
531 530 2 2 0
541 540 2 2 0
551 550 2 2 0
561 560 2 2 0
571 570 2 2 0
581 580 2 2 0
591 590 2 2 0
606 605 1 1 2
651 650 2 2 1
706 705 1 1 2
751 750 2 2 1
806 805 1 1 2
851 850 2 2 1
906 905 1 1 2
951 950 2 2 1
023
24 023 0 1 2
33 032 0 2 1
204 203 1 0 2
231 230 2 0 1
303 302 1 2 0
321 320 2 1 0
690
70 069 1 2 0
97 096 2 1 0
610 609 0 2 1
691 690 0 1 2
907 906 2 0 1
961 960 1 0 2
481
149 148 1 2 0
185 184 2 1 0
419 418 0 2 1
482 481 0 1 2
815 814 2 0 1
842 841 1 0 2
or 6 numbers
well, we get at 3 numbers combinations 0 0 0 2 2 2, at 6 nembers 0 0 0 4 4 4. 2x2x2=8 8x8x8x8x8x8x8 = 2097152, 4x4x4 = 64 64x64x64x64x64x64x64 = 44392781971456
at 3 numbers the spread is large for step by step
at 6 nembers less spread but more combinations 0 0 0 4 4 4.
that is, the search looks like this
970
530 04739 4 2 0
436
516 04639 1 3 2
974
530 04739 4 2 1
005
500 04507 0 0 2
023
584 05236 0 2 3
690
516 04639 2 4 0
481
511 04581 1 3 4
from 500 to 600 step by step 100×100×100×100×100×100×100 = 100000000000000
maybe can get something out of this...
or for 22 pz. lenght 0 1, 10×10×10×10×10×10×10×10×10×10×10= 100000000000
9
50 49 1
60 59 1
7
58 57 1
0
51 50 1
4
50 49 0
55 54 1
3
54 53 1
6
57 56 1
9
50 49 1
60 59 1
7
58 57 1
4
50 49 0
55 54 1
0
51 50 1
0
51 50 1
5
51 50 0
52 51 0
53 52 0
54 53 0
55 54 0
56 55 0
57 56 0
58 57 0
59 58 0
60 59 0
0
51 50 1
2
53 52 1
3
54 53 1
6
57 56 1
9
50 49 1
60 59 1
0
51 50 1
4
50 49 0
55 54 1
8
59 58 1
1
52 51 1
1
52 51 1
if I think correctly 2 positions 0 and 1 for 11 seats calculated as 2×2×2×2×2×2×2×2×2×2×2=2048, by 10 step by step, 10×10×10×10×10×10×10×10×10×10×10= 100000000000, 2048×100000000000 = 204800000000000.
204800000000000
vs
9999999999999999999999
although need 10 hike 22 multiply(( but anyway an interesting result from 50 to 60, from 0 to 10 another result.
111111101111111 119666659114170 32639
111111111111111 191206974700443 32767
111111110111010 409118905032525 32698
111111111111111 611140496167764 32767
1101111010101111 2058769515153876 57007
1111110111111111 4216495639600700 65023
1111111111111111 6763683971478124 65535
1111100111111111 9974455244496707 63999
11110111111111111 30045390491869460 126975
11111111111111010 44218742292676575 131066
111110101111111111 138245758910846492 257023
111111111111111111 199976667976342049 262143
010111111101111111 525070384258266191 98175
1110111101111111111 1135041350219496382 490495
1110111011111101111 1425787542618654982 489455
1111111011011111111 3908372542507822062 521983
1111111111011111111 8993229949524469768 524031
11011111111111111100 30568377312064202855 917500
111111111110111111111 970436974005023690481 2096639
11011111111111111111111 22538323240989823823367 7340031
...
4
50 49 0
55 54 1
...
such can be considered by 1
each of the 3 according to their position...
i.e 2^... trimming and structuring from smallest to largest, so that all sets are completely filled from 1 to 1000, from 1 to 1000000 etc.
