Bitcoin puzzle transaction ~32 BTC prize to who solves it

[CryptoHFs]

It is the same person or group which solve 120 bit.

So it is not kangaroo and not bsgs.

They must know something which we do not know.
The questions is -> what kind of math they use.   And second questions : I'm afraid BTC is not secure any more.

Where does these btc come from? What are they solving? Lost blocks? Cracking lost wallets? Where does these puzzle money come from?

[1715551118]

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[CryptoHFs]

Hi guys,

In continuation to this thread: https://bitcointalk.org/index.php?topic=1305887.0

While playing around with my bot, I found out this mysterious transaction:

https://blockchain.info/tx/08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15

those 32.896 BTC were sent to multiple addresses, all the private keys of those addresses seem to be generated by some kind of formula.

For example:

Address 2:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb
Biginteger PVK value: 3
Hex PVK value: 3

Address 3:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU76rnZwVdz
19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA
Biginteger PVK value: 7
Hex PVK value: 7

Address 4:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU77MfhviY5
1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e
Biginteger PVK value: 8
Hex PVK value: 8

Address 5:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Dq8Au4Pv
1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k
Biginteger PVK value: 21
Hex PVK value: 15

Address 6:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Tmu6qHxS
1PitScNLyp2HCygzadCh7FveTnfmpPbfp8
Biginteger PVK value: 49
Hex PVK value: 31

Address 7:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7hDgvu64y
1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC
Biginteger PVK value: 76
Hex PVK value: 4C

Address 8:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU8xvGK1zpm
1M92tSqNmQLYw33fuBvjmeadirh1ysMBxK
Biginteger PVK value: 224
Hex PVK value: E0

Address 9:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUB3vfDKcxZ
1CQFwcjw1dwhtkVWBttNLDtqL7ivBonGPV
Biginteger PVK value: 467
Hex PVK value: 1d3

Address 10:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUBTL67V6dE
1LeBZP5QCwwgXRtmVUvTVrraqPUokyLHqe
Biginteger PVK value: 514
Hex PVK value: 202

Address 11:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUGxXgtm63M
1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu
Biginteger PVK value: 1155
Hex PVK value: 483

Address 12:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUW5RtS2JN1
1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot
Biginteger PVK value: 2683
Hex PVK value: a7b

Address 13:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUspniiQZds
1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1
Biginteger PVK value: 5216
Hex PVK value: 1460

Address 14:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFVfZyiN5iEG
1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk
Biginteger PVK value: 10544
Hex PVK value: 2930

and so on...

until the addresses 50 (1MEzite4ReNuWaL5Ds17ePKt2dCxWEofwk) it was already cracked by someone.

Any ideas what's the formula behind the generation of these addresses?

Address 2, pvk decimal value: 3
Address 3, pvk decimal value: 7
Address 4, pvk decimal value: 8
Address 5, pvk decimal value: 21
Address 6, pvk decimal value: 49
Address 7, pvk decimal value: 76
Address 8, pvk decimal value: 224
Address 9, pvk decimal value: 467
Address 10, pvk decimal value: 514
Address 11, pvk decimal value: 1155
Address 12, pvk decimal value: 2683
Address 13, pvk decimal value: 5216
Address 14, pvk decimal value: 10544
Address 15 and after, pvk decimal value: ?

The prize would be ~32 BTC

EDIT: If you find the solution feel free to leave a tip 1DPUhjHvd2K4ZkycVHEJiN6wba79j5V1u3

Ain't that a theft?

[ripemdhash]

This is puzzle created by someone from first BTC development. Maybe Satoshi, Maybe Finn,
it is created for people to check that is possible to crack secp256k1 so it is not theft.

But the problem is someone has knowledge - not power as milion GPU's.


it has been cracked by math not by pollard , not by BSGS, not by GPUS.

and it is the problem .

[s.john]

yeah, I'm not confident anymore about the security of Bitcoin and secp256k1, especially when the solvers of these large keys are not willing to share what methods they used to reveal the private keys, the whole purpose of this challenge is to prove the immunity of the math behind bitcoin from any attack, so when someone is able to solve the challenge and not coming out to explain the method, this is basically telling us that someone knows a backdoor.

[ripemdhash]

exactly

[lordfrs]

yeah, I'm not confident anymore about the security of Bitcoin and secp256k1, especially when the solvers of these large keys are not willing to share what methods they used to reveal the private keys, the whole purpose of this challenge is to prove the immunity of the math behind bitcoin from any attack, so when someone is able to solve the challenge and not coming out to explain the method, this is basically telling us that someone knows a backdoor.

