Bitcoin puzzle transaction ~32 BTC prize to who solves it

[Denis_Hitov]

Anyone here knows how to divide a point by 3, 4, 5, 6, 7, 8, 9 and 10 and get a correct result?

Give me a few minutes, you will be amazed, I need to prepare the sample keys on laptop. Stay tuned.😉

Look forward to explanations! Good luck bro!

5- another form of successive subtraction

Code:

import secp256k1 as ice


target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
target = ice.pub2upub(target_public_key)
num = 10 # number of times.
sustract= 1000 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub).hex()
print(res)

Tell me how to make each pub write from a new line, and not all in one line?

[1714650831]

1714650831

[1714650831]

1714650831

[1714650831]

1714650831

[1714650831]

1714650831

[digaran]

I edited my previous post with the results, you could do that for any other number, but you need to know the last digit of the target, otherwise you'd be doing subtraction of 1 to f divided by 10 with your first result.

Ps, I will not study to figure out how to divide by 10m and still have a correct result, if I do, I will not share it, that'd be ECC bent and broken totally.

[artistk]

Hello everyone! is there any topic or a reply with FULL Guide on how to join you Gents in this quest?

I'm a newbie in python but I believe I can follow a guide if someone would like to help me start hunting with you!
I can run a CPU script on a laptop and a GPU script on a Gaming PC.

Anyone can help me Setup or Guide me? Please

[vneos]

Hello everyone! is there any topic or a reply with FULL Guide on how to join you Gents in this quest?

I'm a newbie in python but I believe I can follow a guide if someone would like to help me start hunting with you!
I can run a CPU script on a laptop and a GPU script on a Gaming PC.

Anyone can help me Setup or Guide me? Please

See this site and you will get answer.

https://privatekeys.pw/puzzles/bitcoin-puzzle-tx

[bestie1549]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

everyone is willing to work with you including myself
let us in on what you need us to do and we can work as a team

[kalos15btc]

Hi, so  i found 2 address similar 11 digits of hash160 in extraction of bit 130 that means if you scan on this range and have the private key of those one of two addresses we will split the 13 BTC ,

1st

1AoNf67iZUmz1Ck9eefUfTzWQJMN5pgcPG
02113ba90a97c020ade0f3d8d0369981a723fe2bc4352815df22dd3eafae13c5a5
6b7e582a29a549cc60b591279a963f02eff02f99

pk:00000000000000000000000000000000000de4cfcadfc034c963dd053d719e88

address to search in :116 bit range
1AoNf67iZrwMSYPTDbk3Sh1yXJCARQbD7a
039d0a0241abe2411f64b4f6d29f2e1b6c837b26b6bdded577c3fc93574d3d735c
6b7e582a29a7601b79761f9f153c300c3d988231
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88


2nd
1PtStkm2bWKryHVduVjciPUxVx9UeDcCXG
02dc52ba09b16bc5cbd25aca7c82dd924f81cd31ecf29ecb264fa2cc45393728b9
fb0d9859584e68c24c1698eea4d05d2822fe4b70
pk: ad0f6ba584b355089cf6ce9cc9774

address to search in 116 bit range
1PtStkm2bLM7EK7g1rnTLBxu6aLouVuULV
03ef06cec3b3e35f68ba78618e5a5cf8663cc1a3b685dcfd197c1c0030530b1293
fb0d9859584d782df3fe652d2da5a21c30f137f9
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88




if the pk found of one of those address we will split the prize.
they have the same 11 digits of hash160 that means maybe 90 pourcent is in that range

[ing1996]

Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys.

Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it.

First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations.

Target  pub:

Code:

03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8

Target priv:

Code:

0x800000000000

Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa.

Scalar : aka n/2+1 half n +1 or 1 and half of n.

Code:

7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2

If you multiply our target pub with scalar above, you will get this :
Pub2 :

Code:

02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96

Priv :

Code:

0xc00000000000

Now if we subtract target from pub2, we will get :

Pub3, half of target :

Code:

0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da

Priv :

Code:

0x400000000000

We didn't even use division, *chop chop and good luck diving.😅


* = hurry! Get to work.

hello everyone! You can explain what happens when multiplying

Code:

7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2

by an odd key.

