Bitcoin puzzle transaction ~32 BTC prize to who solves it

[nomachine]

If (someone solves Puzzle 130) -> (they have extreme computational resources).
Not (anyone has extreme computational resources).
Therefore, Not (anyone solved Puzzle 130) -> it may have been manipulated.

If someone solved Puzzle 130, then holy molly, they’ve got a supercomputer the size of a small moon! But wait — no one’s got one of those lying around (unless NASA’s hiding something). So, if no one solved Puzzle 130, holy molly, something fishy’s going on — maybe the puzzle fairies had a little too much coffee and fancy lunches in Dubai! 

Greetings to Everyone here. I hope to add my small contribution and knowledge to these pages, I am ExiledDatum.

Welcome, ExiledDatum! We're glad to have you here. 

[viljy]

I have more and more doubts that some geniuses are solving the puzzle. In my opinion, starting from the 120bit solution, the creator does all this. To keep the interest in his mystery alive. If you compare the sequence of events (dates of decisions, increase in the prize), then everything adds up to a logical picture. Now it is more profitable to use computing power for the inference of neural networks, for their training, rather than for searching for a needle in a haystack with an unknown result. And here you need very large computing power, which will cost a lot to rent.

Honestly, I don't think it was the creator who withdrew the funds. Someone who over the years didn't stop the challenge despite having the private keys - he could have withdrawn them with multi-million profits, and instead deposited dozens of BTC to each address shows that the creator is a person who sleeps on cash and is like oxygen to him - obviously he is and that he will never take it away. So I don't think that someone stopped achieving the ultimate goal of creating this challenge by fawning over change... especially since finding 120,125,130 does not bring anything positive in the perspective of the entire challenge... especially since we don't know of any solutions that he could skillfully announce while making progress towards achieving the goal.
I think the finder is someone who reads us here, but does not share the results or methods of action, because so far she has successfully gained $2 million on it using her excellent method of breaking subsequent thresholds. In other words - for safety's sake she is silent, because she simply -> plans to top up those $2 million with subsequent levels. If we live to see #160, there is a chance that we will learn the solutions.

That's what I thought at first, too. However, it is extremely doubtful that there is any "hidden mathematical genius" in the world.  He would have received more money and recognition if he had made his discoveries public.  I also think it is extremely stupid to use large computing resources to solve the puzzle, when they can be rented out for inference or neural network training and guaranteed to make a profit.
So the version of a great mathematician who solves puzzles in some new way on a single video card is questionable as well as the version of a businessman who prefers to play the lottery instead of a stable business. Then it remains a logical option that the creator of the puzzle (perhaps it is not one person, but an organization) stimulates the "bait" ("look, someone found the key, so it's possible!") to work on testing the security of bitcoin.
There is another option - a quantum computer and Shor's algorithm are used, then it is definitely some organization testing its power.

[kTimesG]

If someone solved Puzzle 130, then holy molly, they’ve got a supercomputer the size of a small moon! But wait — no one’s got one of those lying around (unless NASA’s hiding something). So, if no one solved Puzzle 130, holy molly, something fishy’s going on — maybe the puzzle fairies had a little too much coffee and fancy lunches in Dubai! 

Come on, let's get real. I myself computed that if I had 150k $ lying around I would have solved 130 in less than 90 days. Simple crude math - divide total ops by number of op/s, again by ops/GPU/s, multiply with price/h.

So no supercomputer required here, just $ to pay for the most optimized piece of software to run, to minimize costs.

But now 135 requires 5 times more money to solve, or for GPU prices to go down, or for performance per $ to go up, which WILL happen. What I do not understand, is what the deal is with the 3Emi script address, is it a group, is it some sort of smart contract, why was it not spent already, etc.

[Akito S. M. Hosana]

There is another option - a quantum computer and Shor's algorithm are used, then it is definitely some organization testing its power.

" Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth. "

[brainless]

import os
import random

Target_Puzzle_No = 135
unde-Target_Puzzle_No = 100
k = 0

while True:         
    low  = 2**(Target_Puzzle_No-1)
    high = -1 + 2**(Target_Puzzle_No)
    beween = 2**100
   
    startnumber = random.randrange(low,high)
    check = (startnumber+beween)   
    if startnumber > high :
        break

    st = hex(startnumber)[2:]
    en = hex(check)[2:]
    print (st,file=open("67-checklis.txt", "a"))
    print (st)
    os.system("timeout 20m ./vs-kangaroo-multi -v 1 -gpu -g 128,128 -bits " + st + ":" + en + " 02145d2611c823a396ef6712ce0f712f09b9b4f3135e3e0aa3230fb9b6d08d1e16 ")
    k += 89
# set public key in os command
# set os command as per your kangaroo

[ExiledDatum]

I have more and more doubts that some geniuses are solving the puzzle. In my opinion, starting from the 120bit solution, the creator does all this. To keep the interest in his mystery alive. If you compare the sequence of events (dates of decisions, increase in the prize), then everything adds up to a logical picture. Now it is more profitable to use computing power for the inference of neural networks, for their training, rather than for searching for a needle in a haystack with an unknown result. And here you need very large computing power, which will cost a lot to rent.

Honestly, I don't think it was the creator who withdrew the funds. Someone who over the years didn't stop the challenge despite having the private keys - he could have withdrawn them with multi-million profits, and instead deposited dozens of BTC to each address shows that the creator is a person who sleeps on cash and is like oxygen to him - obviously he is and that he will never take it away. So I don't think that someone stopped achieving the ultimate goal of creating this challenge by fawning over change... especially since finding 120,125,130 does not bring anything positive in the perspective of the entire challenge... especially since we don't know of any solutions that he could skillfully announce while making progress towards achieving the goal.
I think the finder is someone who reads us here, but does not share the results or methods of action, because so far she has successfully gained $2 million on it using her excellent method of breaking subsequent thresholds. In other words - for safety's sake she is silent, because she simply -> plans to top up those $2 million with subsequent levels. If we live to see #160, there is a chance that we will learn the solutions.

That's what I thought at first, too. However, it is extremely doubtful that there is any "hidden mathematical genius" in the world.  He would have received more money and recognition if he had made his discoveries public.  I also think it is extremely stupid to use large computing resources to solve the puzzle, when they can be rented out for inference or neural network training and guaranteed to make a profit.
So the version of a great mathematician who solves puzzles in some new way on a single video card is questionable as well as the version of a businessman who prefers to play the lottery instead of a stable business. Then it remains a logical option that the creator of the puzzle (perhaps it is not one person, but an organization) stimulates the "bait" ("look, someone found the key, so it's possible!") to work on testing the security of bitcoin.
There is another option - a quantum computer and Shor's algorithm are used, then it is definitely some organization testing its power.

Or someone simply noticed the lid does more than keep the liquid in.

I have made more progress by simply eliminating parts that it could not be, logistical analysis of key space and just some plain hands-on common sense. Real-world problem solving applied to reducing the key space by logical means. All keys with repeating prefixes, all with more than two repeating characters to list some already mentioned before. I am actively seeking puzzle solutions personally.

I have only been working on these puzzles for perhaps a year with direction and structured aim. I can only imagine what stage most of you all are at if far beyond my new presence is stumbling blindly through.

I do work with ML, tinker with special custom models in attempts to bridge the rational logic beyond the simple prediction, transformers or Berts.

Someone with the knowledge and means to code bug free model training and compile datasets upon which to aim these models.... now that, seems more likely as those able and with enough understanding could be leveraging this emerging technology which is gaining ground and establishing a stable foundation anyone dedicating the time can begin doing in some garage setup.

I'm sure after removing all those false positives. The reduced key space hardly takes a "hidden mathematical genius".

Though all my work is done, on a single graphics card on a laptop. I believe any solutions happening at this moment is due to the collimation of over four years of continual persistence and well mapped logic in isolated sectors, in calculated sub-ranges.



Here is the output of one of my experiments, breaking the key space of #66 into some 25 million sub-ranges and sampling addresses after removing parts of the key space by a set of rules, coupled with several other gum-shoe insanity.


