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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it (Read 324587 times)

bytcoin

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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

February 15, 2020, 11:17:49 PM

#1361

Quote from: arulbero on February 15, 2020, 11:07:13 PM

Quote from: bytcoin on February 15, 2020, 10:51:12 PM

Quote from: GoldTiger69 on February 15, 2020, 10:05:21 PM

So, w has to values? or maybe is it w1 and w2? That is very confusing.

It got better?

Code:

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = 76470300715912249562689990107401687364194232406198996658976353330269918489458
b = 64658408237276871767689061520961436408509493287485285377611016482361694763299

b_inv = pow(b, p-2, p)
w1 = a*b_inv % p
w1
12447032699845648078645791161909514142990644957498005805208944683777961822095

a_inv = pow(a, p-2, p)
w2 = b*a_inv % p
w2
66620152837833785920928131416087065201280002472666144035333386572317622196480

I'm not good at math ... But I don't think it's that easy! Can you create a tool that I can test? If you really did it ... I'll send you 3 BTC

arulbero

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Merit: 2267

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

February 15, 2020, 11:22:39 PM

#1362

Quote from: bytcoin on February 15, 2020, 11:17:49 PM

I'm not good at math ... But I don't think it's that easy! Can you create a tool that I can test? If you really did it ... I'll send you 3 BTC

If you use python:

copy this file "test.py"

Code:

#!/usr/bin/env python

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = 76470300715912249562689990107401687364194232406198996658976353330269918489458
b = 64658408237276871767689061520961436408509493287485285377611016482361694763299

b_inv = pow(b, p-2, p)
w1 = a*b_inv % p
print (w1)

a_inv = pow(a, p-2, p)
w2 = b*a_inv % p
print (w2)

and then in the terminal:

python test.py

Code:

$ python test.py
12447032699845648078645791161909514142990644957498005805208944683777961822095
66620152837833785920928131416087065201280002472666144035333386572317622196480

bytcoin

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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

February 15, 2020, 11:49:42 PM

#1363

Quote from: arulbero on February 15, 2020, 11:22:39 PM

Quote from: bytcoin on February 15, 2020, 11:17:49 PM

I'm not good at math ... But I don't think it's that easy! Can you create a tool that I can test? If you really did it ... I'll send you 3 BTC

If you use python:

copy this file "test.py"

Code:

and then in the terminal:

python test.py

Code:

$ python test.py
12447032699845648078645791161909514142990644957498005805208944683777961822095
66620152837833785920928131416087065201280002472666144035333386572317622196480

I've never used python ... I'm not good at programming or math.
But I really appreciate your suggestion ... If I achieved my goal ... I will send you 1 BTC

bytcoin

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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

February 16, 2020, 12:15:19 AM
Last edit: February 16, 2020, 12:33:07 AM by bytcoin

#1364

Quote from: arulbero on February 15, 2020, 11:22:39 PM

Quote from: bytcoin on February 15, 2020, 11:17:49 PM

I'm not good at math ... But I don't think it's that easy! Can you create a tool that I can test? If you really did it ... I'll send you 3 BTC

If you use python:

copy this file "test.py"

Code:

and then in the terminal:

python test.py

Code:

$ python test.py
12447032699845648078645791161909514142990644957498005805208944683777961822095
66620152837833785920928131416087065201280002472666144035333386572317622196480

One question ... how did you discover the value of a and b?

Discover this!

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = Huh

??
b =

??

w = GF(p)

w (((a b)))
w (((b a)))

w= 41689968412277401815192940146121615515073176981576112778517871276338175969527
w= 94599322545067982882578206403521588574602656593708975301033069273870448067861

I will send you 1 BTC if you find the value a and b

bytcoin

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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

February 16, 2020, 01:36:12 AM

#1365

Quote from: j2002ba2 on February 15, 2020, 11:06:14 PM

Quote from: bytcoin on February 15, 2020, 09:37:25 PM

Sorry for the post off topic ... but can someone develop a tool to solve this?

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = Huh

?
b =

?

w = GF(p)

w (((a b)))
w (((b a)))

w= 12447032699845648078645791161909514142990644957498005805208944683777961822095
w= 66620152837833785920928131416087065201280002472666144035333386572317622196480

If you can develop any tool that does this ... I will send you 3 BTC

Well, that's easy:

Code:

0 < b < p
a = b * 12447032699845648078645791161909514142990644957498005805208944683777961822095 (mod p)

So, an obvious solution is a=12447032699845648078645791161909514142990644957498005805208944683777961822095, b=1

It's not that easy!
The correct is:
a = 76470300715912249562689990107401687364194232406198996658976353330269918489458
b = 64658408237276871767689061520961436408509493287485285377611016482361694763299

I need a tool that finds the a and b value correctly

arulbero

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Merit: 2267

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

February 16, 2020, 06:55:09 AM
Last edit: February 16, 2020, 07:08:13 AM by arulbero

#1366

Quote from: bytcoin on February 16, 2020, 12:15:19 AM

One question ... how did you discover the value of a and b?

Simply you posted that values and I have seen them before you obfuscated.

Quote from: bytcoin on February 16, 2020, 12:15:19 AM

Discover this!

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = Huh

??
b =

If you mean: retrieve 'a' and 'b' you used from 'w1' and 'w2' is not possible, because given w1 and w2 there is no only a couple (a,b) but many many couples with a/b = w1 and b/a = w2 mod p

For example, for each value of a:

b = w2 * a mod p because b = (b/a) * a mod p

Code:

a = 1 b = 94599322545067982882578206403521588574602656593708975301033069273870448067861
a = 2 b = 73406555852819770341585427798355269296367748908343046219460975406222734641385
a = 10 b = 19656511552150265437214183965712622923326051704490517949489387606559188723914
a = 50 b = 98282557760751327186070919828563114616630258522452589747446938032795943619570
a = 99999999999999 b = 56876268607704945280095310365193269978753984617363817957513687854617877809261
a = 10000000000000000000000000000000000000000000000000000000000000000000000000000 b = 10939142339967927822460418038112020591422658251099346849357313458001062196732

For all these couples (a,b) you have that a/b = w1 mod p and b/a = w2 mod p.

Besides if you provide w1, you don't need to provide w2 too, because w2 is (w1)^-1 mod p --> w2 = 1/w1

w1 = 41689968412277401815192940146121615515073176981576112778517871276338175969527
w2 = pow(w1, p-2, p)
w2
94599322545067982882578206403521588574602656593708975301033069273870448067861

You have to provide w1 and a or b , but from w1 and w2 you cannot get a specific 'a' and 'b'.

In other terms you are providing only a single information (w1) and you want to know 2 values, it is like to resolve a single equation with 2 variables, like y = 2 * x (in your case b = w2 * a mod p)

there is no a single solution, but infinite couples (1,2), (2,4), (3,6) ..... there is no way to guess the couple you used!

j2002ba2

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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

February 16, 2020, 01:13:12 PM

#1367

Quote from: bytcoin on February 16, 2020, 01:36:12 AM

Quote from: j2002ba2 on February 15, 2020, 11:06:14 PM

Quote from: bytcoin on February 15, 2020, 09:37:25 PM

Sorry for the post off topic ... but can someone develop a tool to solve this?

