cuddlefish


July 19, 2011, 09:17:30 AM 

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


July 19, 2011, 09:18:47 AM 

Oh, the address with the 5 BTC is 1HqtKWKCLTs4eUuvTDPpJC4AaBYMiiXWok





cuddlefish


July 20, 2011, 05:22:09 AM 

>>> 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_


July 20, 2011, 06:49:59 AM 

>>> 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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Activity: 1260


July 20, 2011, 07:11:54 AM 

>>> 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


July 20, 2011, 02:19:23 PM 

Ooops.
>>> random.seed(3) >>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e1] ... 39 e 89 4 60 B 98 4 >>>




anthony_


July 20, 2011, 08:49:17 PM 

Ooops.
>>> random.seed(3) >>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e1] ... 39 e 89 4 60 B 98 4 >>>
it appears we get the first 8 digits then =P




cuddlefish


July 21, 2011, 02:14:27 AM 

Ooops.
>>> random.seed(3) >>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e1] ... 39 e 89 4 60 B 98 4 >>>
it appears we get the first 8 digits then =P No, those are the 0indexed positions.




anthony_


July 21, 2011, 05:05:23 AM 

Ooops.
>>> random.seed(3) >>> for i in range(4): e = random.randint(1, len(r)); print e; print r[e1] ... 39 e 89 4 60 B 98 4 >>>
it appears we get the first 8 digits then =P No, those are the 0indexed positions. We already got the next 4 numbers for that seed aswell is what i'm saying




mc_lovin
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www.bitcointrading.com


July 21, 2011, 09:06:22 PM 

interesting little game.