Code:

import subprocess
import time
tr=21267647932558650000000000000000000000
with open(r"C:\Users\S\Desktop\Collider-1.7.8\kl.txt","w") as f:
   for x in range(2**65):
      kl = ice.point_addition(pub, ice.point_negation(ice.scalar_multiplication(tr))).hex()
      tr += 10000000000000000000000
      f.write(kl)
      if x % 2000000 == 0:
          args = ['C:\\Users\\S\\Desktop\\Collider-1.7.8\\Collider.exe', '-t', '512', '-b', '72', '-p', '306', '-pk', '10000000000000000', '-pke', '1ffffffffffffffff', '-w', '28', '-htsz', '27', '-infile', 'kl.txt']
          proc = subprocess.run(args, cwd=r'C:\Users\S\Desktop\Collider-1.7.8')

Simple math. The only difference is how fast you can apply it.

[ripemdhash]

not to belive.

even if someone has reduct down to 100 bit, you have 2**25 of pubkeys to check for pollardrho or bsgs -> it takes time check not one or two monts.

and how you knowed that you should check  this part MSB of privatekey? 2126764793255865 + 000000000000

[MrFreeDragon]

yeah, I'm not confident anymore about the security of Bitcoin and secp256k1, especially when the solvers of these large keys are not willing to share what methods they used to reveal the private keys, the whole purpose of this challenge is to prove the immunity of the math behind bitcoin from any attack, so when someone is able to solve the challenge and not coming out to explain the method, this is basically telling us that someone knows a backdoor.

For #120 and #125 puzzle addresses it was "random" luck, and not some special knowledge. So, keep calm and be confident in secp256k1 and bitcoin security.

[ripemdhash]

yeah, I'm not confident anymore about the security of Bitcoin and secp256k1, especially when the solvers of these large keys are not willing to share what methods they used to reveal the private keys, the whole purpose of this challenge is to prove the immunity of the math behind bitcoin from any attack, so when someone is able to solve the challenge and not coming out to explain the method, this is basically telling us that someone knows a backdoor.

For #120 and #125 puzzle addresses it was "random" luck, and not some special knowledge. So, keep calm and be confident in secp256k1 and bitcoin security.

for 120 it can be random luck, but for two privatekeys -> not possibility.

[albert0bsd]

for 120 it can be random luck, but for two privatekeys -> not possibility.

Agree with you, this make think a lot

[GR Sasa]

I just got the information, but don't tell me where i got them from.

The solvers of #120 and #125 were the same persons and they were miners, they always had a mining farm which they minted dogecoins and Ethereum and other low altcoins.

They had access to huge GPU power, and most probably used Jean's Kangaroo for bruteforcing.

Most importantly; They are CURRENTLY NOW running and trying to hunt for #130 again! I hope we can be able to stop them and gain the 130's reward before them and before the end of the year. Based on calculations from the miners 3Emiwzxme7Mrj4d89uqohXNncnRM15YESs The estimated time for #130 to be hunted from the miners, should be around January 2024, February 2024, and if lucky already in December 2023!

So summary of the story: You have guys time before February 2024 to get #130's reward before the miners solve the third puzzle and get #120, #125, and #130.

[MrFreeDragon]

So, it is not some special knowledge (hidden from the public). It was a luck for #120 and #125 - by the luck I also mean that they used Kangaroo method and Jean's GPU code (it also need some luck). The #120 and #125 winners just have huge GPU power.

[bill32767]

This is puzzle created by someone from first BTC development. Maybe Satoshi, Maybe Finn,
it is created for people to check that is possible to crack secp256k1 so it is not theft.

But the problem is someone has knowledge - not power as milion GPU's.


it has been cracked by math not by pollard , not by BSGS, not by GPUS.

and it is the problem .

You are right. Its definitely maths

[bill32767]

yeah, I'm not confident anymore about the security of Bitcoin and secp256k1, especially when the solvers of these large keys are not willing to share what methods they used to reveal the private keys, the whole purpose of this challenge is to prove the immunity of the math behind bitcoin from any attack, so when someone is able to solve the challenge and not coming out to explain the method, this is basically telling us that someone knows a backdoor.

Code:

import subprocess
import time
tr=21267647932558650000000000000000000000
with open(r"C:\Users\S\Desktop\Collider-1.7.8\kl.txt","w") as f:
   for x in range(2**65):
      kl = ice.point_addition(pub, ice.point_negation(ice.scalar_multiplication(tr))).hex()
      tr += 10000000000000000000000
      f.write(kl)
      if x % 2000000 == 0:
          args = ['C:\\Users\\S\\Desktop\\Collider-1.7.8\\Collider.exe', '-t', '512', '-b', '72', '-p', '306', '-pk', '10000000000000000', '-pke', '1ffffffffffffffff', '-w', '28', '-htsz', '27', '-infile', 'kl.txt']
          proc = subprocess.run(args, cwd=r'C:\Users\S\Desktop\Collider-1.7.8')

Simple math. The only difference is how fast you can apply it.