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = Huh

?
b =

Well, that's easy:

Code:

0 < b < p
a = b * 12447032699845648078645791161909514142990644957498005805208944683777961822095 (mod p)

So, an obvious solution is a=12447032699845648078645791161909514142990644957498005805208944683777961822095, b=1

The values I provided for a and b are correct, given the input. According to the puzzle there are a bit less than 2²⁵⁶ possible pairs.

The screenshot provides more information, one could see the tops of the numbers, and restrict a and b. A quick look makes it obvious where 7 5 4 1 6 are.

Kostelooscoin

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Merit: 16

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

March 12, 2020, 09:34:20 AM

#1368

Quote from: supika on September 08, 2019, 09:02:08 AM

Hello,

If I have the below script, is possible to add a line like addr2 = ... for uncompressed addresses?
I have little experience with python and that's why I ask for help.
Thanks!

import random
from bit import *
#from PyRandLib import *
#rand = FastRand63()
#random.seed(rand())

c1 = str (random.choice("1"))
b1 = "01111111111111111111111111111111111111111111111111111111"
b2 = "00111111111111111111111111111111111111111111111111111111"
b3 = "00011111111111111111111111111111111111111111111111111111"
b4 = "00001111111111111111111111111111111111111111111111111111"
b5 = "00000111111111111111111111111111111111111111111111111111"
b6 = "00000011111111111111111111111111111111111111111111111111"
b7 = "00000001111111111111111111111111111111111111111111111111"
b8 = "00000000111111111111111111111111111111111111111111111111"
b9 = "00000000011111111111111111111111111111111111111111111111"
b10 = "00000000001111111111111111111111111111111111111111111111"
b11 = "00000000000111111111111111111111111111111111111111111111"
b12 = "00000000000011111111111111111111111111111111111111111111"
b13 = "00000000000001111111111111111111111111111111111111111111"
b14 = "00000000000000111111111111111111111111111111111111111111"
b15 = "00000000000000011111111111111111111111111111111111111111"
b16 = "00000000000000001111111111111111111111111111111111111111"
b17 = "00000000000000000111111111111111111111111111111111111111"
b18 = "00000000000000000011111111111111111111111111111111111111"
b19 = "00000000000000000001111111111111111111111111111111111111"
b20 = "00000000000000000000111111111111111111111111111111111111"
b21 = "00000000000000000000011111111111111111111111111111111111"
b22 = "00000000000000000000001111111111111111111111111111111111"
b23 = "00000000000000000000000111111111111111111111111111111111"
b24 = "00000000000000000000000011111111111111111111111111111111"
b25 = "00000000000000000000000001111111111111111111111111111111"
b26 = "00000000000000000000000000111111111111111111111111111111"
b27 = "00000000000000000000000000011111111111111111111111111111"
b28 = "00000000000000000000000000001111111111111111111111111111"
b29 = "00000000000000000000000000000111111111111111111111111111"
b30 = "00000000000000000000000000000011111111111111111111111111"
b31 = "00000000000000000000000000000001111111111111111111111111"
b32 = "00000000000000000000000000000000111111111111111111111111"
b33 = "00000000000000000000000000000000011111111111111111111111"
b34 = "00000000000000000000000000000000001111111111111111111111"
b35 = "00000000000000000000000000000000000111111111111111111111"
b36 = "00000000000000000000000000000000000011111111111111111111"
b37 = "00000000000000000000000000000000000001111111111111111111"
b38 = "00000000000000000000000000000000000000111111111111111111"
b39 = "00000000000000000000000000000000000000011111111111111111"
b40 = "00000000000000000000000000000000000000001111111111111111"
b41 = "00000000000000000000000000000000000000000111111111111111"
b42 = "00000000000000000000000000000000000000000011111111111111"
b43 = "00000000000000000000000000000000000000000001111111111111"
b44 = "00000000000000000000000000000000000000000000111111111111"
b45 = "00000000000000000000000000000000000000000000011111111111"
b46 = "00000000000000000000000000000000000000000000001111111111"
b47 = "00000000000000000000000000000000000000000000000111111111"
b48 = "00000000000000000000000000000000000000000000000011111111"
b49 = "00000000000000000000000000000000000000000000000001111111"
b50 = "00000000000000000000000000000000000000000000000000111111"
b51 = "00000000000000000000000000000000000000000000000000011111"
b52 = "00000000000000000000000000000000000000000000000000001111"
b53 = "00000000000000000000000000000000000000000000000000000111"
b54 = "00000000000000000000000000000000000000000000000000000011"
b55 = "00000000000000000000000000000000000000000000000000000001"
#xx = "00000000000000000000000000000000000000000000000000000001"

while True:
spisok = [b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13,b14,b15,b16,b17,b18,b19,b20,b21,b22,b23,b24,b25,b26,b27,b28,b29,b30,b31,b32,b33,b34,b35,b36,b37,b38,b39,b40,b41,b42,b43,b44,b45,b46,b47,b48,b49,b50,b51,b52,b53,b54,b55]
for element in (spisok):
for spisok in range(1000):
s = element
d = ''.join(random.sample(s,len(s)))
bina = (c1+d)
b = int(c1+d,2)
key = Key.from_int(b)
addr = key.address
if addr == "1LzhS3k3e9Ub8i2W1V8xQFdB8n2MYCHPCa":
print ("found!!!",b,addr)
s1 = str(b)
s2 = addr
f=open(u"C:/a.txt","a")
f.write(s1)
f.write(s2)
f.close()
pass
else:
print (s,bina,b,addr)
pass

Hello is it possible to search for an address from a list?

sssergy2705

Copper Member
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Merit: 1

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

March 12, 2020, 09:48:20 AM

#1369

Code:

import random
from bit import *

list = ["15c9mPGLku1HuW9LRtBf4jcHVpBUt8txKz","1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf","1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt",
"1Kn5h2qpgw9mWE5jKpk8PP4qvvJ1QVy8su","1AVJKwzs9AskraJLGHAZPiaZcrpDr1U6AB","1Me6EfpwZK5kQziBwBfvLiHjaPGxCKLoJi",
"1NpYjtLira16LfGbGwZJ5JbDPh3ai9bjf4","16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN","18ZMbwUFLMHoZBbfpCjUJQTCMCbktshgpe",
"13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so","1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9","1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ",
"19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG","19YZECXj3SxEZMoUeJ1yiPsw8xANe7M7QR","1PWo3JeB9jrGwfHDNpdGK54CRas7fsVzXU",
"1JTK7s9YVYywfm5XUH7RNhHJH1LshCaRFR","12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4","1FWGcVDK3JGzCC3WtkYetULPszMaK2Jksv",
"1J36UjUByGroXcCvmj13U6uwaVv9caEeAt","1DJh2eHFYQfACPmrvpyWc8MSTYKh7w9eRF","1Bxk4CQdqL9p22JEtDfdXMsng1XacifUtE",
"15qF6X51huDjqTmF9BJgxXdt1xcj46Jmhb","1ARk8HWJMn8js8tQmGUJeQHjSE7KRkn2t8","1BCf6rHUW6m3iH2ptsvnjgLruAiPQQepLe",
"15qsCm78whspNQFydGJQk5rexzxTQopnHZ","13zYrYhhJxp6Ui1VV7pqa5WDhNWM45ARAC","14MdEb4eFcT3MVG5sPFG4jGLuHJSnt1Dk2",
"1CMq3SvFcVEcpLMuuH8PUcNiqsK1oicG2D","1Kh22PvXERd2xpTQk3ur6pPEqFeckCJfAr","1K3x5L6G57Y494fDqBfrojD28UJv4s5JcK",
"1PxH3K1Shdjb7gSEoTX7UPDZ6SH4qGPrvq","16AbnZjZZipwHMkYKBSfswGWKDmXHjEpSf","19QciEHbGVNY4hrhfKXmcBBCrJSBZ6TaVt",
"1L12FHH2FHjvTviyanuiFVfmzCy46RRATU","1EzVHtmbN4fs4MiNk3ppEnKKhsmXYJ4s74","1AE8NzzgKE7Yhz7BWtAcAAxiFMbPo82NB5",
"17Q7tuG2JwFFU9rXVj3uZqRtioH3mx2Jad","1K6xGMUbs6ZTXBnhw1pippqwK6wjBWtNpL","19eVSDuizydXxhohGh8Ki9WY9KsHdSwoQC",
"15ANYzzCp5BFHcCnVFzXqyibpzgPLWaD8b","18ywPwj39nGjqBrQJSzZVq2izR12MDpDr8","1CaBVPrwUxbQYYswu32w7Mj4HR4maNoJSX",
"1JWnE6p6UN7ZJBN7TtcbNDoRcjFtuDWoNL","1KCgMv8fo2TPBpddVi9jqmMmcne9uSNJ5F","1CKCVdbDJasYmhswB6HKZHEAnNaDpK7W4n",
"1PXv28YxmYMaB8zxrKeZBW8dt2HK7RkRPX","1AcAmB6jmtU6AiEcXkmiNE9TNVPsj9DULf","1EQJvpsmhazYCcKX5Au6AZmZKRnzarMVZu",
"1CMjscKB3QW7SDyQ4c3C3DEUHiHRhiZVib","18KsfuHuzQaBTNLASyj15hy4LuqPUo1FNB","15EJFC5ZTs9nhsdvSUeBXjLAuYq3SWaxTc",
"1HB1iKUqeffnVsvQsbpC6dNi1XKbyNuqao","1GvgAXVCbA8FBjXfWiAms4ytFeJcKsoyhL","12JzYkkN76xkwvcPT6AWKZtGX6w2LAgsJg",
"1824ZJQ7nKJ9QFTRBqn7z7dHV5EGpzUpH3","18A7NA9FTsnJxWgkoFfPAFbQzuQxpRtCos","1NeGn21dUDDeqFQ63xb2SpgUuXuBLA4WT4",
"174SNxfqpdMGYy5YQcfLbSTK3MRNZEePoy","1NLbHuJebVwUZ1XqDjsAyfTRUPwDQbemfv","1MnJ6hdhvK37VLmqcdEwqC3iFxyWH2PHUV",
"1KNRfGWw7Q9Rmwsc6NT5zsdvEb9M2Wkj5Z","1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6","1GuBBhf61rnvRe4K8zu8vdQB3kHzwFqSy7",
"17s2b9ksz5y7abUm92cHwG8jEPCzK3dLnT","1GDSuiThEV64c166LUFC9uDcVdGjqkxKyh","1Me3ASYt5JCTAK2XaC32RMeH34PdprrfDx",
"1CdufMQL892A69KXgv6UNBD17ywWqYpKut","1BkkGsX9ZM6iwL3zbqs7HWBV7SvosR6m8N","1PXAyUB8ZoH3WD8n5zoAthYjN15yN5CVq5",
"1AWCLZAjKbV1P7AHvaPNCKiB7ZWVDMxFiz","1G6EFyBRU86sThN3SSt3GrHu1sA7w7nzi4","1MZ2L1gFrCtkkn6DnTT2e4PFUTHw9gNwaj",
"1Hz3uv3nNZzBVMXLGadCucgjiCs5W9vaGz","1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua","16zRPnT8znwq42q7XeMkZUhb1bKqgRogyy",
"1KrU4dHE5WrW8rhWDsTRjR21r8t3dsrS3R","17uDfp5r4n441xkgLFmhNoSW1KWp6xVLD","13A3JrvXmvg5w9XGvyyR4JEJqiLz8ZySY3",
"16RGFo6hjq9ym6Pj7N5H7L1NR1rVPJyw2v","1UDHPdovvR985NrWSkdWQDEQ1xuRiTALq","15nf31J46iLuK1ZkTnqHo7WgN5cARFK3RA",
"1Ab4vzG6wEQBDNQM1B2bvUz4fqXXdFk2WT","1Fz63c775VV9fNyj25d9Xfw3YHE6sKCxbt","1QKBaU6WAeycb3DbKbLBkX7vJiaS8r42Xo",
"1CD91Vm97mLQvXhrnoMChhJx4TP9MaQkJo","15MnK2jXPqTMURX4xC3h4mAZxyCcaWWEDD","13N66gCzWWHEZBxhVxG18P8wyjEWF9Yoi1",
"1NevxKDYuDcCh1ZMMi6ftmWwGrZKC6j7Ux","19GpszRNUej5yYqxXoLnbZWKew3KdVLkXg","1M7ipcdYHey2Y5RZM34MBbpugghmjaV89P",
"18aNhurEAJsw6BAgtANpexk5ob1aGTwSeL","1FwZXt6EpRT7Fkndzv6K4b4DFoT4trbMrV","1CXvTzR6qv8wJ7eprzUKeWxyGcHwDYP1i2",
"1MUJSJYtGPVGkBCTqGspnxyHahpt5Te8jy","13Q84TNNvgcL3HJiqQPvyBb9m4hxjS3jkV","1LuUHyrQr8PKSvbcY1v1PiuGuqFjWpDumN",
"18192XpzzdDi2K11QVHR7td2HcPS6Qs5vg","1NgVmsCCJaKLzGyKLFJfVequnFW9ZvnMLN","1AoeP37TmHdFh8uN72fu9AqgtLrUwcv2wJ",
"1FTpAbQa4h8trvhQXjXnmNhqdiGBd1oraE","14JHoRAdmJg3XR4RjMDh6Wed6ft6hzbQe9","19z6waranEf8CcP8FqNgdwUe1QRxvUNKBG",
"14u4nA5sugaswb6SZgn5av2vuChdMnD9E5","1NBC8uXJy1GiJ6drkiZa1WuKn51ps7EPTv"]