Address Range: 19FaJ7mnZ1o2bRdLWNPswW829NZ3KtkUve to 1DJjP7WMvJNZSa1nURTfPMXQP4Pjx4CZVJ
Subrange 167575:
  Decimal Range: 223088402697171072894127398419995557888 to 223089728030944213908759676992704479231
  Hexadecimal Range: 00000000000000000000000000000000a7d54374bc6a7ab8a000000000000000 to 00000000000000000000000000000000a7d584ccccccc88b8fffffffffffffff
  Address Range: 1HPxAfdbeShoNQoLyGrMYjwX8KE3JDTgqP to 196jyzz3omwHiWx7SAUiRfD7DLeEd6dpqN
Subrange 167576:
  Decimal Range: 223089728030944213908759676992704479232 to 223091053364717354923391955565413400575
  Hexadecimal Range: 00000000000000000000000000000000a7d584ccccccc88b9000000000000000 to 00000000000000000000000000000000a7d5c624dd2f165e7fffffffffffffff

[citb0in]

Or someone simply noticed the lid does more than keep the liquid in.

I have made more progress by simply eliminating parts that it could not be, logistical analysis of key space and just some plain hands-on common sense. Real-world problem solving applied to reducing the key space by logical means. All keys with repeating prefixes, all with more than two repeating characters to list some already mentioned before. I am actively seeking puzzle solutions personally.

I have only been working on these puzzles for perhaps a year with direction and structured aim. I can only imagine what stage most of you all are at if far beyond my new presence is stumbling blindly through.

I do work with ML, tinker with special custom models in attempts to bridge the rational logic beyond the simple prediction, transformers or Berts.

Someone with the knowledge and means to code bug free model training and compile datasets upon which to aim these models.... now that, seems more likely as those able and with enough understanding could be leveraging this emerging technology which is gaining ground and establishing a stable foundation anyone dedicating the time can begin doing in some garage setup.

I'm sure after removing all those false positives. The reduced key space hardly takes a "hidden mathematical genius".

Though all my work is done, on a single graphics card on a laptop. I believe any solutions happening at this moment is due to the collimation of over four years of continual persistence and well mapped logic in isolated sectors, in calculated sub-ranges.

not true for #110 and #115

[brainless]

Can some one explain about forks, inside Bitcoin puzzle  how to cash out or convert to bitcoin

[albert0bsd]

Can some one explain about forks, inside Bitcoin puzzle  how to cash out or convert to bitcoin

Usually those forks are worthless but in any case you can make some tens bucks with them. You can use Guarda wallet (web app) to move them to another wallet. And you can use changenow.io to exchange them to bitcoin.

[brainless]

Can some one explain about forks, inside Bitcoin puzzle  how to cash out or convert to bitcoin

Usually those forks are worthless but in any case you can make some tens bucks with them. You can use Guarda wallet (web app) to move them to another wallet. And you can use changenow.io to exchange them to bitcoin.

Thankx for info
I were looking something like strong wallet, where load privkeys and all crypto loads in one click and convert them

[GR Sasa]

Just curious, So is the solver of #120, #125, #130 the same solvers or#130 is a new guy?

[citb0in]

Just curious, So is the solver of #120, #125, #130 the same solvers or#130 is a new guy?

nobody knows that except the solver himself and anyone who claims the opposite is lying, icluding me 

[albert0bsd]

Thank you to Alberto, NoMachine, and many others who have contributed so much to these endeavors. I have learned a great deal from many of you who so willingly shared your code, time and very essence while working on this puzzle.

Thank you very much, I also receive your email, it really means a lot to me.

Welcome and have an excellent day.

[PrimeOrder]

Hello all. Been lurking for a while, finally joined.

I've been playing with bitcoin keys etc. for a while but only discovered the puzzle a few months ago. I've written my own python scripts and have been playing around with the various other ones mentioned on here (thanks to the creators for their work).

I have a question if anyone is familiar with Iceland2k41's scripts. I was playing around with his key division and subtraction scripts. I've visited his github but it seems questions are not always answered very clearly. Anyway, my question ..

I know how to recreate the original key with his PubDiv.py script but I can't find how to do it with the PubSub.py one. A couple of people asked on his github but weren't really answered.

Thanks for any help you can give.

