Bitcoin puzzle transaction ~32 BTC prize to who solves it

[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.😉

[nc50lc]

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  

When they say "HASH160 of the Public Key" it means: RIPEMD160[SHA256(PubKey)]
Now to get from Address to Public key, you need to reverse RIPEMD160 first then SHA256.

If they found a way to do it, they wouldn't just claimed the puzzle reward, but exploit Bitcoin Mining to get constant block rewards.
Depending on the attack, they could also claim a couple of related bounties.

[digaran]

Is there any way to figure out the first character for #160 private key?, I figured since working more than 3 months on #125 resulted in a failure, it's better to work on something which nobody would go near for years. Lol

[batareyka]

Is there any way to figure out the first character for #160 private key?, I figured since working more than 3 months on #125 resulted in a failure, it's better to work on something which nobody would go near for years. Lol

If there is a way to find out the first one, it will be possible to find out all the other symbols.
Which will lead to a complete Bitcoin hack.

[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  

When they say "HASH160 of the Public Key" it means: RIPEMD160[SHA256(PubKey)]
Now to get from Address to Public key, you need to reverse RIPEMD160 first then SHA256.

If they found a way to do it, they wouldn't just claimed the puzzle reward, but exploit Bitcoin Mining to get constant block rewards.
Depending on the attack, they could also claim a couple of related bounties.

So its basically impossible cuz in cryptography (crypto world) reversal is impossible.

[rosengold]

Congratulations to the solver of #125

12.5 BTC this was the greatest prize found by far on this puzzle. US $377.000 as today rates  

Please share the keys to update info about puzzle

[rosengold]

Is there any way to figure out the first character for #160 private key?, I figured since working more than 3 months on #125 resulted in a failure, it's better to work on something which nobody would go near for years. Lol

#160 is between:

7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

and

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

so the first character of #160 pvkey is from 8 to F in hex (8, 9, A, B, C, D, E, F)

most of priv keys starting with this same characters where found starting with E or F but this is completely impossible to know if it's the same for this.

[JamiePoc]

Hi all, I have not been here for a very long time and was hoping I could get some information on this thread.

First
There is a list in this thread showing all the addresses from 1 - 160
Do I understand this correctly?
Address 66 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so still needs resolving
I ask as I see in the post people congratulating others on finds in much higher ranges, is this one not worth the effort since it’s the easiest remaining.

And secondly
I wrote some custom software tonight to have a go at the above address, the software is currently checking slightly above 45,700 addresses per second, am I wasting my time? and how does this equate to the hash rates I see mentioned here.

Sorry if these seem dumb questions, answers would be very much appreciated.

So I did a little digging, is this correct
A hash will always be the same number of characters used so 1 Bitcoin address checked per second would equal a hash rate of 64?
If my above statement is correct then the software I wrote is running at 2,924,800 Hash's per second or 2.924 Mh/s, again am I wasting my time?

[casperas20]

u need to live   like   1 milion years to solve it with that slow speed