[ing1996]

Anyone here knows how to divide a point by 3, 4, 5, 6, 7, 8, 9 and 10 and get a correct result?

Give me a few minutes, you will be amazed, I need to prepare the sample keys on laptop. Stay tuned.😉

Look forward to explanations! Good luck bro!

5- another form of successive subtraction

Code:

import secp256k1 as ice


target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
target = ice.pub2upub(target_public_key)
num = 10 # number of times.
sustract= 1000 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub).hex()
print(res)

Tell me how to make each pub write from a new line, and not all in one line?

I don't really know python myself! bro for such simple tasks, you can use ChatGTP, he can write such tasks in code + with explanations!

[zahid888]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

im ready to work with you

As previously mentioned, I have a total of 588 ranges, where is a private key of puzzle 66. Each range requires a 48-bit computation (I am referring to GPU computation). and I am seeking an individual who is capable of performing these computations within few minutes or at least an hour.

As for our collaboration approach, I am currently developing a script to scan all 588 ranges. Once the private key is obtained, the script will automatically perform the following actions: 2% of the amount will be deducted as fees, 59% will be sent to the individual performing the computations, and 39% will be sent to me. my bitcoin address and the computation performer's bitcoin address will be pre-entered into the script. The reason for allocating 59% to the computation performer is due to their resource usage, whether through GPU rental or electricity consumption. I intend to share the entire script with the individual to ensure transparency, while keeping the ranges known only to me. Once the script initiates, neither of us will have any access to it. But when scanning of one range completed, it will be saved along with the proof of work. Certainly, it will take me some time to perform all these processes.

So why not continue searching for that person in the meantime?
(This range is an example of 588 ranges: 3b07a000000000000:3b07affffffffffff, and this address is for proof of work: 1GYpdGZBaqRDKkpFWbaRezrVeGy6g6UNfE, 1Ngdkik5LLGzmzYUWKiPYDQ6CRRGWCPo5L, 19XbiMqyqeaJLDSvHLzhJkxngDYsBPXTQo, 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so)
The person who will perform the counting of this range needs to be scanned and to post here with proof of work, or they can personally DM me as well.  In this interaction, I will assess their operational speed in completing this 48-bit range. For scanning, they can use any software, but my suggestion is to use 'VanBitCrackenS v1.0' because here they can take advantage of multi-GPU. If the puzzle address falls within this range, then it will be yours. 👍

[digaran]

hello everyone! You can explain what happens when multiplying

Code:

7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2

by an odd key.

Download and run this java calculator, it's very slow and heavy but it's good for manual calculations which is why you can only do manual calc with it.
https://github.com/MrMaxweII/Secp256k1-Calculator

Select scalar on both sides, and place private keys in hex format, then select * , + , - , ÷ , and see for yourself what would be the result.

What you should consider when dividing an odd key, dividing any odd key by 2 will always give you .5 ( point 5, half ), anything on the left side of the point . Is your actuall result no matter if you are dividing by 2 or any other number, but for example we use 2, and anything on the right side of the point . Is a 2^255 + key, we don't want that, so we subtract that from our result to get the actual answer.

3 divided by 2 = 1.>5 <  this 5 here means half of n, so if we subtract it from 1.5, we will get 1, our actual result.

Now lets make it a bit difficult, let us divide 7 by 3 = 2.33333333333333333333333333333333333333333333333333333333333333333333333333333 3333333333333333333333333333333333333333333333333333333333333333333333333333333 3333333333333333333333333333333333333333333

Now to get the n/33333........... more 333333..... etc, no need to do any complicated calculation, we just divide 7 by 3 mod n to quickly get the result, then we subtract it from the result to get our key.


Never mind all the above, I have something to twist your minds, take the following key and double, divide, do many other things with it to get really confused about how EC works. 😂

Introducing to you 2^256 of secp256k1

Code:

14551231950b75fc4402da1732fc9bebf

Try multiplying it by 2, 3, 4, 5 etc as well as dividing it, this little sucker is hiding it's half under the ground!

