FiatKiller
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February 26, 2014, 07:32:01 PM |
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Sh&^! I ve got it k = 254634957345987934598347597261842936423946893247. will I get the price? please? nope, because your solution is wrong: ======================================================================= 0x9772c99f2db406a441000d9288ba9cd7ed05e7a0ae02e09615b7cd39cb8b1cedL not matching 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798L I got it! 55066263022277343669578718895168534326250603453777594175500187360389116729240 nope, not that I see - but thanks for scaring me lol
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Kyraishi
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February 26, 2014, 07:52:47 PM |
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what is that k for ?
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serje
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Activity: 1232
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February 26, 2014, 07:56:44 PM |
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what is that k for ?
If you find the K you will find the private key to the address written in that script! so you can import that address to your wallet unfortunately it has 0 BTC
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Space for rent if its still trending
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Skoupi
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Skoupi the Great
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February 26, 2014, 10:28:02 PM |
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Guys seriously,
OP is either drop dead idiot or just amuses himself with everyone around that tries to come up with a "solution".
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Kiki112
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February 26, 2014, 10:41:44 PM |
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4210
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serje
Legendary
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February 26, 2014, 10:51:06 PM |
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4210
1893710265418723563212490
Guys this is not lotto! you have to run the script from the first page and replace K with your number!
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Space for rent if its still trending
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serje
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February 26, 2014, 11:15:47 PM |
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Answer is either 47, 69,4371, 8331, 9832, 184833321, or 4532231082
My Bitcoin Address: 1MYpNKj25HRBFpv22YpuZsuz2zZHKBLUur
Thanks.
the number can be this long 115,792,089,237,316,195,423,570,985,008,690,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 are you sure your answers are correct? did you even bother checking them?
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Space for rent if its still trending
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FiatKiller
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February 27, 2014, 04:05:58 PM |
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EK, we need to confirm the looping code as I pointed out in my last post, otherwise we are REALLY wasting time and even a 1000 years will not solve it. Don't you need ppp=challenge.point in the loop for each new value of k? Also, didn't you leave out the "k*" in the loop? thanks Is their anyway to make the python script simply loop and keep adding + 1 to k? Or choose a random number for k?
Yes, this would be the following code. This will however (at least I think so) not be very promising: #! /usr/bin/env python
import random import array import cPickle import struct
class CurveFp( object ): def __init__( self, p, a, b ): self.__p = p self.__a = a self.__b = b
def p( self ): return self.__p
def a( self ): return self.__a
def b( self ): return self.__b
def contains_point( self, x, y ): return ( y * y - ( x * x * x + self.__a * x + self.__b ) ) % self.__p == 0
class Point( object ): def __init__( self, curve, x, y, order = None ): self.__curve = curve self.__x = x self.__y = y self.__order = order if self.__curve: assert self.__curve.contains_point( x, y ) if order: assert self * order == INFINITY def __add__( self, other ): if other == INFINITY: return self if self == INFINITY: return other assert self.__curve == other.__curve if self.__x == other.__x: if ( self.__y + other.__y ) % self.__curve.p() == 0: return INFINITY else: return self.double()
p = self.__curve.p() l = ( ( other.__y - self.__y ) * \ inverse_mod( other.__x - self.__x, p ) ) % p x3 = ( l * l - self.__x - other.__x ) % p y3 = ( l * ( self.__x - x3 ) - self.__y ) % p return Point( self.__curve, x3, y3 )
def negative (self): negative_self = Point( self.__curve, self.__x, -self.__y, self.__order ) return negative_self
def __mul__( self, other ): def leftmost_bit( x ): assert x > 0 result = 1L while result <= x: result = 2 * result return result / 2
e = other if self.__order: e = e % self.__order if e == 0: return INFINITY if self == INFINITY: return INFINITY assert e > 0 e3 = 3 * e negative_self = Point( self.__curve, self.__x, -self.__y, self.__order ) i = leftmost_bit( e3 ) / 2 result = self while i > 1: result = result.double() if ( e3 & i ) != 0 and ( e & i ) == 0: result = result + self if ( e3 & i ) == 0 and ( e & i ) != 0: result = result + negative_self i = i / 2 return result
def __rmul__( self, other ): return self * other
def __str__( self ): if self == INFINITY: return "infinity" return "(%d,%d)" % ( self.__x, self.__y )
def double( self ): if self == INFINITY: return INFINITY
p = self.__curve.p() a = self.__curve.a() l = ( ( 3 * self.__x * self.__x + a ) * \ inverse_mod( 2 * self.__y, p ) ) % p x3 = ( l * l - 2 * self.__x ) % p y3 = ( l * ( self.__x - x3 ) - self.__y ) % p return Point( self.__curve, x3, y3 )
def halve( self ): if self == INFINITY: return INFINITY
p = self.__curve.p() a = self.__curve.a() # next three lines must be reverted somehow, then I am a multi millionaire :-) # as a=0 in this case, I have eliminated it! l = ( ( 3 * self.__x * self.__x ) * inverse_mod( 2 * self.__y, p ) ) % p x3 = ( l * l - 2 * self.__x ) % p y3 = ( l * ( self.__x - x3 ) - self.__y ) % p
return Point( self.__curve, x3, y3 )
def x( self ): return self.__x
def y( self ): return self.__y
def curve( self ): return self.__curve def order( self ): return self.__order INFINITY = Point( None, None, None )
def inverse_mod( a, m ): if a < 0 or m <= a: a = a % m c, d = a, m uc, vc, ud, vd = 1, 0, 0, 1 while c != 0: q, c, d = divmod( d, c ) + ( c, ) uc, vc, ud, vd = ud - q*uc, vd - q*vc, uc, vc assert d == 1 if ud > 0: return ud else: return ud + m
_p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2FL _r = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141L _b = 0x0000000000000000000000000000000000000000000000000000000000000007L _a = 0x0000000000000000000000000000000000000000000000000000000000000000L _Gx = 0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798L _Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8L
class Public_key( object ): def __init__( self, generator, point ): self.curve = generator.curve() self.generator = generator self.point = point n = generator.order() if not n: raise RuntimeError, "Generator point must have order." if not n * point == INFINITY: raise RuntimeError, "Generator point order is bad." if point.x() < 0 or n <= point.x() or point.y() < 0 or n <= point.y(): raise RuntimeError, "Generator point has x or y out of range."