2 numbers
970
436
974
005
6 05 0 0 1
51 50 1 1 0
023
690
481
3 numbers
970
80 079 2 1 0
98 097 1 2 0
710 709 2 0 1
791 790 1 0 2
908 907 0 2 1
971 970 0 1 2
436
347 346 1 0 2
365 364 2 0 1
437 436 0 1 2
464 463 0 2 1
635 634 2 1 0
644 643 1 2 0
974
480 479 2 1 0
498 497 1 2 0
750 749 2 0 1
795 794 1 0 2
948 947 0 2 1
975 974 0 1 2
005
6 005 0 0 2
16 015 0 0 2
26 025 0 0 2
36 035 0 0 2
46 045 0 0 2
51 050 0 0 1
52 051 0 0 1
53 052 0 0 1
54 053 0 0 1
55 054 0 0 1
56 055 0 0 1
57 056 0 0 1
58 057 0 0 1
59 058 0 0 1
60 059 0 0 1
66 065 0 0 2
76 075 0 0 2
86 085 0 0 2
96 095 0 0 2
106 105 1 1 2
151 150 2 2 1
206 205 1 1 2
251 250 2 2 1
306 305 1 1 2
351 350 2 2 1
406 405 1 1 2
451 450 2 2 1
501 500 1 1 0
502 501 1 1 0
503 502 1 1 0
504 503 1 1 0
505 504 1 1 0
506 505 1 1 0
507 506 1 1 0
508 507 1 1 0
509 508 1 1 0
510 509 1 1 0
511 510 2 2 0
521 520 2 2 0
531 530 2 2 0
541 540 2 2 0
551 550 2 2 0
561 560 2 2 0
571 570 2 2 0
581 580 2 2 0
591 590 2 2 0
606 605 1 1 2
651 650 2 2 1
706 705 1 1 2
751 750 2 2 1
806 805 1 1 2
851 850 2 2 1
906 905 1 1 2
951 950 2 2 1
023
24 023 0 1 2
33 032 0 2 1
204 203 1 0 2
231 230 2 0 1
303 302 1 2 0
321 320 2 1 0
690
70 069 1 2 0
97 096 2 1 0
610 609 0 2 1
691 690 0 1 2
907 906 2 0 1
961 960 1 0 2
481
149 148 1 2 0
185 184 2 1 0
419 418 0 2 1
482 481 0 1 2
815 814 2 0 1
842 841 1 0 2
or 6 numbers
Quote
count = 0
v = []
S=[]
Nn =['970', '436', '974', '005', '023', '690', '481'] #970 436 974 005 023 690 481
for elem in Nn:
h1 = elem #970436974005023690481 108717677802902655490863978809014617508706260176761067916188909553243434167589
print(" ")
print(elem)
g1 = ([h1[i:i + 1] for i in range(0, len(h1), 1)])
g = g1
i = 256
while i <= 10000:
a=pow(2,i)
b = str(a)[1:6]
v.append(b)
i=i+1
j = set(v)
jj = sorted(j)
for elem in jj:
count += 1
if g[0] in elem:
if g[1] in elem:
if g[2] in elem:
print(count,elem,elem.index(g[0]),elem.index(g[1]),elem.index(g[2]))
j = (elem,elem.index(g[0]),elem.index(g[1]),elem.index(g[2]))
S.append(j)
#print(sorted(S))
#print(elem)
count = 0
v = []
S=[]
Nn =['970', '436', '974', '005', '023', '690', '481'] #970 436 974 005 023 690 481
for elem in Nn:
h1 = elem #970436974005023690481 108717677802902655490863978809014617508706260176761067916188909553243434167589
print(" ")
print(elem)
g1 = ([h1[i:i + 1] for i in range(0, len(h1), 1)])
g = g1
i = 256
while i <= 10000:
a=pow(2,i)
b = str(a)[1:6]
v.append(b)
i=i+1
j = set(v)
jj = sorted(j)
for elem in jj:
count += 1
if g[0] in elem:
if g[1] in elem:
if g[2] in elem:
print(count,elem,elem.index(g[0]),elem.index(g[1]),elem.index(g[2]))
j = (elem,elem.index(g[0]),elem.index(g[1]),elem.index(g[2]))
S.append(j)
#print(sorted(S))
#print(elem)
count = 0
well, we get at 3 numbers combinations 0 0 0 2 2 2, at 6 nembers 0 0 0 4 4 4. 2x2x2=8 8x8x8x8x8x8x8 = 2097152, 4x4x4 = 64 64x64x64x64x64x64x64 = 44392781971456
at 3 numbers the spread is large for step by step
at 6 nembers less spread but more combinations 0 0 0 4 4 4.