[bestie1549]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

im ready to work with you

As previously mentioned, I have a total of 588 ranges, where is a private key of puzzle 66. Each range requires a 48-bit computation (I am referring to GPU computation). and I am seeking an individual who is capable of performing these computations within few minutes or at least an hour.

As for our collaboration approach, I am currently developing a script to scan all 588 ranges. Once the private key is obtained, the script will automatically perform the following actions: 2% of the amount will be deducted as fees, 59% will be sent to the individual performing the computations, and 39% will be sent to me. my bitcoin address and the computation performer's bitcoin address will be pre-entered into the script. The reason for allocating 59% to the computation performer is due to their resource usage, whether through GPU rental or electricity consumption. I intend to share the entire script with the individual to ensure transparency, while keeping the ranges known only to me. Once the script initiates, neither of us will have any access to it. But when scanning of one range completed, it will be saved along with the proof of work. Certainly, it will take me some time to perform all these processes.

So why not continue searching for that person in the meantime?
(This range is an example of 588 ranges: 3b07a000000000000:3b07affffffffffff, and this address is for proof of work: 1GYpdGZBaqRDKkpFWbaRezrVeGy6g6UNfE, 1Ngdkik5LLGzmzYUWKiPYDQ6CRRGWCPo5L, 19XbiMqyqeaJLDSvHLzhJkxngDYsBPXTQo, 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so)
The person who will perform the counting of this range needs to be scanned and to post here with proof of work, or they can personally DM me as well.  In this interaction, I will assess their operational speed in completing this 48-bit range. For scanning, they can use any software, but my suggestion is to use 'VanBitCrackenS v1.0' because here they can take advantage of multi-GPU. If the puzzle address falls within this range, then it will be yours. 👍

what would the scanner get for wasting time and resources if the keys needed don't fall within these ranges?

[Tepan]

Hi Guys, I'm back, and i found the "p2sh-segwit" of 130 # puzzle address "1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua"


"p2sh-segwit": "3J9ANKe3NQgDvtS33JX1v4CB1tyDZAYgrS"

maybe there's some clue or something to work with p2sh ?

because i see the pattern of the puzzle, if someone not spend the fund, the P2SH not generated on privatekeys.pw

[bestie1549]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

what would the scanner get for wasting time and resources if the keys needed don't fall within these ranges?

I hope you know that addresses don't reflect on other addresses, they have no business whatsoever... if we are talking about pubkeys then It is pretty understandable, and I suggest you have the public key for number 66 to be able to have such confidence about the range to be scanned and if I were you and I had the pubkey then I wouldn't hesitate using kangaroo or keyhunt bsgs where it wont take up to 10 minutes to get your key

[lordfrs]

Hi Guys, I'm back, and i found the "p2sh-segwit" of 130 # puzzle address "1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua"


"p2sh-segwit": "3J9ANKe3NQgDvtS33JX1v4CB1tyDZAYgrS"

maybe there's some clue or something to work with p2sh ?

because i see the pattern of the puzzle, if someone not spend the fund, the P2SH not generated on privatekeys.pw

(p2wsh) =bc1qjtgt74dxanh4pcmwnfs9usskcg8n33cxxhp0kcnaa82qg6ye26eqrjqvzm
p2wpkh) = bc1q5fyj9pfq2x5sqt4lfjry54dvkadmfnm4sdlecz

ETH= 0x10E2d3289af14a510c88e5107B11d69c47Dd5683
Tron = TBWVYCE2x7fsszXZ16iXDEmfr8QvrAEuU2



Knowing the public key is enough to generate addresses.

[ing1996]

hello everyone! You can explain what happens when multiplying

Code:

7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2

by an odd key.

Download and run this java calculator, it's very slow and heavy but it's good for manual calculations which is why you can only do manual calc with it.
https://github.com/MrMaxweII/Secp256k1-Calculator

Select scalar on both sides, and place private keys in hex format, then select * , + , - , ÷ , and see for yourself what would be the result.

What you should consider when dividing an odd key, dividing any odd key by 2 will always give you .5 ( point 5, half ), anything on the left side of the point . Is your actuall result no matter if you are dividing by 2 or any other number, but for example we use 2, and anything on the right side of the point . Is a 2^255 + key, we don't want that, so we subtract that from our result to get the actual answer.

