Private key for address with 5 BTC.

[cuddlefish]

For each 0.025 BTC that shows up at 1NFXQfoqGJBNuEXMSYJwUYGt5bJF6jbFrg past the date of this post, I will reveal a random digit (as shown by Python's random.randint(), seeded with the post number.) of the key.

It's in PEM format.

-----BEGIN EC PRIVATE KEY-----
****************************************************************
****************/***********************************************
******************************==
-----END EC PRIVATE KEY-----

As an extra wrinkle, I'll give you the sha256 hash of the private key, for your bruteforcing pleasure.

>>> hashlib.sha256(data).hexdigest()
'0d2c8b5e5aa8ed0b2c10ad8d0940ee24e5df5e48160fe09b31efeaa8e72f6083'

This is the hash of ONLY the inner part (not ---BEGIN EC PRIVATE KEY--- or ---END EC PRIVATE KEY---)
There are no newlines at the ends, only between lines.

[1715226321]

1715226321

[1715226321]

1715226321

[1715226321]

1715226321

[cuddlefish]

Oh, the address with the 5 BTC is 1HqtKWKCLTs4eUuvTDPpJC4AaBYMiiXWok

[anthony_]

Sent 0.1BTC for the first four characters to get things started.
http://blockexplorer.com/tx/5bc74811af3886fbb4ef551fa6e28cabe02ee8470559bcbbebf155c858aae361

[cuddlefish]

Sent 0.1BTC for the first four characters to get things started.
http://blockexplorer.com/tx/5bc74811af3886fbb4ef551fa6e28cabe02ee8470559bcbbebf155c858aae361

>>> random.seed(3)
>>> for i in range(4): print random.randint(1, len(data))
...
39
89
60
98
>>> for i in range(4): print r[random.randint(1, len(data))]
...
f
7
C
0

[anthony_]

Sent 0.1BTC for the first four characters to get things started.
http://blockexplorer.com/tx/5bc74811af3886fbb4ef551fa6e28cabe02ee8470559bcbbebf155c858aae361

>>> random.seed(3)
>>> for i in range(4): print random.randint(1, len(data))
...
39
89
60
98
>>> for i in range(4): print r[random.randint(1, len(data))]
...
f
7
C
0

from that output it doesn't show that you type random.seed the second time, so it would have continued from the last point in the PRNG and the f 7 C 0 would be wrong

[Maged]

Sent 0.1BTC for the first four characters to get things started.
http://blockexplorer.com/tx/5bc74811af3886fbb4ef551fa6e28cabe02ee8470559bcbbebf155c858aae361

>>> random.seed(3)
>>> for i in range(4): print random.randint(1, len(data))
...
39
89
60
98
>>> for i in range(4): print r[random.randint(1, len(data))]
...
f
7
C
0

from that output it doesn't show that you type random.seed the second time, so it would have continued from the last point in the PRNG and the f 7 C 0 would be wrong

If that's true, then this should be the correct assignment...

>>> random.seed(3)
>>> for i in range(4): print random.randint(1, 162)
...
39
89
60
98
>>> for i in range(4): print random.randint(1, 162)
...
102
11
3
136

[cuddlefish]

Ooops.


>>> random.seed(3)
>>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e-1]
...
39
e
89
4
60
B
98
4
>>>

[anthony_]

Ooops.


>>> random.seed(3)
>>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e-1]
...
39
e
89
4
60
B
98
4
>>>

it appears we get the first 8 digits then =P

[cuddlefish]

Ooops.


>>> random.seed(3)
>>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e-1]
...
39
e
89
4
60
B
98
4
>>>

it appears we get the first 8 digits then =P

No, those are the 0-indexed positions.

[anthony_]

Ooops.


>>> random.seed(3)
>>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e-1]
...
39
e
89
4
60
B
98
4
>>>

it appears we get the first 8 digits then =P

No, those are the 0-indexed positions.

We already got the next 4 numbers for that seed as-well is what i'm saying

[mc_lovin]

interesting little game.

[PawGo]

ho ho ho
What a puzzle 11 year old.
I see owner was not active during last 4 years - strange that coins are still there.

It gives known part:

Code:

MHQCAQEEI**7***************************e************oAcGBSuBBAAKoUQDQgAE********/********4*******4****f*********************************0*********************==

so it is rather not possible to bruteforce for a given hash...

[Seraphimjm]

Would have probably taken the coins sent to reveal and up to the point where it was almost easily solvable just move the coins then declare found/solved.

classic raffle ticket scheme with a little slight of hand.  The first 0.1btc sent was the bait, but there were no fish... 

[paapa1268]

ho ho ho
What a puzzle 11 year old.
I see owner was not active during last 4 years - strange that coins are still there.

It gives known part:

Code:

MHQCAQEEI**7***************************e************oAcGBSuBBAAKoUQDQgAE********/********4*******4****f*********************************0*********************==

so it is rather not possible to bruteforce for a given hash...

the only solution is to use the same version script he used and repeate process until got hit the version used by him can be seen https://lapo.it/asn1js decoder see his other post he posted random pem key decode using above mention site and in option it will give all version info used