Author

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 193. (Read 244595 times)

jr. member
Activity: 75
Merit: 5
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
newbie
Activity: 26
Merit: 8
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
newbie
Activity: 5
Merit: 0
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
copper member
Activity: 1330
Merit: 899
🖤😏
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.
newbie
Activity: 49
Merit: 0
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! Wink




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?
copper member
Activity: 1330
Merit: 899
🖤😏
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
newbie
Activity: 25
Merit: 2
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?
newbie
Activity: 10
Merit: 0
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.
hero member
Activity: 630
Merit: 731
Bitcoin g33k
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.
newbie
Activity: 49
Merit: 0
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.
newbie
Activity: 26
Merit: 8
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?
jr. member
Activity: 50
Merit: 1
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
member
Activity: 282
Merit: 20
the right steps towerds the goal
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.
member
Activity: 239
Merit: 53
New ideas will be criticized and then admired.
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.
member
Activity: 239
Merit: 53
New ideas will be criticized and then admired.


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)
jr. member
Activity: 75
Merit: 5
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.  Wink

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.
jr. member
Activity: 79
Merit: 1
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.  Wink

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.
jr. member
Activity: 75
Merit: 5
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.  Wink

Awesome bro... We'd be waiting for the updates from you. share let us all learn. thanks
jr. member
Activity: 79
Merit: 1
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.  Wink

eg. i need to learn.

copper member
Activity: 1330
Merit: 899
🖤😏


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. 😉
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