[masoudtanhha]

Hello, what do you think about this idea?

https://iancoleman.io/bip39/

https://ibb.co/hRcp93x

Code:

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Ability repeat every 11 row              (Start From Puzzel 4)
Able repeat ever 33 & 11 & 33 ...     (Start From Puzzel 5)
About Repeat Ever 11 & 33 & 11 ...  (Start From Puzzel 16)
above Repeat Ever 11 & 11...           (Start from puzzel 39)
------------------------------------------------------------------
And From Puzzel 4 look at clomn  after last abandon word ever 7 row word changed to (b) and Then (c)

https://ibb.co/vzdB5nP

With this prediction of puzzle 67, the first word after Abandon starts with the letter C

abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon c . . . . .


start Range Puzzel 67
0000000000000000000000000000000000000000000000040000000000000000
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End Range Puzzel 67
000000000000000000000000000000000000000000000007ffffffffffffffff
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But because according to the prediction, it starts with the letter C and the first word that starts with Cabbage
Abandon is not repeated in all lines after mixing the words

In my opinion, the beginning of Range can be from here on

https://ibb.co/DVSYY1F


Python Code For Image

Code:

import customtkinter as ctk
from tkinter import filedialog, messagebox
from mnemonic import Mnemonic
import hashlib
import ecdsa
import base58
import pyperclip

# تنظیمات اولیه تم
ctk.set_appearance_mode("Dark")
ctk.set_default_color_theme("blue")

# تابع تبدیل کلید عمومی به آدرس فشرده P2PKH
def pubkey_to_address(public_key):
    sha256_result = hashlib.sha256(public_key).digest()
    ripemd160 = hashlib.new('ripemd160')
    ripemd160.update(sha256_result)
    hashed_public_key = ripemd160.digest()
    versioned_key = b'\x00' + hashed_public_key
    checksum = hashlib.sha256(hashlib.sha256(versioned_key).digest()).digest()[:4]
    address_bytes = versioned_key + checksum
    address = base58.b58encode(address_bytes).decode()
    return address

# تابع تولید آدرس فشرده P2PKH از entropy
def generate_address_from_entropy():
    entropy_hex = entropy_entry.get()
    if not entropy_hex:
        messagebox.showerror("Error", "Please generate or enter the entropy first.")
        return

    try:
        private_key_bytes = bytes.fromhex(entropy_hex)
        private_key = ecdsa.SigningKey.from_string(private_key_bytes, curve=ecdsa.SECP256k1)
        public_key = private_key.verifying_key.to_string("compressed")
        address = pubkey_to_address(public_key)
        p2pkh_entry.delete(0, ctk.END)
        p2pkh_entry.insert(0, address)
    except Exception as e:
        messagebox.showerror("Error", f"Invalid entropy or private key: {str(e)}")

# تابع تولید چک‌سام (کلمه 24) و محاسبه entropy
def generate_checksum():
    selected_words = [combo.get() for combo in combos[:23]]
    if '' in selected_words:
        messagebox.showerror("Error", "Please select all 23 words.")
        return

    mnemo = Mnemonic("english")
    valid_checksums = []

    for word in mnemo.wordlist:
        mnemonic = selected_words + [word]
        try:
            entropy = mnemo.to_entropy(' '.join(mnemonic))
            valid_checksums.append(word)
        except:
            continue

    if valid_checksums:
        combos[23].configure(values=valid_checksums)
        combos[23].set(valid_checksums[0])
    else:
        messagebox.showerror("Error", "No valid checksum word found.")

# تابع برای نمایش entropy
def show_entropy():
    selected_words = [combo.get() for combo in combos[:23]]  # دریافت 23 کلمه
    if '' in selected_words:
        messagebox.showerror("Error", "Please select all words.")
        return

    checksum_word = combos[23].get()  # دریافت کلمه چک‌سام (کلمه 24)
    if not checksum_word:
        messagebox.showerror("Error", "Please generate or select a checksum word.")
        return

    selected_words.append(checksum_word)  # اضافه کردن چک‌سام به لیست

    mnemo = Mnemonic("english")
    try:
        entropy = mnemo.to_entropy(' '.join(selected_words))
        entropy_entry.delete(0, ctk.END)
        entropy_entry.insert(0, entropy.hex())
    except Exception as e:
        messagebox.showerror("Error", f"Invalid mnemonic: {str(e)}")