c1 = str (random.choice("1"))
b1 = "01111111111111111111111111111111111111111111111111111111"
b2 = "00111111111111111111111111111111111111111111111111111111"
b3 = "00011111111111111111111111111111111111111111111111111111"
b4 = "00001111111111111111111111111111111111111111111111111111"
b5 = "00000111111111111111111111111111111111111111111111111111"
b6 = "00000011111111111111111111111111111111111111111111111111"
b7 = "00000001111111111111111111111111111111111111111111111111"
b8 = "00000000111111111111111111111111111111111111111111111111"
b9 = "00000000011111111111111111111111111111111111111111111111"
b10 = "00000000001111111111111111111111111111111111111111111111"
b11 = "00000000000111111111111111111111111111111111111111111111"
b12 = "00000000000011111111111111111111111111111111111111111111"
b13 = "00000000000001111111111111111111111111111111111111111111"
b14 = "00000000000000111111111111111111111111111111111111111111"
b15 = "00000000000000011111111111111111111111111111111111111111"
b16 = "00000000000000001111111111111111111111111111111111111111"
b17 = "00000000000000000111111111111111111111111111111111111111"
b18 = "00000000000000000011111111111111111111111111111111111111"
b19 = "00000000000000000001111111111111111111111111111111111111"
b20 = "00000000000000000000111111111111111111111111111111111111"
b21 = "00000000000000000000011111111111111111111111111111111111"
b22 = "00000000000000000000001111111111111111111111111111111111"
b23 = "00000000000000000000000111111111111111111111111111111111"
b24 = "00000000000000000000000011111111111111111111111111111111"
b25 = "00000000000000000000000001111111111111111111111111111111"
b26 = "00000000000000000000000000111111111111111111111111111111"
b27 = "00000000000000000000000000011111111111111111111111111111"
b28 = "00000000000000000000000000001111111111111111111111111111"
b29 = "00000000000000000000000000000111111111111111111111111111"
b30 = "00000000000000000000000000000011111111111111111111111111"
b31 = "00000000000000000000000000000001111111111111111111111111"
b32 = "00000000000000000000000000000000111111111111111111111111"
b33 = "00000000000000000000000000000000011111111111111111111111"
b34 = "00000000000000000000000000000000001111111111111111111111"
b35 = "00000000000000000000000000000000000111111111111111111111"
b36 = "00000000000000000000000000000000000011111111111111111111"
b37 = "00000000000000000000000000000000000001111111111111111111"
b38 = "00000000000000000000000000000000000000111111111111111111"
b39 = "00000000000000000000000000000000000000011111111111111111"
b40 = "00000000000000000000000000000000000000001111111111111111"
b41 = "00000000000000000000000000000000000000000111111111111111"
b42 = "00000000000000000000000000000000000000000011111111111111"
b43 = "00000000000000000000000000000000000000000001111111111111"
b44 = "00000000000000000000000000000000000000000000111111111111"
b45 = "00000000000000000000000000000000000000000000011111111111"
b46 = "00000000000000000000000000000000000000000000001111111111"
b47 = "00000000000000000000000000000000000000000000000111111111"
b48 = "00000000000000000000000000000000000000000000000011111111"
b49 = "00000000000000000000000000000000000000000000000001111111"
b50 = "00000000000000000000000000000000000000000000000000111111"
b51 = "00000000000000000000000000000000000000000000000000011111"
b52 = "00000000000000000000000000000000000000000000000000001111"
b53 = "00000000000000000000000000000000000000000000000000000111"
b54 = "00000000000000000000000000000000000000000000000000000011"
b55 = "00000000000000000000000000000000000000000000000000000001"
#xx = "00000000000000000000000000000000000000000000000000000001"

while True:
spisok = [b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13,b14,b15,b16,b17,b18,b19,b20,b21,b22,b23,b24,b25,b26,b27,b28,b29,b30,b31,b32,b33,b34,b35,b36,b37,b38,b39,b40,b41,b42,b43,b44,b45,b46,b47,b48,b49,b50,b51,b52,b53,b54,b55]
for element in (spisok):
for spisok in range(1000):
s = element
d = ''.join(random.sample(s,len(s)))
bina = (c1+d)
b = int(c1+d,2)
key = Key.from_int(b)
addr = key.address
if addr in list:
print ("found!!!",b,addr)
s1 = str(b)
s2 = addr
f=open(u"C:/a.txt","a")
f.write(s1)
f.write(s2)
f.close()
pass
else:
print (s,bina,b,addr)
pass

It will be something like this. I did not change the range, since it is very difficult to find anything in this way.

Kostelooscoin

Member

Offline

Activity: 206
Merit: 16

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

March 12, 2020, 10:03:30 AM

#1370

Quote from: sssergy2705 on March 12, 2020, 09:48:20 AM

Code:

It will be something like this. I did not change the range, since it is very difficult to find anything in this way.

thank you very much

Texgen

Newbie

Offline

Activity: 9
Merit: 0

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

April 05, 2020, 05:42:35 PM

#1371

As I understand it, you cant leave links to other resources here? I saw a channel on Patreone called Bitcoin Riches, where they talk about a script that you can solve a puzzle.

bigvito19

Full Member

Offline

Activity: 732
Merit: 112

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

April 06, 2020, 04:35:54 PM

#1372

Quote from: Texgen on April 05, 2020, 05:42:35 PM

As I understand it, you cant leave links to other resources here? I saw a channel on Patreone called Bitcoin Riches, where they talk about a script that you can solve a puzzle.

Inbox me the links so I can see what you spamming about

spider703

Full Member

Offline

Activity: 1918
Merit: 149

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

April 19, 2020, 09:02:43 AM

#1373

hi all
I have a question - how do you get Biginteger PVK? Of what? what you need to take to get Biginteger PVK
example:
Address 6:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Tmu6qHxS
1PitScNLyp2HCygzadCh7FveTnfmpPbfp8
Biginteger PVK value: 49 - where did this number come from?
Hex PVK value: 31

BTC 1Hof999zuqUKpifmzrSABv7tNr4nRaoJKM LTC Lf2L6DTBr2gXT38d7cVRqDQiHMndtXQyNW or write me in https://t.me/spider703

brainless

Member

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Activity: 398
Merit: 35

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

April 22, 2020, 02:04:22 PM

#1374

Quote from: spider703 on April 19, 2020, 09:02:43 AM

Jeanluc GPU Kangaroo version launched, Enjoy high speed
https://github.com/JeanLucPons/Kangaroo

13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF

Jean_Luc

Sr. Member

Offline

Activity: 462
Merit: 737

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

April 24, 2020, 03:00:05 PM

#1375

Quote

Jeanluc GPU Kangaroo version launched, Enjoy high speed
https://github.com/JeanLucPons/Kangaroo

Hi there,
If someone find something with this tool, please inform us Wink

New release 1.3 is ready.
If you have any idea on how to optimize it, I would appreciate any tips.
Thanks
JL

Edit: There is a description of the method used on the REAME.

Andzhig

Jr. Member

Offline

Activity: 184
Merit: 3

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

April 24, 2020, 08:53:25 PM
Last edit: May 07, 2020, 07:02:17 PM by Andzhig

#1376

Could you do such a method on gpu https://bt.irlbtc.com/view/1306983.msg53447044#msg53447044
there really is enough hemorrhoids too

6 digits as an example 123456 into 2 parts 123 456 and the number of options for "scanning" from 500 to 3000.