3 divided by 2 = 1.>5 <  this 5 here means half of n, so if we subtract it from 1.5, we will get 1, our actual result.

Now lets make it a bit difficult, let us divide 7 by 3 = 2.33333333333333333333333333333333333333333333333333333333333333333333333333333 3333333333333333333333333333333333333333333333333333333333333333333333333333333 3333333333333333333333333333333333333333333

Now to get the n/33333........... more 333333..... etc, no need to do any complicated calculation, we just divide 7 by 3 mod n to quickly get the result, then we subtract it from the result to get our key.


Never mind all the above, I have something to twist your minds, take the following key and double, divide, do many other things with it to get really confused about how EC works. 😂

Introducing to you 2^256 of secp256k1

Code:

14551231950b75fc4402da1732fc9bebf

Try multiplying it by 2, 3, 4, 5 etc as well as dividing it, this little sucker is hiding it's half under the ground!

Thanks for the info bro. I use this calculator, it works great! And how do we subtract half of n from the point X and Y coordinates?, or do we need to get the coordinates x, y from half of n? (if there is, then what are the coordinates of half of n). Or was the subtraction just an example to show so that I would understand?.

OK, tomorrow I'll try to test this key))

[Tepan]

Hi Guys, I'm back, and i found the "p2sh-segwit" of 130 # puzzle address "1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua"


"p2sh-segwit": "3J9ANKe3NQgDvtS33JX1v4CB1tyDZAYgrS"

maybe there's some clue or something to work with p2sh ?

because i see the pattern of the puzzle, if someone not spend the fund, the P2SH not generated on privatekeys.pw

(p2wsh) =bc1qjtgt74dxanh4pcmwnfs9usskcg8n33cxxhp0kcnaa82qg6ye26eqrjqvzm
p2wpkh) = bc1q5fyj9pfq2x5sqt4lfjry54dvkadmfnm4sdlecz

ETH= 0x10E2d3289af14a510c88e5107B11d69c47Dd5683
Tron = TBWVYCE2x7fsszXZ16iXDEmfr8QvrAEuU2



Knowing the public key is enough to generate addresses.

It's not a rule either to look for the address of a known public key, haha.


I can search for anything, and I don't necessarily have to use a public key. For example, I can find the P2SH for puzzle number 66.

THIS
13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so # P2PKH

"bc1qyr2956nky56hqr8fuzepdccejse4mw994lyftn" BECH32
"3PdQoXyQwWmerpt3SbF7Hbh3aukXXXXXXX" P2SH ( sorry if i do "X" end address, it might be easier to cashout when you can do something with this main/test net addresses.

WHO KNOWS THE PUBLIC KEY 66? I CAN EVEN FIND THE P2SH ADDRESS, HAHA!  

[kalos15btc]

Alright, after numerous calculations, I've come to the point where I need someone who is interested in puzzle 66 and capable of counting 48 bits within a few minutes. I have a total of 588 ranges for them to work on, with the opportunity to claim the prize of 6.6 BTC. Alternatively, if the author of the puzzle – the one who created this challenge to test the difficulty of increasing bits – has come across my post, I challenge them to assist me in improving my counting speed. I am confident in my ability to crack puzzle 66 in a matter of days.

im ready to work with you

As previously mentioned, I have a total of 588 ranges, where is a private key of puzzle 66. Each range requires a 48-bit computation (I am referring to GPU computation). and I am seeking an individual who is capable of performing these computations within few minutes or at least an hour.

As for our collaboration approach, I am currently developing a script to scan all 588 ranges. Once the private key is obtained, the script will automatically perform the following actions: 2% of the amount will be deducted as fees, 59% will be sent to the individual performing the computations, and 39% will be sent to me. my bitcoin address and the computation performer's bitcoin address will be pre-entered into the script. The reason for allocating 59% to the computation performer is due to their resource usage, whether through GPU rental or electricity consumption. I intend to share the entire script with the individual to ensure transparency, while keeping the ranges known only to me. Once the script initiates, neither of us will have any access to it. But when scanning of one range completed, it will be saved along with the proof of work. Certainly, it will take me some time to perform all these processes.