# تابع برای ذخیره اطلاعات در فایل
def save_results():
    selected_words = [combo.get() for combo in combos]
    entropy_hex = entropy_entry.get()
    address = p2pkh_entry.get()

    if not entropy_hex or not address:
        messagebox.showerror("Error", "Please generate all necessary fields before saving.")
        return

    file_path = filedialog.asksaveasfilename(defaultextension=".txt", filetypes=[("Text files", "*.txt")])
    if file_path:
        with open(file_path, 'w') as file:
            file.write("BIP39 Mnemonic Words:\n")
            file.write(", ".join(selected_words) + "\n\n")
            file.write(f"Entropy (Hex): {entropy_hex}\n")
            file.write(f"P2PKH (Compressed Address): {address}\n")
        messagebox.showinfo("Saved", f"Results saved to {file_path}")

# تابع کپی به کلیپ‌بورد
def copy_to_clipboard(entry_widget):
    pyperclip.copy(entry_widget.get())
    messagebox.showinfo("Copied", "Copied to clipboard")

# تنظیمات صفحه اصلی
root = ctk.CTk()
root.geometry("1200x700")
root.title("BIP39 Mnemonic Generator")

combos = []

# ایجاد 24 کادر کشویی برای کلمات
for i in range(24):
    label = ctk.CTkLabel(root, text=f"Word {i+1}")
    label.grid(row=i//4, column=(i % 4) * 2, padx=10, pady=5, sticky="w")

    combo = ctk.CTkComboBox(root, values=Mnemonic("english").wordlist)
    combo.grid(row=i//4, column=(i % 4) * 2 + 1, padx=10, pady=5, sticky="w")
    combos.append(combo)

# بخش پایین صفحه برای نمایش entropy و سایر دکمه‌ها
entropy_label = ctk.CTkLabel(root, text="Entropy (Hex):")
entropy_label.grid(row=8, column=0, padx=10, pady=10, sticky="w")

entropy_entry = ctk.CTkEntry(root, width=500)
entropy_entry.grid(row=8, column=1, columnspan=3, padx=10, pady=10, sticky="w")

checksum_button = ctk.CTkButton(root, text="Generate Checksum", command=generate_checksum)
checksum_button.grid(row=9, column=1, padx=10, pady=5, sticky="w")

entropy_button = ctk.CTkButton(root, text="Show Entropy", command=show_entropy)
entropy_button.grid(row=9, column=2, padx=10, pady=5, sticky="w")

p2pkh_label = ctk.CTkLabel(root, text="P2PKH (Compressed):")
p2pkh_label.grid(row=12, column=0, padx=10, pady=10, sticky="w")

p2pkh_entry = ctk.CTkEntry(root, width=500)
p2pkh_entry.grid(row=12, column=1, columnspan=3, padx=10, pady=10, sticky="w")

generate_address_button = ctk.CTkButton(root, text="Generate Address", command=generate_address_from_entropy)
generate_address_button.grid(row=13, column=1, padx=10, pady=5, sticky="w")

save_button = ctk.CTkButton(root, text="Save Results", command=save_results)
save_button.grid(row=13, column=2, padx=10, pady=5, sticky="w")

# اجرای برنامه
root.mainloop()

I hope it is useful and I am waiting for your comments

[benjaniah]

I had calculated 2628a3ad7d90cd20f to 2728ad7d90cd20f. I had sitting in a bat file created and last ran 5/13/2024.

Same here, I had a script that scanned ranges at given intervals, here is part of that script:

Code:

run_command "./vanitysearch -t 0 -gpu -gpuId 0,1,2,3 -o 283prefixes.txt --continue 10 --keyspace 28326F10000000000:+10000000000 13zb1hQbW" 10
run_command "./vanitysearch -t 0 -gpu -gpuId 0,1,2,3 -o 283prefixes.txt --continue 11 --keyspace 2832AF10000000000:+10000000000 13zb1hQbW" 11
run_command "./vanitysearch -t 0 -gpu -gpuId 0,1,2,3 -o 283prefixes.txt --continue 12 --keyspace 2832EF10000000000:+10000000000 13zb1hQbW" 12
run_command "./vanitysearch -t 0 -gpu -gpuId 0,1,2,3 -o 283prefixes.txt --continue 13 --keyspace 28332F10000000000:+10000000000 13zb1hQbW" 13

I finished this script run on Sept. 8. So I just skipped over 66's range... no good luck there.  I also spent a decent chunk of money, and decided to stop throwing good money after bad. I'll just be an observer rather than a participant for the rest of the puzzles. Good luck to people out there still searching. Although you'd probably be better off investing your money rather than gambling it on the puzzles. As the saying goes, Only spend what you're willing to lose!