123456 > 6250000 variants,combinations 2500x2500

for each number its own number of combinations

100000 6250000 variants,combinations 2500x2500
100001 6250000 variants,combinations 2500x2500
100002 6250000 variants,combinations 2500x2500
100003 6250000 variants,combinations 2500x2500
...
999999 6250000 variants,combinations 2500x2500

I ran 2 different numbers 6 digits with 2^10000 to 2^20000000 get out 4500000 for each. Identical docking +- 10 combinations (which is also interesting, they must also exist for 2500x2500x2500 etc x4 x5 x6 x7 ...).

2^10000 to 2^20000000

335 667
335 668

Equal element found: (242, 1568)
Equal element found: (225, 772)
Equal element found: (228, 285)
Equal element found: (865, 258)
Equal element found: (1042, 27)
Equal element found: (28, 663)
Equal element found: (455, 451)
Equal element found: (377, 309)

or

525 070
384 258

Equal element found: (494, 241)
Equal element found: (71, 147)
Equal element found: (1281, 877)
Equal element found: (357, 14)
Equal element found: (82, 2316)
Equal element found: (684, 191)
Equal element found: (124, 1070)
Equal element found: (97, 2212)
Equal element found: (457, 1512)

its hard to say for small numbers they fly out when scanning, but the idea is to take all the combinations from the number and weed them out of all the options and in what is left to find the right i.e 6250000 - 4500000...

and checking on a 9-digit number is problematic 15625000000 2500x2500x2500 because we can in area 2^100000000 work generating more is problematic (of course can simply take combinations 000,001,002 ... 998,999 instead of 2^ ... but with 2^ ... more interesting)

Well, in general, on our

pz 67 147573952589676412927
pz 68 295147905179352825855
pz 69 590295810358705651711

610351562500000000000000 2500×2500×2500×2500×2500×2500×2500
or
7812500000000000000 500x500x500x500x500x500x500

out of 2^20000000 squeezed out somewhere combinations 275936334 several runs, weighs 10 gb txt file (2^20000000 60 gb).

2759363340‬ 100gb
5518726680‬ 200gb
8278090020‬ 300gb
11037453360‬ 400gb
13796816700 500gb
‬27593633400‬ 1tb
275936334000‬ 10tb
2759363340000‬ 100tb

apparently not a lot

610351562500000000000000
2759363340‬ 100gb

here we are only interested that the number we need has the same number of combinations i.e 610351562500000000000000
100000000000000000000 610351562500000000000000 2500×2500×2500×2500×2500×2500×2500
100000000000000000001 610351562500000000000000 2500×2500×2500×2500×2500×2500×2500
100000000000000000002 610351562500000000000000 2500×2500×2500×2500×2500×2500×2500
100000000000000000003 610351562500000000000000 2500×2500×2500×2500×2500×2500×2500
...
999999999999999999999 610351562500000000000000 2500×2500×2500×2500×2500×2500×2500

the question is how to catch him in our 2^20000000

Quote

4157908 541469525002647428772 21 148YXXg36qvrpPHDpvthJ1D3SztddfyPoG 1E4QSTebYmecEDx2vAZobhGjZSQnpx1GNU 37 49 6 17 273 166 11 0
4157909 308093305400329685154 21 1GdrBrwhL7D6TYQZ3yos9twAufVSBhks1U 1Msemrtc5AutsD7CGvqf6KX6JGxMDBe51c 35 49 6 17 273 166 11 0
4157910 616187610800658371309 21 1LPE1cga2UrMzEhdXrzDrTVapc5aw821eY 1KujerTnZVjjXmMia8PRVUKkqGtfEG57KV 37 49 6 17 273 166 11 0
4157911 232375220601317742618 21 19M7eKcifwsz5S1wwUN2NADjYXLg1Nauem 1Nuh7y8jmLBy4jwDapZif5QHoz7KKVJuuQ 32 49 6 17 273 166 11 0
4157912 846875644320863148123 21 1t7YaYRkcu9R22DbTHW2Q83HcEZtQU7zj 1MZH8GHDSznTyiXk6SxRmNdY4REPPSncKM 36 49 6 17 273 166 11 0
4157913 693750288640727297247 21 16sT1krctJXDaikFWbP3P1tnJfh2qexsAj 19gEe9Jc6eCHRNADwKoQVwpRUmbwCH1DTc 30 49 6 17 273 166 11 0
4157914 386500576281454594494 21 192f2iuQyp6ajofJ2YFfAnk3cKMeFuQVoj 1QJYiNdhnwfaDJGmBb6Kc8txfkjqZdCZd1 30 49 6 17 273 166 11 0
4157915 772001152562908188989 21 16wSSXBL5eBBwoxXKB9FWbLYFwUXCVc33S 1QGW5m7vqACRpaCugZ5rk2Lmgcu79KDytJ 40 49 6 17 273 166 11 0
4157916 354000630912581637197 21 14oGh4BjKghfMsXbzKxNAkGFE2Qbz5Je2T 1GLuHp8hvji3HpuAwb7kJp9D45wBbemZBD 39 49 6 17 273 166 11 0
4157917 709000261825163275395 21 1FRcqUKWE3fUFpvsFEccRwedpntMspmczX 1eo3vDej7WAe7zazuFgede9rcrPZm7ypV 30 49 6 17 273 166 11 0
4157918 418000522650326550791 21 131yqPZiNbUihaEgmnt88iGDHUJvfBZEFV 1DxVFDN9pun8wdhkkwbgLbPxH1t6yMFizD 39 49 6 17 273 166 11 0
4157919 83200904330465110158 20 1B3u7wpQbfAQsK1A2askEbP4VgFf4oFh7M 13sxV6Fd7xuzoEPm5eVzr7HiX1YeA6HCXv 34 49 6 17 273 166 11 0
4157920 167400808660930220316 21 1AHnMREer6qgHw6CL4mqj28Trh1xE9EAxX 1HnGKXhDb32sbAhJk9rSg8cMczggvrbLK8 39 49 6 17 273 166 11 0
4157921 334800617320861440633 21 1Df39ARvaFY2StBxheEZwnuWXpc347qUSW 12eF3btGKGND5Rz1jRzLUdacXCF2tTGSrB 35 49 6 17 273 166 11 0
4157922 866160723464772488126 21 1GTdhbLwKdbnYCNzaes6xwLn1R2Rn8iSRR 1LHDoQwiH1oprg9EY26VJHRhGSxLAmiqQt 37 49 6 17 273 166 11 0
4157923 733320447928544976253 21 1H98PzMj3HLvytgXtnJ2s3VuTvR3UdB3b5 1NAbygLAGprJhR26a6VP28zT7xdaDmeGww 29 49 6 17 273 166 11 0
4157924 467640894856089952506 21 1535WFNGcFfexPZM7tAvBmJjC7hsWRZppM 1K4p65gCgyNNmt17NkWDR2X9FcHEi7yjpe 36 49 6 17 273 166 11 0
4157925 293328178371417590101 21 12hmHZRgpt1eaFH1Tj25LGAANgXZxVb3df 19C3aEWDsjwb4kdd1CMdTFFn8NxNqzJyg8 35 49 6 17 273 166 11 0