How is it simple maths? are you among those that got #120 & #125? why is those with exposed PubKey easier than others like #66

PS: Maybe the creator of this puzzle can spend from other addresses (like randomly the odd numbers from 100 below) to see how fast our BSGS can go and to make the puzzle less difficult. We have been here since 2015 or so and alot of us has put in so much to get to here 

[GR Sasa]

As @Etar said, we need to unite to stop the miners or he will take all prizes for himself.

[lordfrs]

yeah, I'm not confident anymore about the security of Bitcoin and secp256k1, especially when the solvers of these large keys are not willing to share what methods they used to reveal the private keys, the whole purpose of this challenge is to prove the immunity of the math behind bitcoin from any attack, so when someone is able to solve the challenge and not coming out to explain the method, this is basically telling us that someone knows a backdoor.

Code:

import subprocess
import time
tr=21267647932558650000000000000000000000
with open(r"C:\Users\S\Desktop\Collider-1.7.8\kl.txt","w") as f:
   for x in range(2**65):
      kl = ice.point_addition(pub, ice.point_negation(ice.scalar_multiplication(tr))).hex()
      tr += 10000000000000000000000
      f.write(kl)
      if x % 2000000 == 0:
          args = ['C:\\Users\\S\\Desktop\\Collider-1.7.8\\Collider.exe', '-t', '512', '-b', '72', '-p', '306', '-pk', '10000000000000000', '-pke', '1ffffffffffffffff', '-w', '28', '-htsz', '27', '-infile', 'kl.txt']
          proc = subprocess.run(args, cwd=r'C:\Users\S\Desktop\Collider-1.7.8')

Simple math. The only difference is how fast you can apply it.

How is it simple maths? are you among those that got #120 & #125? why is those with exposed PubKey easier than others like #66

PS: Maybe the creator of this puzzle can spend from other addresses (like randomly the odd numbers from 100 below) to see how fast our BSGS can go and to make the puzzle less difficult. We have been here since 2015 or so and alot of us has put in so much to get to here 

If the pubkey of 66 puzzles is known, it will be solved in a few seconds.

[CryptoHFs]

As @Etar said, we need to unite to stop the miners or he will take all prizes for himself.

Then have my question answered

[bill32767]

Or maybe they can share us the hint how lucky they were (if it was luck) with the addresses with exposed PubKey or the maths formula/pattern they used. We are eagar to learn   Also, maybe, there's a way to get the Pubkey of an address, maybe --> HAS160 (+ some codes) --> RPMID (+ maybe some codes) --> PubKey. I mean what could a harmless Pubkey do  

Now everything just difficult in this present time  
Maybe easier in the next millennium  

Think about it! They said "its difficult to get the private key of any wallet" but here we are, before our own eyes, alot of know wallets has been hacked cracked. Would you now doubt the possibility of getting calculating the PubKey of an unspent wallet? I doubt not. Get the PubKey and #66 to #100 will be swiped (maybe) in 60sec with BSGS CPU of just 8GB RAM.
Think about it! Lets take this puzzle to another level.

[bill32767]

yeah, I'm not confident anymore about the security of Bitcoin and secp256k1, especially when the solvers of these large keys are not willing to share what methods they used to reveal the private keys, the whole purpose of this challenge is to prove the immunity of the math behind bitcoin from any attack, so when someone is able to solve the challenge and not coming out to explain the method, this is basically telling us that someone knows a backdoor.

Code:

import subprocess
import time
tr=21267647932558650000000000000000000000
with open(r"C:\Users\S\Desktop\Collider-1.7.8\kl.txt","w") as f:
   for x in range(2**65):
      kl = ice.point_addition(pub, ice.point_negation(ice.scalar_multiplication(tr))).hex()
      tr += 10000000000000000000000
      f.write(kl)
      if x % 2000000 == 0:
          args = ['C:\\Users\\S\\Desktop\\Collider-1.7.8\\Collider.exe', '-t', '512', '-b', '72', '-p', '306', '-pk', '10000000000000000', '-pke', '1ffffffffffffffff', '-w', '28', '-htsz', '27', '-infile', 'kl.txt']
          proc = subprocess.run(args, cwd=r'C:\Users\S\Desktop\Collider-1.7.8')

Simple math. The only difference is how fast you can apply it.

How is it simple maths? are you among those that got #120 & #125? why is those with exposed PubKey easier than others like #66

PS: Maybe the creator of this puzzle can spend from other addresses (like randomly the odd numbers from 100 below) to see how fast our BSGS can go and to make the puzzle less difficult. We have been here since 2015 or so and alot of us has put in so much to get to here 

If the pubkey of 66 puzzles is known, it will be solved in a few seconds.

Then your simple maths is not useful here, uless you have to explain further how vital it can be in solving these puzzle.

[digaran]

Nice one and congratulation, I was working to post some more hints, but it appears it has been already solved. It would be nice to know the private key if that's OK.😉