[nomachine]

Hello, what do you think about this idea?

Alright, let’s break this down:

We’ve got

2048×2048×2048×2048×2048×2048×2048
= 2048 ^ 7


which equals a staggering 151115727451828646838272 permutations. Yep, that’s a lot!

Now, if our Python script can crank out 350,000 permutations a day, we’re looking at approximately 1182092323862940 years to brute-force our way through them all.

So, if you’re thinking about getting rich by brute force mnemonic , here’s my advice: save yourself the headache! Grab some fried chicken and beers, plop down on the couch, and binge-watch Netflix. Trust me, it’s way more fun than waiting eons for a lucky break!

[ExiledDatum]

I had calculated 2628a3ad7d90cd20f to 2728ad7d90cd20f. I had sitting in a bat file created and last ran 5/13/2024.

Same here, I had a script that scanned ranges at given intervals, here is part of that script:

Code:

run_command "./vanitysearch -t 0 -gpu -gpuId 0,1,2,3 -o 283prefixes.txt --continue 10 --keyspace 28326F10000000000:+10000000000 13zb1hQbW" 10
run_command "./vanitysearch -t 0 -gpu -gpuId 0,1,2,3 -o 283prefixes.txt --continue 11 --keyspace 2832AF10000000000:+10000000000 13zb1hQbW" 11
run_command "./vanitysearch -t 0 -gpu -gpuId 0,1,2,3 -o 283prefixes.txt --continue 12 --keyspace 2832EF10000000000:+10000000000 13zb1hQbW" 12
run_command "./vanitysearch -t 0 -gpu -gpuId 0,1,2,3 -o 283prefixes.txt --continue 13 --keyspace 28332F10000000000:+10000000000 13zb1hQbW" 13

I finished this script run on Sept. 8. So I just skipped over 66's range... no good luck there.  I also spent a decent chunk of money, and decided to stop throwing good money after bad. I'll just be an observer rather than a participant for the rest of the puzzles. Good luck to people out there still searching. Although you'd probably be better off investing your money rather than gambling it on the puzzles. As the saying goes, Only spend what you're willing to lose!

I will have to speak with you, that looks like a modified version and I can only speculate you fell upon the growth factor of 18 to 20. Since the puzzle is already solved and the factor only helps find #66 I don't mind sharing the math behind my similar conclusion. I can not upload the image\screenshot of the conversation with my custom, privately hosted model. I can share the text within a small section of it.


So let me disclose this first and foremost. I have a custom model trained and fine-tuned by myself for personal research and assistance. This is not any publicly avaliable chat model. I am lazy and use it for rapid calculations and assistance in holding large hex values and decimal values in a memory pool to save on computations and energy load.
------------------------------------------------------------------------------------------------------------------------------------------------

Model said:
Thank you for providing the details for #64 and #65 puzzle keys. Let's calculate #64 puzzle key and #65 puzzle key * 2 - #64 key.

#64 Puzzle Key:

HEX: 000000000000000000000000000000000000000000000000f7051f27b09112d4
#65 Puzzle Key * 2 - #64 Key:

(1a838b13505b26867 * 2) - 000000000000000000000000000000000000000000000000f7051f27b09112d4 = 1a838b13505b26867
Now, let's perform the steps to get the Bitcoin address and public key for each of these puzzle keys:

For #64 Puzzle Key:

Convert the private key from hex to bytes: 000000000000000000000000000000000000000000000000f7051f27b09112d4

Create a signing key from the private key bytes using the SECP256k1 elliptic curve.

Get the corresponding public key.

Serialize the public key in compressed format (33 bytes).

Calculate the SHA-256 hash of the public key.

Calculate the RIPEMD-160 hash of the SHA-256 hash.

Add the version byte (0x00 for mainnet) to the RIPEMD-160 hash.

Calculate the double SHA-256 checksum.

Append the checksum to the extended RIPEMD-160 hash.

Convert the bytes to a base58-encoded Bitcoin address.

For #65 Puzzle Key * 2 - #64 Key:

Convert the private key from hex to bytes: 1a838b13505b26867

Create a signing key from the private key bytes using the SECP256k1 elliptic curve.

Get the corresponding public key.

Serialize the public key in compressed format (33 bytes).