[/size]

"combination detector"

Quote

j=[]

ii = 1
while ii <= 1:
   i = 8192 # 2^ start
   while i <= 16384: # 2^ end #1024 2048 4096 8192 16384 32768 65536 131072

   a = pow(2,i)
   n = str(a)

   nn1= "335" # desired number
   nn2= "667" # desired number
   #nn3= "364"

   if nn1 in n:
   v1 = n.index(nn1)
   gg1 = int(v1)
   if gg1 <= 2500: #500-3000
   jj = j.append(gg1)

   if nn2 in n:
   v2 = n.index(nn2)
   gg2 = int(v2)
   if gg2 <= 2500: #500-3000
   jj = j.append(gg2)

   #if nn3 in n:
   #v3 = n.index(nn3)
   #gg3 = int(v3)
   #if gg3 <= 2500:
   #jj = j.append(gg3)

   print(gg1,gg2," ","2^",i)
   else:
   pass
   i=i+1
   ii=ii+1

***

Well, then look for this kangaroo, similar to bitcrack.
We take the number divided by 2 to 6 characters (they are easier to sort out step by step).
Then we hinder to the approximate (displacement due to non-integer numbers).
30045390491869460
30045254740606976
135751262484

   861921
   625564
   958885
   571654
   735234
95pz 675663 527A792B183C7F64A0E8B1F4
100pz 718175 AF55FC59C335C8EC67ED24826
105pz 751783 16F14FC2054CD87EE6396B33DF3

Quote

:2...

29083230144918045706788529192435
14541615072459022853394264596217
7270807536229511426697132298108
3635403768114755713348566149054
1817701884057377856674283074527
908850942028688928337141537263
454425471014344464168570768631
227212735507172232084285384315
113606367753586116042142692157
56803183876793058021071346078
28401591938396529010535673039
14200795969198264505267836519
7100397984599132252633918259
3550198992299566126316959129
1775099496149783063158479564
887549748074891531579239782
443774874037445765789619891
221887437018722882894809945
110943718509361441447404972
55471859254680720723702486
27735929627340360361851243
13867964813670180180925621
6933982406835090090462810
3466991203417545045231405
1733495601708772522615702
866747800854386261307851
433373900427193130653925
216686950213596565326962
108343475106798282663481
54171737553399141331740
27085868776699570665870
13542934388349785332935
6771467194174892666467
3385733597087446333233
1692866798543723166616
846433399271861583308
423216699635930791654
211608349817965395827
105804174908982697913
52902087454491348956
26451043727245674478
13225521863622837239
6612760931811418619
3306380465905709309
1653190232952854654
826595116476427327
413297558238213663
206648779119106831
103324389559553415
51662194779776707
25831097389888353
12915548694944176
6457774347472088
3228887173736044
1614443586868022
807221793434011
403610896717005
201805448358502
100902724179251
50451362089625
25225681044812
12612840522406
6306420261203
3153210130601
1576605065300
788302532650
394151266325
197075633162
98537816581
49268908290
24634454145
12317227072
6158613536
3079306768
1539653384
769826692
384913346
192456673
96228336
48114168
24057084
12028542
6014271
3007135
1503567
751783
375891
187945
93972
46986
23493
11746
5873
2936
1468
734
367
183
91
45
22
11
5
2
1

x2...

751783
1503566
3007132
6014264
12028528
24057056
48114112
96228224
192456448
384912896
769825792
1539651584
3079303168
6158606336
12317212672
24634425344
49268850688
98537701376
197075402752
394150805504
788301611008
1576603222016
3153206444032
6306412888064
12612825776128
25225651552256
50451303104512
100902606209024
201805212418048
403610424836096
807220849672192
1614441699344384
3228883398688768
6457766797377536
12915533594755072
25831067189510144
51662134379020288
103324268758040576
206648537516081152
413297075032162304
826594150064324608
1653188300128649216
3306376600257298432
6612753200514596864
13225506401029193728
26451012802058387456
52902025604116774912
105804051208233549824
211608102416467099648
423216204832934199296
846432409665868398592
1692864819331736797184
3385729638663473594368
6771459277326947188736
13542918554653894377472
27085837109307788754944
54171674218615577509888
108343348437231155019776
216686696874462310039552
433373393748924620079104
866746787497849240158208
1733493574995698480316416
3466987149991396960632832
6933974299982793921265664
13867948599965587842531328
27735897199931175685062656
55471794399862351370125312
110943588799724702740250624
221887177599449405480501248
443774355198898810961002496
887548710397797621922004992
1775097420795595243844009984
3550194841591190487688019968
7100389683182380975376039936
14200779366364761950752079872
28401558732729523901504159744
56803117465459047803008319488
113606234930918095606016638976
227212469861836191212033277952
454424939723672382424066555904
908849879447344764848133111808
1817699758894689529696266223616
3635399517789379059392532447232
7270799035578758118785064894464
14541598071157516237570129788928
29083196142315032475140259577856

[/size]

last 105pz 751783 multiply by 2 to 1298074214633706907132624082305023, we get 29083196142315032475140259577856 minus the desired 105pz 29083230144918045706788529192435 at the exit 34002603013231648269614579. 29083196142315032475140259577856-34002603013231648269614579=29083162139712019243491989963277.
29083196142315032475140259577856+34002603013231648269614579=29083230144918045706788529192435 (here you can add numbers on top).
And the range is

000000000000000000000000000000000000016f14c3dfab3278119c694cc20d
000000000000000000000000000000000000016f14fc2054cd87ee6396b33df3
03bcf7ce887ffca5e62c9cabbdb7ffa71dc183c52c04ff4ee5ee82e0c55c39d77b

and for each check from 500000 to 999999, it will take a ~week for ones. 500000 weeks. 110pz take ~2 weeks for 1 scan.
well, and the approximate slip difference pick up. Or use in calculations after the decimal point (which does not reduce the hemorrhoids). And can forget about the 2^256 search altogether (or can use some sophistication in computing).

***
Need this on gpu.

Quote

import random
from bit import Key
from bit.format import bytes_to_wif
import time

list = ["16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN","13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so","1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9",
   "1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ","19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG"]

#**************************************************************************************************************

print("loop start...")