sex = CurveFp( _p, _a, _b ) ass = Point( sex, _Gx, _Gy, _r ) g = ass
if __name__ == "__main__": print '=======================================================================' ### generate privkey challenge = Public_key(g, Point( sex, 0x4641b45737ee8e11ae39899060160507d61a30928b0d3e37b6aede29b4ed807bL, 0xb61b706b81dbb5512c556dfd16815cced84e2fa12b5c8b6440057355f0df2a12L)) ppp=challenge.point
# find the correct k k=random.randrange(1,2**255) # !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ppp=ppp + k*g
while True: ppp=ppp+g k=k+1 if ppp.x() == g.x(): print "found!!!!!!! k=" + hex(k) else: print hex(ppp.x()) + " not matching " + hex(g.x())
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playcoders
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February 27, 2014, 08:00:01 PM |
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found!!!!!!! k=0x0 1ooo75KPdpQyJAGByr5Atm2S8H9DrrhyZ sent to me 1000BTC
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andulolika
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February 28, 2014, 10:09:55 AM |
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This is posible to do only with computing power? Or you can do it by simply math. Edit: if its a random number why don't simply use a bot starting with 1,2,3,4,5,......999,1000,1001....
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joksim299
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Bitdice is scam scam scammmmmmmmmmmmmmmmmmmmmmmmmm
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February 28, 2014, 11:47:54 AM |
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This is posible to do only with computing power? Or you can do it by simply math. Edit: if its a random number why don't simply use a bot starting with 1,2,3,4,5,......999,1000,1001....
Read posts above. Number can be from 1 to 2^256 it may take 1000 years to solve this in that way.
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Cryptofreak82
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March 03, 2014, 08:12:34 PM |
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All I can read in the scrpt is : sex = CurveFp( _p, _a, _b ) ass = Point( sex, _Gx, _Gy, _r ) Sex and Ass? ? Now how should I understand this? Evil?
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playcoders
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March 04, 2014, 12:09:46 AM |
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lol funny
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BlockChainLottery
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March 04, 2014, 11:10:47 AM |
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Using a python script is not a very efficient method of brute forcing it. I have a feeling that a lot of total noobs are seriously thinking that it is possible this way. Then again chance, randomness, and probabilities are very counter intuitive. It is just not possible to grasp the numbers involved.
Does anybody know how much k's can be checked per second with that python script?
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roslinpl
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March 04, 2014, 12:26:50 PM |
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Using a python script is not a very efficient method of brute forcing it. I have a feeling that a lot of total noobs are seriously thinking that it is possible this way. Then again chance, randomness, and probabilities are very counter intuitive. It is just not possible to grasp the numbers involved.
Does anybody know how much k's can be checked per second with that python script?
not a lot not enough to find one hehehe
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roslinpl
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March 04, 2014, 12:27:25 PM |
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lol funny isn't ? for me this was funny as well kind regards!
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FiatKiller
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March 04, 2014, 01:31:31 PM |
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Using a python script is not a very efficient method of brute forcing it. I have a feeling that a lot of total noobs are seriously thinking that it is possible this way. Then again chance, randomness, and probabilities are very counter intuitive. It is just not possible to grasp the numbers involved.
Does anybody know how much k's can be checked per second with that python script?
Winning the lottery is very unlikely also, but someone has to win eventually. If you use random numbers, the very next number chosen COULD be it. What the hell. :-D If you are checking a particular range and it's not in that range, then you are truly wasting time. The only shot you had was in picking that range and no way to know if it is correct beforehand. I'm checking maybe 3 mill a day on one fast PC. But I'm still not convinced the code is correct since EK is not answering.
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BlockChainLottery
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March 04, 2014, 02:40:38 PM |
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Winning the lottery is very unlikely also, but someone has to win eventually. If you use random numbers, the very next number chosen COULD be it. ...
Sure, but you have to include the fact that the number space is huge. The number of tickets in a lottery are at least contained in a pool from which you know a winner is drawn. That is different than looking for a number in a huge range of integers. Simple crude calculation: 2 160 bitcoin addresses 2 billion computers worldwide 13 million addresses checked per day on a computer 22 160 / ( 2e9 * 3e6 * 365 ) = 667e27 years Timeline: big bang sun was born now sun dies existing stars burn out you finding k all matter evaporated |-------------|---------------|--------------|---------------|-----------------------|-----------------| 0 9.2e9 13.8e9 19e9 1e15 667e27 1e34 Entropy is a bitch.
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FiatKiller
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March 04, 2014, 02:59:05 PM |
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dreamkiller... lol
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