Bitcoin puzzle transaction ~32 BTC prize to who solves it

[bestie1549]

Hey guys, I just think to learn the script how this address are calculated.

because bitcoin has its own native scripting language,
While funds are usually locked to a public key alone, they can also be secured with a script.

we just need to know how the address calculated, no reverse math in here.

soon i'll update what i learn about the pubkey and the address.

because this puzzle was working on single signature (P2PKH).

Peace. 

Awesome bro... We'd be waiting for the updates from you. share let us all learn. thanks

Because i read more and figure the ways.
bitcoin is never moved, is just locked and unlocked.
The only change is who can spend that amount.

that is very correct and I buy that too. Bitcoin is just like the ocean of waves
Only you who go to the sea can take the water home but if you take a portion from the waves, it doesn't mean that it will keep waving in your container.
the wallet addresses are the containers, the public addresses are the waves and the private keys are the ocean of water that makes the waves.

[mcdouglasx]

for x in range(1,65536):
         f= (ters(pub,x))
         data= open(“halfpub.txt”,”a”)
         data.write(f+”\n”)
         data.close()

It seems I was using ice secp256k1 files from 2022, after downloading the latest one from github I managed to get it running, but first public key I provided the script only generated 780 keys and returned error no 13, permission denied to open the txt file, my second attempt with another key generated 65535 keys and stopped, now what happened with the first run, why would it want to open the txt file?

Btw, how can I change the number of subtraction? For example which ones should I change if I wanted to subtract 50 instead of 1 and then divide?


k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))


def ters(Qx,Scalar):
     ScalarBin = str(bin(Scalar))[2:]
     le=len(ScalarBin)
     for i in range (1,le+1):
        if ScalarBin[le-i] == "0":
            Qx=ice.point_multiplication(k,Qx)
        else:
            Qx=ice.point_addition(Qx,neg1)
            Qx=ice.point_multiplication(k,Qx)
     return ice.point_to_cpub(Qx)


Should I replace all 1s with 50? Lol, I'm noob.

Your help is appreciated.👍

2nd edit:
A hint for interested parties, 2^255, find it's associated private key and double it mod n. Cheers. 😉

1-this is the script of a simple consecutive subtraction.

Code:

   
target =1361123746758658236458661571245234340160
#target_pub= "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
for i in range (1,3):
    h = int(str(i)+"000000000000000000000000000000000000000")
    b = (target - h)                  
    print(b)
    data = open("sustract.txt","a")
    data.write(str(i)+"= "+str(b)+"\n")
    data.close()

data open, python creates a text file if it doesn't exist where the results are saved.
maybe you have to grant write permissions.
run cmd as administrator.

Code:

for Na in range(start,end):

1,2,3,4,5,6,7,8,9,10
I use it for the number of successive subtractions.

Code:

h = int(str(i)+"000000000000000000000000000000000000000")

Code:

str(i)+"000000000000000000000000000000000000000")

# str(i)
It will be replaced according to the sequence in the range, the heap of zero will be added at the end, this to automate the amount to be subtracted like:

Code:

1000000000000000000000000000000000000000
2000000000000000000000000000000000000000
3000000000000000000000000000000000000000
4000000000000000000000000000000000000000

2-this script is the same but using pubkeys.

Code:

import secp256k1 as ice

target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
target_upub = ice.pub2upub(target_public_key)

for i in range (1, 11):
    
    char1 = int(str(i)+"000000000000000000000000000000000000000")
    pubchar1 = ice.scalar_multiplication(char1)
    resultx= ice.point_subtraction(target_upub, pubchar1)
    resultmut=resultx.hex()
    fh=ice.to_cpub(resultmut)
    f=(str(i)+"="+fh)
    print(f)
    data = open("subtractpub.txt","a")
    data.write(f+"\n")
    data.close()

3-this script subtracts the custom amount, 9999 times.
change the range to your liking.
the amount to subtract (b1) you customize according to your pubkey

Code:

import secp256k1 as ice

target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
f = ice.pub2upub(target_public_key).hex()

for i in range (10000):

    #B1= amount subtracted every round
    B1= 10000000000000000000000000000000000000
    B= ice.scalar_multiplication(B1)
    A=[]
    A.append(f)
    o=str(A)[2:-2]
    T=ice.to_cpub(o)
    upub= ice.pub2upub(o)
    f = ice.point_subtraction(upub, B).hex()
    A.append(f)
    fh2=ice.to_cpub(f)
    pubstart= fh2
    fin=(str(i)+"="+str(fh2))
    print(fin)
    data = open("sustract-x.txt","a")
    data.write(fin+"\n")
    data.close()

Example if you are looking for a pub in a range: 10000:20000
you could choose b1= 100 and for i in range (201)
This would subtract 100 in 100, 200 times from your target.

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)

[mcdouglasx]

Reducing a pubkey in terms of bsgs depends on how many pubkeys can be searched simultaneously using a .txt file, for example 1 million pubs. How many can you search without slowing down your system.

[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.

[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

[vneos]

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.

How can I cooperate with you?

[Denis_Hitov]

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.

Hello.
Write to me, we will calculate the ranges.

[citb0in]

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.

you would have to give more information, otherwise nothing can be imagined. Do you need computing power? Does your "partner" have to search a single 48bit range that you provide to him, how do you imagine the billing in case of finding the correct privkey, etc... everything else is just speculation and leaves room for wrong assumptions.

[abdenn0ur]

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.

Would be real nice of you if you could share the ranges here.

[tptkimikaze]

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.

i am also still trying to use Bit because I see that's the most highest probability of eliminating less likely keys and ranges. I have integrate and complete a whole Python code which I can predetermine the permutations of first 4 digits, skip keys that have a maximum consecutive "0" and "1", setting the weight on "0" and "1". This has eliminated a lot of keys but still, it needs a lot of computation power. I can easily find 30 up to 40 puzzle easily, but when it comes to 66 then headache already. I possibly know how you come out with the 48 bits ranges. Most likely you are trying to run a lower bits of predetermine weight 0 and 1 permutation to get the most likely range, am I correct?

[digaran]

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.😉



Here you go


2147483648  last number  of private key in decimal is 8, that's why we need to divide 8 by 10.

035318f9b1a2697010c5ac235e9af475a8c7e5419f33d47b18d33feeb329eb99a4

0000000000000000000000000000000000000000000000000000000080000000

divide by 10
99999999999999999999999999999998d668eaf0cf91f9bd7317d25489ba2727

8/10
99999999999999999999999999999998d668eaf0cf91f9bd7317d2547ced5a5b

subtract both above, result
0260b63888a1c67e22939b795af570f36d455cdc002f3ee475517fac728aa5ce24

000000000000000000000000000000000000000000000000000000000ccccccc

[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?

[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. 👍