count = 0

def func():
   f1 = random.randint(0,100) #(0,100)
   f2 = random.randint(0,1000) #(0,1000)
   f3 = random.randint(0,1500) #(0,1500)
   f4 = random.randint(0,2000) #(0,2000)
   f5 = random.randint(0,2500) #(0,2500)
   aaa = random.choice([f1,f2,f3,f4,f5,f5,f4,f3,f2,f1]) #([f1,f2,f3,f4,f5])
   return aaa

while True:
   k1 = func() #0
   k2 = func()
   k3 = func()
   k4 = func()
   k5 = func()
   k6 = func()
   k7 = func()
   F=(k1,k2,k3,k4,k5,k6,k7) #(k1,k2,k3,k4,k5,k6,k7)
   print(F)
# L3 = open(u"C:/pow65_2.txt", "r") # 2^10000 start
# for line in L3:
   i = 8192 # 2^start
   while i <= 32768: # 2^end 512 1024 2048 4096 8192 16384 32768 65536 131072

   power = pow(2,i)
   powerstr = str(power)[0:3000]
   powerint = powerstr
   hh = powerint

   ff1 = str(hh)[k1:k1+3]
   ff2 = str(hh)[k2:k2+3]
   ff3 = str(hh)[k3:k3+3]
   ff4 = str(hh)[k4:k4+3]
   ff5 = str(hh)[k5:k5+3]
   ff6 = str(hh)[k6:k6+3]
   ff7 = str(hh)[k7:k7+3]

   ran = int(ff1+ff2+ff3+ff4+ff5+ff6+ff7) #(ff1+ff2+ff3+ff4+ff5+ff6+ff7)
   if ran != 1: # 115792089237316195423570985008687907853269984665640564039457584007913129639936
   baba = len(str(ran))
   if baba >= 0: # num length
   bina = bin(ran)[2:]
   ed = bina.count("0")
   if ed >= 0: # zeros
   if ed <= 150: # zeros
   key1 = Key.from_int(ran)
   wif = bytes_to_wif(key1.to_bytes(), compressed=False)
   key2 = Key(wif)
   key1 == key2
   addr1 = key1.address
   addr2 = key2.address
   if addr1 in list:
   print (k1,k2,k3,k4,k5,k6,k7,ran,"found!!!",ran)
   s5 = str(ran)
   f=open(u"C:/a.txt","a")
   f.write(s5 + '\n')
   f.close()
   break
   if addr2 in list:
   print (k1,k2,k3,k4,k5,k6,k7,ran,"found!!!",ran)
   s5 = str(ran)
   f=open(u"C:/a.txt","a")
   f.write(s5 + '\n')
   f.close()
   break
   else:
   count += 1
   print(count,ran,baba,addr1,addr2," ",ed," ",k1,k2,k3,k4,k5,k6,k7)
   i=i+1
   count = 0
   print ("!!!loop end!!!")
   time.sleep(10.0)
   pass

[/size]

And here already need to think about how to catch 0,0,0,0,0,0,0-2500,2500,2500,2500,2500,2500,2500<>2500,2500,2500,2500,2500,2500,2500-0,0,0,0,0,0,0

It seems that the distribution of numbers is uniform 2^... (000...999) hence the distribution is uniform (uniformly scattered "random"). 2^10000-2^250000 https://imgur.com/whAbSK6 in the near range, apparently this is the case further .i.e 2^200000000 ,can take random and add 1 to each per run.

879 829 1172 726 1000 912 36
...
880 830 1173 727 1001 913 37
...
881 831 1174 728 1002 914 38
...

***

And a small "feature". To run multiple copies at once

Create a text file and rename its extension in cmd. *.txt to *.cmd
And in it how many copies you need to run.

Quote

@echo off

cd "C:\Users\***\Desktop"
startmin youprog.py
startmin youprog.py
startmin youprog.py
startmin youprog.py
startmin youprog.py
startmin youprog.py
startmin youprog.py
startmin youprog.py
startmin youprog.py

exit

***

Some observations. Generating 2^... in pari/gp ((default(parisize, 40000000000) 38146 mb)) 2^10000000000 2.80 gb on hdd length 3010299958 num, time generation of 30-40 minutes. Finds a lot of 10 digit numbers but not all. Repeats of 3 digits vary from degree to degree. For "111" 1003262. from degree to degree (2^bla1 2^bla2 2^bla3...) of number in the number they jump but according to the number of the numbers themselves they stay within certain limits, i.e. number of 6 places within ~1000 (and themselves 3 significant as you know 1000... "000,001,002...998,999"). The option to discard all places for "111" does not work.

places "111" in 2^10000000000

Quote

202
448
986
2808
3631
3881
4331
7296
8205
11137
12333
13028
13744
14143
20527
21346
24379
24569
26271
27160
27850
28446
28807
28831
29264
29408
29759
32041
32413
33980
34273
35114
35588
35846
36514
36658
40175
40909
42581
42637
42743
42963
43290
44341
47022
47277
49682
50763
50849
52280
52436
53004
55473
57240
57390
57454
58304
59735
60500
61763
62065
63876
64006
64219
65047
67221
67252
67916
68726
69048
69440
71504
72214
73126
73741
73947
74701
75159
75393
76991
78194
78841
79843
80196
80933
81841
83949
84443
85061
85271
85784
86123
86788
87576
87825
90369
91001
91740
91911
93419
94114
97539
97786
97844
98599
100109 > (~1000 for "111" in each 2^10000000000, 2^10000000001, 2^10000000002, 2^10000000003, etc... )
100470
100746
103501
104697
105874
107724
108451
109754
110345
111158
112031
...
988898
988989
989070
989498
992442
992752
993564
994025
994800
995639
996291
996338
996449
997440
997470
998390 < (~1000 for "111" in each 2^10000000000, 2^10000000001, 2^10000000002, 2^10000000003, etc... )
1001064
1001350
1003314
1003668
1005157
1005865
1007372
1007495
1007841
1008797
1009059
1010724
1013903
1014250
1014292
1014359
1014895
1015828
1017144
1018056
1018407
1019228
1020424
1020788
1022083
1022121
1025545
1028358
1030442
1032906
...
1003406097
1003406649
1003408898
1003410251
1003410297
1003414072
1003414605
1003418188
1003418417
1003424139
1003425377
1003427268
1003427716
1003430037
1003430666
1003430736
1003430954
1003431379
1003431677
1003433263

[/size]

***
~ 1000 for "000-999" this he reserves apparently this space for all sets 1000x1000=1000000. the next are probably 4 digits "0000-9999" etc...

***

what will happen if we substitute a set "111" of one 2^1000003 (781, 1347, 1554, 2868, 3175, 3300, 3724) for 2^1000000-1000005

