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Author Topic: Private key for address with 5 BTC.  (Read 2661 times)
cuddlefish
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July 19, 2011, 09:17:30 AM
 #1

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.

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cuddlefish
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July 19, 2011, 09:18:47 AM
 #2

Oh, the address with the 5 BTC is 1HqtKWKCLTs4eUuvTDPpJC4AaBYMiiXWok
anthony_
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July 20, 2011, 03:49:25 AM
 #3

Sent 0.1BTC for the first four characters to get things started.
http://blockexplorer.com/tx/5bc74811af3886fbb4ef551fa6e28cabe02ee8470559bcbbebf155c858aae361
cuddlefish
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July 20, 2011, 05:22:09 AM
 #4

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_
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July 20, 2011, 06:49:59 AM
 #5

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
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July 20, 2011, 07:11:54 AM
 #6

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
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July 20, 2011, 02:19:23 PM
 #7

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_
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July 20, 2011, 08:49:17 PM
 #8

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
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July 21, 2011, 02:14:27 AM
 #9

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_
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July 21, 2011, 05:05:23 AM
 #10

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 Wink
mc_lovin
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July 21, 2011, 09:06:22 PM
 #11

interesting little game.

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