So why not continue searching for that person in the meantime?
(This range is an example of 588 ranges: 3b07a000000000000:3b07affffffffffff, and this address is for proof of work: 1GYpdGZBaqRDKkpFWbaRezrVeGy6g6UNfE, 1Ngdkik5LLGzmzYUWKiPYDQ6CRRGWCPo5L, 19XbiMqyqeaJLDSvHLzhJkxngDYsBPXTQo, 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so)
The person who will perform the counting of this range needs to be scanned and to post here with proof of work, or they can personally DM me as well.  In this interaction, I will assess their operational speed in completing this 48-bit range. For scanning, they can use any software, but my suggestion is to use 'VanBitCrackenS v1.0' because here they can take advantage of multi-GPU. If the puzzle address falls within this range, then it will be yours. 👍

im sorry i didint see the bits, i cant cuz im using cpu
but i have 1 question for you,, why your searching by address and not hash160 ? its 2 time faster than address, try it with keyhunt the last update its work faster two times than address,

[lordfrs]

Hi Guys, I'm back, and i found the "p2sh-segwit" of 130 # puzzle address "1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua"


"p2sh-segwit": "3J9ANKe3NQgDvtS33JX1v4CB1tyDZAYgrS"

maybe there's some clue or something to work with p2sh ?

because i see the pattern of the puzzle, if someone not spend the fund, the P2SH not generated on privatekeys.pw

(p2wsh) =bc1qjtgt74dxanh4pcmwnfs9usskcg8n33cxxhp0kcnaa82qg6ye26eqrjqvzm
p2wpkh) = bc1q5fyj9pfq2x5sqt4lfjry54dvkadmfnm4sdlecz

ETH= 0x10E2d3289af14a510c88e5107B11d69c47Dd5683
Tron = TBWVYCE2x7fsszXZ16iXDEmfr8QvrAEuU2



Knowing the public key is enough to generate addresses.

It's not a rule either to look for the address of a known public key, haha.


I can search for anything, and I don't necessarily have to use a public key. For example, I can find the P2SH for puzzle number 66.

THIS
13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so # P2PKH

"bc1qyr2956nky56hqr8fuzepdccejse4mw994lyftn" BECH32
"3PdQoXyQwWmerpt3SbF7Hbh3aukXXXXXXX" P2SH ( sorry if i do "X" end address, it might be easier to cashout when you can do something with this main/test net addresses.

WHO KNOWS THE PUBLIC KEY 66? I CAN EVEN FIND THE P2SH ADDRESS, HAHA!  

3PdQoXyQwWmerpt3SbF7Hbh3aukC5w28GP

[digaran]

Thanks for the info bro. I use this calculator, it works great! And how do we subtract half of n from the point X and Y coordinates?, or do we need to get the coordinates x, y from half of n? (if there is, then what are the coordinates of half of n). Or was the subtraction just an example to show so that I would understand?.

OK, tomorrow I'll try to test this key))

You could use offline tools or online such as secret scan dot org or privatekeys dot pw in crypto calculator section.

With java one I linked, you can't calculate addition subtraction of a vector ( public key ) with scalar ( private key ), you can only multiply and divide.
Also my calculation above is not accurate, if we divide 7 by 3 we need to divide n by 3 and subtract the both results to get our answer which is 2. Remember when you do sub, add, if you subtract a -n point from a +n point, you will be adding them, learn - and + ops first.

[mcdouglasx]

5- another form of successive subtraction

Code:

import secp256k1 as ice


target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
target = ice.pub2upub(target_public_key)
num = 10 # number of times.
sustract= 1000 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub).hex()
print(res)

Tell me how to make each pub write from a new line, and not all in one line?

Code:

import secp256k1 as ice


target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
target = ice.pub2upub(target_public_key)
num = 10 # number of times.
sustract= 1000 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub)
for t in range (num+1):
    h= res[t*65:t*65+65]
    data = open("loop-subtrac.txt","a")
    data.write(str(h.hex())+"\n")
    data.close()