Quote

=====================================
[476, 1188, 1449, 2051, 5058, 7687, 8113, 8683, 9735, 9860, 10102, 10613, 11094, 12206, 12321, 12343, 12632, 13468, 16041, 18286, 20340, 20955, 25393, 26821, 27533, 27773, 30050, 30582, 30620, 31461, 32343, 32366, 34221, 35148, 38133, 38951, 39565, 41044, 41704, 42304, 43053, 44128, 45126, 46981, 47085, 48728, 49604, 49752, 50484, 51901, 52319, 52781, 53250, 53506, 54661, 54809, 54944, 55972, 57706, 58053, 58142, 58428, 59002, 62637, 62705, 63883, 64974, 66263, 66833, 67647, 70452, 71945, 73604, 74540, 74851, 76370, 77860, 78849, 79562, 80329, 80610, 81327, 82211, 82598, 82849, 84168, 88316, 88535, 88767, 89857, 90070, 90596, 91796, 92027, 93269, 93414, 95107, 97537, 98366, 98532, 99115, 100165]
389 639 639 139 139 638 889
[5391, 5785, 6318, 9873, 11540, 11636, 11952, 12403, 12947, 14369, 14566, 17470, 17903, 18144, 18563, 19288, 19726, 19848, 22098, 22111, 24239, 24260, 24616, 25039, 25367, 26960, 27787, 28233, 28652, 30775, 31231, 31901, 32059, 32630, 32885, 32886, 33393, 34685, 35583, 36001, 36546, 36782, 38503, 38821, 39300, 40227, 40410, 42314, 46367, 47060, 47715, 48670, 48994, 49066, 49557, 49763, 50800, 51525, 51584, 52183, 53012, 53124, 54189, 59652, 60578, 63245, 63910, 64185, 64481, 65241, 66199, 67119, 70052, 71860, 75170, 75411, 75939, 76721, 77680, 77873, 77879, 78039, 79758, 79785, 80130, 81097, 82233, 82728, 87128, 87427, 87922, 88511, 89251, 93319, 93753, 94817, 96135, 97590, 98548, 99668]
277 527 527 027 027 527 777
[614, 760, 1638, 4893, 5106, 5814, 8233, 8371, 9640, 10212, 10253, 11604, 11886, 12073, 12443, 13031, 14776, 14869, 16332, 17224, 19299, 19565, 20183, 22598, 22786, 22963, 23858, 24015, 24099, 26347, 27005, 28062, 29478, 29974, 30441, 32227, 32961, 33669, 34281, 35354, 36160, 36583, 38069, 43327, 45377, 47538, 47897, 47938, 48466, 49682, 51320, 52508, 53058, 54679, 54767, 55397, 55595, 56339, 56718, 58294, 58395, 59044, 61945, 65231, 66136, 68086, 69955, 70187, 70340, 71936, 72234, 73322, 73779, 77151, 79612, 80048, 81428, 82761, 83051, 83736, 84944, 85398, 85521, 85768, 86319, 86410, 86661, 87351, 93001, 93719, 95189, 95419, 95679, 96335, 96933, 99328]
555 055 055 055 055 055 555
[186, 781, 1347, 1554, 2868, 3175, 3300, 3724, 4799, 5491, 5665, 6234, 7764, 7944, 8418, 8970, 10146, 10147, 12179, 12624, 14385, 14957, 15853, 16425, 18470, 19737, 20630, 21541, 22794, 23567, 26528, 26537, 26745, 29376, 29560, 31384, 34738, 36025, 37175, 38333, 40472, 41525, 42146, 42500, 46209, 46880, 47365, 48621, 49264, 50697, 51747, 53786, 54277, 56566, 56626, 56699, 56700, 57307, 58825, 61071, 63309, 66686, 67875, 69289, 70111, 70721, 70802, 72656, 72691, 73019, 74076, 74144, 75920, 79441, 81636, 81679, 83109, 83159, 84991, 87916, 88853, 89202, 90139, 94166, 94426, 96162, 96974, 97449, 97457, 98698]
111 111 111 111 111 111 111
[261, 2041, 4444, 4460, 4658, 5082, 5417, 5468, 6039, 6099, 8204, 9105, 9216, 9489, 9793, 10142, 11845, 14251, 15994, 17704, 17935, 18134, 18683, 18948, 22493, 23121, 24270, 25462, 25500, 26819, 34546, 36151, 41210, 41305, 44003, 45552, 45579, 49700, 50488, 50742, 51701, 52154, 52305, 53682, 54519, 55808, 58518, 58897, 59727, 60021, 63169, 64523, 65056, 65073, 65100, 65304, 65372, 67674, 68145, 68309, 68478, 69311, 70807, 71614, 72560, 72639, 72958, 75508, 75776, 76175, 78594, 78822, 79049, 79607, 80440, 80727, 81193, 82609, 84041, 84150, 85627, 87721, 89326, 89553, 90499, 90930, 92072, 92079, 92570, 93113, 95126, 95427, 96141, 96540, 97439, 98879, 99132, 99881]
422 622 222 022 022 822 422
[153, 194, 1073, 1096, 1115, 1336, 2296, 2391, 5165, 6577, 6887, 9209, 10185, 13467, 14683, 15103, 15655, 15847, 16407, 16767, 17206, 17356, 17514, 18548, 19656, 22500, 24577, 24665, 26881, 30392, 32270, 33040, 33345, 34170, 34355, 35025, 36078, 37029, 37036, 38050, 38085, 40933, 41751, 41876, 45164, 46116, 47439, 48227, 48281, 48309, 48465, 49242, 51211, 53042, 53373, 54933, 55700, 57112, 58325, 60471, 60975, 61352, 66761, 66840, 70853, 71353, 74001, 74056, 74224, 74803, 75319, 75836, 75993, 79279, 80211, 80302, 80660, 80969, 82597, 82731, 85330, 85648, 87116, 87762, 89848, 89998, 90014, 91160, 92681, 92924, 93410, 94132, 94162, 97551, 97739, 97923]
844 244 444 044 044 644 844
>>>

[/size]

Quote

a = "111"

i = 1000002
while i <= 1000003:

   f = pow(2,i)
   ff = str(f)
   s = ff
   v = ([s[i:i + 3] for i in range(0, len(s), 3)])
   x = [i for i, ltr in enumerate(v) if ltr == a]
   print(v,x)
   print(v[781],v[1347],v[1554],v[2868],v[3175],v[3300],v[3724])

   i=i+1

withmuse

Newbie

Offline

Activity: 1
Merit: 0

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

May 09, 2020, 03:46:48 PM

#1377

0x64 - CC C0 83 23 FA CC 7D 51 Cool

sssergy2705

Copper Member
Jr. Member

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Merit: 1

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

May 09, 2020, 04:19:26 PM

#1378

Quote from: withmuse on May 09, 2020, 03:46:48 PM

0x64 - CC C0 83 23 FA CC 7D 51 Cool

Are you bored? Why is it here?
If you want to communicate, there are many specialized forums and sites.

arulbero

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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

May 09, 2020, 04:21:13 PM

#1379

Quote from: withmuse on May 09, 2020, 03:46:48 PM

0x64 - CC C0 83 23 FA CC 7D 51 Cool

It is a joke. It is not the key 64.

Private key : 000000000000000000000000000000000000000000000000ccc08323facc7d51
Public key : 9b89e4472155b856c6d3f1907df38dc5d1a174eff43f9fddfc0e9d69ee915ca0 35891d2316ac33777ff5e855731c1a759a068c214e5d68e1fa96b8a0f59bbce8

PrKey WIF c.: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZ2jmYDi8fbPYBXgRnsX
Address c. : 3ee41340bd651edeba9f7f24be99607f36e38652
Address c. : 16jY7qXWg98MChdopNJ6fLgBkA5Mfi8X7T

but this is the address #64: 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN

https://www.blockchain.com/btc/address/16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN

MrFreeDragon

Sr. Member

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Activity: 443
Merit: 350

Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

May 09, 2020, 06:24:47 PM

#1380

Quote from: arulbero on May 09, 2020, 04:21:13 PM

Quote from: withmuse on May 09, 2020, 03:46:48 PM

0x64 - CC C0 83 23 FA CC 7D 51 Cool

It is a joke. It is not the key 64.
-snip-

Lol ) It was no joke as I see. The key is 64bit Wink

But another one, not from the target address

Visual Private Key Generator - https://btckeygen.com

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