cuddlefish (OP)
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July 19, 2011, 09:17:30 AM Last edit: July 19, 2011, 09:28:05 AM by cuddlefish |
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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 (OP)
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July 19, 2011, 09:18:47 AM |
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Oh, the address with the 5 BTC is 1HqtKWKCLTs4eUuvTDPpJC4AaBYMiiXWok
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anthony_
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July 20, 2011, 03:49:25 AM |
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cuddlefish (OP)
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July 20, 2011, 05:22:09 AM |
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>>> 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
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anthony_
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July 20, 2011, 06:49:59 AM |
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>>> 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
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Maged
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July 20, 2011, 07:11:54 AM Last edit: July 20, 2011, 07:54:08 PM by Maged |
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>>> 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
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cuddlefish (OP)
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July 20, 2011, 02:19:23 PM |
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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 >>>
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anthony_
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July 20, 2011, 08:49:17 PM |
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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
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cuddlefish (OP)
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July 21, 2011, 02:14:27 AM |
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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.
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anthony_
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July 21, 2011, 05:05:23 AM |
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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
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mc_lovin
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www.bitcointrading.com
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July 21, 2011, 09:06:22 PM |
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interesting little game.
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PawGo
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April 23, 2022, 05:05:04 PM |
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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: MHQCAQEEI**7***************************e************oAcGBSuBBAAKoUQDQgAE********/********4*******4****f*********************************0*********************==
so it is rather not possible to bruteforce for a given hash...
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Seraphimjm
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April 23, 2022, 05:13:52 PM |
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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...
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paapa1268
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April 01, 2024, 11:44:33 AM |
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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: 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
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