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    Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 324898 times)
    Feron
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    April 15, 2022, 09:28:02 AM
    Last edit: April 15, 2022, 11:48:31 AM by Feron
     #1841

    mathematics multiples scanner
    Code:
    from math import ceil,floor
    import secp256k1 as ice
    #from bit import Key
    import random
    def times(x,y):
     if x < 10 and y < 10:
      return x*y
     xxx = max(len(str(x)),len(str(y)))
     vvv = ceil(xxx/2)
     kxk = floor(x 10**vvv)
     sxs = x % (10**vvv)
     kyk = floor(y 10**vvv)
     sys = y % (10**vvv)
     aaa = times(kxk,kyk)
     bbb = times(sxs,sys)
     ccc = times(kxk + sxs,kyk + sys) - aaa - bbb
     return int(aaa*(10**(vvv*2)) + ccc*(10**vvv) + bbb)
    for xx in range(1,1000000000000000):
     #ra = random.randint(4611686018427387904,9223372036854775808)
     vv = times(xx,xx)
     ke = ice.privatekey_to_h160(0,True,vv).hex()
     if (ke).endswith("48a4"):
      print(xx,"x",xx,"=",vv,ke)
     if (ke) == "3ee4133d991f52fdf6a25c9834e0745ac74248a4":
      f=open("test.txt","a")
      f.write(str(vv)+"-"+(ke)+"\n")
      f.close()
    you can use a random multiples stack
    random mode
    Code:
    from math import ceil,floor
    import secp256k1 as ice
    #from bit import Key
    import random
    def times(x,y):
     if x < 10 and y < 10:
      return x*y
     xxx = max(len(str(x)),len(str(y)))
     vvv = ceil(xxx/2)
     kxk = floor(x 10**vvv)
     sxs = x % (10**vvv)
     kyk = floor(y 10**vvv)
     sys = y % (10**vvv)
     aaa = times(kxk,kyk)
     bbb = times(sxs,sys)
     ccc = times(kxk + sxs,kyk + sys) - aaa - bbb
     return int(aaa*(10**(vvv*2)) + ccc*(10**vvv) + bbb)
    for xx in range(1,1000000000):
     ra = random.randint(3040000000,4300000000)
     vv = times(ra,ra)
     ke = ice.privatekey_to_h160(0,True,vv).hex()
     if (ke).endswith("48a4"):
      print(ra,"x",ra,"=",vv,ke)
     if (ke) == "3ee4133d991f52fdf6a25c9834e0745ac74248a4":
      f=open("test.txt","a")
      f.write(str(vv)+"-"+(ke)+"\n")
      f.close()
    random calculator mode
    Code:
    from math import ceil,floor
    import secp256k1 as ice
    #from bit import Key
    import random
    def times(x,y):
     if x < 10 and y < 10:
      return x*y
     xxx = max(len(str(x)),len(str(y)))
     vvv = ceil(xxx/2)
     kxk = floor(x 10**vvv)
     sxs = x % (10**vvv)
     kyk = floor(y 10**vvv)
     sys = y % (10**vvv)
     aaa = times(kxk,kyk)
     bbb = times(sxs,sys)
     ccc = times(kxk + sxs,kyk + sys) - aaa - bbb
     return int(aaa*(10**(vvv*2)) + ccc*(10**vvv) + bbb)
    for xx in range(1,1000000000):
     ra = random.randint(3040000000,4300000000)
     rb = random.randint(3040000000,4300000000)
     vv = times(ra,rb)
     ke = ice.privatekey_to_h160(0,True,vv).hex()
     if (ke).endswith("48a4"):
      print(ra,"x",rb,"=",vv,ke)
     if (ke) == "3ee4133d991f52fdf6a25c9834e0745ac74248a4":
      f=open("test.txt","a")
      f.write(str(vv)+"-"+(ke)+"\n")
      f.close()
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