that is, the search looks like this
970
530 04739 4 2 0
436
516 04639 1 3 2
974
530 04739 4 2 1
005
500 04507 0 0 2
023
584 05236 0 2 3
690
516 04639 2 4 0
481
511 04581 1 3 4
from 500 to 600 step by step 100×100×100×100×100×100×100 = 100000000000000
maybe can get something out of this...
or for 22 pz. lenght 0 1, 10×10×10×10×10×10×10×10×10×10×10= 100000000000
Quote
count = 0
v = []
S=[]
h2 = ("9704369740050236904811") #970436974005023690481 108717677802902655490863978809014617508706260176761067916188909553243434167589
g2 = ([h2[i:i + 1] for i in range(0, len(h2), 1)])
Nn =g2 #970 436 974 005 023 690 481
for elem in Nn:
h1 = elem #970436974005023690481 108717677802902655490863978809014617508706260176761067916188909553243434167589
print(" ")
print(elem)
g1 = ([h1[i:i + 1] for i in range(0, len(h1), 1)])
g = g1
i = 256
while i <= 10000:
a=pow(2,i)
b = str(a)[1:3]
v.append(b)
i=i+1
j = set(v)
jj = sorted(j)
for elem in jj:
count += 1
if g[0] in elem:
if count >= 50:
if count <= 60:
print(count,elem,elem.index(g[0]))
j = (elem,elem.index(g[0]))
S.append(j)
#print(sorted(S))
#print(elem)
count = 0
v = []
S=[]
h2 = ("9704369740050236904811") #970436974005023690481 108717677802902655490863978809014617508706260176761067916188909553243434167589
g2 = ([h2[i:i + 1] for i in range(0, len(h2), 1)])
Nn =g2 #970 436 974 005 023 690 481
for elem in Nn:
h1 = elem #970436974005023690481 108717677802902655490863978809014617508706260176761067916188909553243434167589
print(" ")
print(elem)
g1 = ([h1[i:i + 1] for i in range(0, len(h1), 1)])
g = g1
i = 256
while i <= 10000:
a=pow(2,i)
b = str(a)[1:3]
v.append(b)
i=i+1
j = set(v)
jj = sorted(j)
for elem in jj:
count += 1
if g[0] in elem:
if count >= 50:
if count <= 60:
print(count,elem,elem.index(g[0]))
j = (elem,elem.index(g[0]))
S.append(j)
#print(sorted(S))
#print(elem)
count = 0
9
50 49 1
60 59 1
7
58 57 1
0
51 50 1
4
50 49 0
55 54 1
3
54 53 1
6
57 56 1
9
50 49 1
60 59 1
7
58 57 1
4
50 49 0
55 54 1
0
51 50 1
0
51 50 1
5
51 50 0
52 51 0
53 52 0
54 53 0
55 54 0
56 55 0
57 56 0
58 57 0
59 58 0
60 59 0
0
51 50 1
2
53 52 1
3
54 53 1
6
57 56 1
9
50 49 1
60 59 1
0
51 50 1
4
50 49 0
55 54 1
8
59 58 1
1
52 51 1
1
52 51 1
if I think correctly 2 positions 0 and 1 for 11 seats calculated as 2×2×2×2×2×2×2×2×2×2×2=2048, by 10 step by step, 10×10×10×10×10×10×10×10×10×10×10= 100000000000, 2048×100000000000 = 204800000000000.
204800000000000
vs
9999999999999999999999
although need 10 hike 22 multiply(( but anyway an interesting result from 50 to 60, from 0 to 10 another result.
111111101111111 119666659114170 32639
111111111111111 191206974700443 32767
111111110111010 409118905032525 32698
111111111111111 611140496167764 32767
1101111010101111 2058769515153876 57007
1111110111111111 4216495639600700 65023
1111111111111111 6763683971478124 65535
1111100111111111 9974455244496707 63999
11110111111111111 30045390491869460 126975
11111111111111010 44218742292676575 131066
111110101111111111 138245758910846492 257023
111111111111111111 199976667976342049 262143
010111111101111111 525070384258266191 98175
1110111101111111111 1135041350219496382 490495
1110111011111101111 1425787542618654982 489455
1111111011011111111 3908372542507822062 521983
1111111111011111111 8993229949524469768 524031
11011111111111111100 30568377312064202855 917500
111111111110111111111 970436974005023690481 2096639
11011111111111111111111 22538323240989823823367 7340031
...
4
50 49 0
55 54 1
...
such can be considered by 1
Precisely it will be:
<8000000000000000/7cce5efdaccf6808, ffffffffffffffff/7cce5efdaccf6808>
;-)
<8000000000000000/7cce5efdaccf6808, ffffffffffffffff/7cce5efdaccf6808>
;-)
For those who do not believe the keys are random: what would be the next value?
< 3.5
For those who do not believe the keys are random: what would be the next value?
< 2.5
For those who do not believe the keys are random: what would be the next value?
rosengold, is it available to test freely.. or not yet.? thanks
not yet, I want to show a minimum working script in the bits +50, I'll share when I finish it.
Hi guys,
In continuation to this thread: https://bitcointalksearch.org/topic/brute-force-on-bitcoin-addresses-video-of-the-action-1305887
While playing around with my bot, I found out this mysterious transaction:
https://blockchain.info/tx/08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15
those 32.896 BTC were sent to multiple addresses, all the private keys of those addresses seem to be generated by some kind of formula.
For example:
Address 2:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb
Biginteger PVK value: 3
Hex PVK value: 3
Address 3:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU76rnZwVdz
19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA
Biginteger PVK value: 7
Hex PVK value: 7
Address 4:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU77MfhviY5
1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e
Biginteger PVK value: 8
Hex PVK value: 8
Address 5:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Dq8Au4Pv
1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k
Biginteger PVK value: 21
Hex PVK value: 15
Address 6:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Tmu6qHxS
1PitScNLyp2HCygzadCh7FveTnfmpPbfp8
Biginteger PVK value: 49
Hex PVK value: 31
Address 7:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7hDgvu64y
1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC
Biginteger PVK value: 76
Hex PVK value: 4C
Address 8:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU8xvGK1zpm
1M92tSqNmQLYw33fuBvjmeadirh1ysMBxK
Biginteger PVK value: 224
Hex PVK value: E0
Address 9:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUB3vfDKcxZ
1CQFwcjw1dwhtkVWBttNLDtqL7ivBonGPV
Biginteger PVK value: 467
Hex PVK value: 1d3
Address 10:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUBTL67V6dE
1LeBZP5QCwwgXRtmVUvTVrraqPUokyLHqe
Biginteger PVK value: 514
Hex PVK value: 202
Address 11:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUGxXgtm63M
1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu
Biginteger PVK value: 1155
Hex PVK value: 483
Address 12:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUW5RtS2JN1
1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot
Biginteger PVK value: 2683
Hex PVK value: a7b
Address 13:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUspniiQZds
1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1
Biginteger PVK value: 5216
Hex PVK value: 1460
Address 14:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFVfZyiN5iEG
1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk
Biginteger PVK value: 10544
Hex PVK value: 2930
and so on...
until the addresses 50 (1MEzite4ReNuWaL5Ds17ePKt2dCxWEofwk) it was already cracked by someone.
Any ideas what's the formula behind the generation of these addresses?
Address 2, pvk decimal value: 3
Address 3, pvk decimal value: 7
Address 4, pvk decimal value: 8
Address 5, pvk decimal value: 21
Address 6, pvk decimal value: 49
Address 7, pvk decimal value: 76
Address 8, pvk decimal value: 224
Address 9, pvk decimal value: 467
Address 10, pvk decimal value: 514
Address 11, pvk decimal value: 1155
Address 12, pvk decimal value: 2683
Address 13, pvk decimal value: 5216
Address 14, pvk decimal value: 10544
Address 15 and after, pvk decimal value: ?
The prize would be ~32 BTC
EDIT: If you find the solution feel free to leave a tip 1DPUhjHvd2K4ZkycVHEJiN6wba79j5V1u3
In continuation to this thread: https://bitcointalksearch.org/topic/brute-force-on-bitcoin-addresses-video-of-the-action-1305887
While playing around with my bot, I found out this mysterious transaction:
https://blockchain.info/tx/08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15
those 32.896 BTC were sent to multiple addresses, all the private keys of those addresses seem to be generated by some kind of formula.
For example:
Address 2:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb
Biginteger PVK value: 3
Hex PVK value: 3
Address 3:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU76rnZwVdz
19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA
Biginteger PVK value: 7
Hex PVK value: 7
Address 4:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU77MfhviY5
1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e
Biginteger PVK value: 8
Hex PVK value: 8
Address 5:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Dq8Au4Pv
1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k
Biginteger PVK value: 21
Hex PVK value: 15
Address 6:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Tmu6qHxS
1PitScNLyp2HCygzadCh7FveTnfmpPbfp8
Biginteger PVK value: 49
Hex PVK value: 31
Address 7:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7hDgvu64y
1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC
Biginteger PVK value: 76
Hex PVK value: 4C
Address 8:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU8xvGK1zpm
1M92tSqNmQLYw33fuBvjmeadirh1ysMBxK
Biginteger PVK value: 224
Hex PVK value: E0
Address 9:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUB3vfDKcxZ
1CQFwcjw1dwhtkVWBttNLDtqL7ivBonGPV
Biginteger PVK value: 467
Hex PVK value: 1d3
Address 10:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUBTL67V6dE
1LeBZP5QCwwgXRtmVUvTVrraqPUokyLHqe
Biginteger PVK value: 514
Hex PVK value: 202
Address 11:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUGxXgtm63M
1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu
Biginteger PVK value: 1155
Hex PVK value: 483
Address 12:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUW5RtS2JN1
1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot
Biginteger PVK value: 2683
Hex PVK value: a7b
Address 13:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUspniiQZds
1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1
Biginteger PVK value: 5216
Hex PVK value: 1460
Address 14:
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFVfZyiN5iEG
1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk
Biginteger PVK value: 10544
Hex PVK value: 2930
and so on...
until the addresses 50 (1MEzite4ReNuWaL5Ds17ePKt2dCxWEofwk) it was already cracked by someone.
Any ideas what's the formula behind the generation of these addresses?
Address 2, pvk decimal value: 3
Address 3, pvk decimal value: 7
Address 4, pvk decimal value: 8
Address 5, pvk decimal value: 21
Address 6, pvk decimal value: 49
Address 7, pvk decimal value: 76
Address 8, pvk decimal value: 224
Address 9, pvk decimal value: 467
Address 10, pvk decimal value: 514
Address 11, pvk decimal value: 1155
Address 12, pvk decimal value: 2683
Address 13, pvk decimal value: 5216
Address 14, pvk decimal value: 10544
Address 15 and after, pvk decimal value: ?
The prize would be ~32 BTC
EDIT: If you find the solution feel free to leave a tip
1DPUhjHvd2K4ZkycVHEJiN6wba79j5V1u3i do not see the possibility in solving this complex technology with the help of this detail as it is almost not to be computed by humans but by a computer. although its an old post but then it is simply a computerized puzzle.
rosengold, is it available to test freely.. or not yet.? thanks
At this time we have two possibilities well tested to solve #64 or #120:
for #64 once we don't know this public key we need to find solutions by brute force searching the range in a faster way that we can, (maybe using steps ? looking for patterns ?).
for #120 and others known public keys you can try (by your own luck) for example the Jean_Luc solution for server, rent some GPU power at Vast.ai and TRY IT, of course there's no warranty.
I'm working on something useful in python but it is very slow, that's why I convinced that the future of this puzzle is the computer distributed solutions by GPU.
for #64 once we don't know this public key we need to find solutions by brute force searching the range in a faster way that we can, (maybe using steps ? looking for patterns ?).
for #120 and others known public keys you can try (by your own luck) for example the Jean_Luc solution for server, rent some GPU power at Vast.ai and TRY IT, of course there's no warranty.
I'm working on something useful in python but it is very slow, that's why I convinced that the future of this puzzle is the computer distributed solutions by GPU.
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