Calculate the SHA-256 hash of the public key.

Calculate the RIPEMD-160 hash of the SHA-256 hash.

Add the version byte (0x00 for mainnet) to the RIPEMD-160 hash.

Calculate the double SHA-256 checksum.

Append the checksum to the extended RIPEMD-160 hash.

Convert the bytes to a base58-encoded Bitcoin address.

Please give me a moment to perform these calculations.

I have performed the calculations for both #64 Puzzle Key and #65 Puzzle Key * 2 - #64 Key to obtain their respective Bitcoin addresses and public keys:

For #64 Puzzle Key:

Bitcoin Address: 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN
Public Key: (This key can be derived from the private key but is not shown here.)
For #65 Puzzle Key * 2 - #64 Key:

Bitcoin Address: 18ZMbwUFLMHoZBbfpCjUJQTCMCbktshgpe
Public Key: 02b21a6b1590b145841a0dabbe71ea01e29ed60f0e468cff36445a9c92eb3a6375
These addresses and public keys have been generated following the standard Bitcoin address generation rules. If you have any further questions or if there's anything else I can assist you with, please feel free to ask.


You said:
generate #66

Model said:
To generate #66, we'll follow the same steps as before. Here are the details:

#66 Puzzle Key:

HEX: 000000000000000000000000000000000000000000000000f7051f27b09112d4
Now, let's perform the steps to get the Bitcoin address and public key for #66 Puzzle Key:

Convert the private key from hex to bytes: 000000000000000000000000000000000000000000000000f7051f27b09112d4

Create a signing key from the private key bytes using the SECP256k1 elliptic curve.

Get the corresponding public key.

Serialize the public key in compressed format (33 bytes).

Calculate the SHA-256 hash of the public key.

Calculate the RIPEMD-160 hash of the SHA-256 hash.

Add the version byte (0x00 for mainnet) to the RIPEMD-160 hash.

Calculate the double SHA-256 checksum.

Append the checksum to the extended RIPEMD-160 hash.

Convert the bytes to a base58-encoded Bitcoin address.

Please give me a moment to perform these calculations for #66.

I have generated the Bitcoin address for #66 Puzzle Key using the standard Bitcoin address generation rules:

Bitcoin Address: 1DejBqA9KXMpRoJjTJtsSyZm1N1uqWv2Dr
Please note that I cannot generate the corresponding public key as it requires access to the private key, which I don't have. If you have any further questions or if there's anything else I can assist you with, please feel free to ask.


You said:
Generate #65 *3 + #64 puzzle key and generate #65 *3 - #64

2/2

Model said:

To generate #65 * 3 + #64 Puzzle Key, we'll follow these steps:

Take the #65 Puzzle Key: 000000000000000000000000000000000000000000000001a838b13505b26867
Multiply it by 3.
Add the #64 Puzzle Key: 000000000000000000000000000000000000000000000000f7051f27b09112d4
Let's calculate this:

1a838b13505b26867 * 3 = 00000000000000000000000000000000000000000000000514158040b76d9e1 Now, add the #64 Puzzle Key:

00000000000000000000000000000000000000000000000514158040b76d9e1 + 000000000000000000000000000000000000000000000000f7051f27b09112d4 = 00000000000000000000000000000000000000000000000665d70945c5e4b05

So, the result of #65 * 3 + #64 Puzzle Key is:

HEX: 00000000000000000000000000000000000000000000000665d70945c5e4b05

Now, let's calculate #65 * 3 - #64 Puzzle Key:

1a838b13505b26867 * 3 = 00000000000000000000000000000000000000000000000514158040b76d9e1 Now, subtract the #64 Puzzle Key:

00000000000000000000000000000000000000000000000514158040b76d9e1 - 000000000000000000000000000000000000000000000000f7051f27b09112d4 = 00000000000000000000000000000000000000000000000443e3d1d5ae38ad So, the result of #65 * 3 - #64 Puzzle Key is:

HEX: 00000000000000000000000000000000000000000000000443e3d1d5ae38ad

If you need further assistance or have any more calculations to perform, please let me know.


You said:
Take the previous two results and subtract #65 from them

Model said:

Sure, let's subtract the #65 Puzzle Key (1a838b13505b26867) from the two results we obtained previously:

Result of #65 * 3 + #64: 00000000000000000000000000000000000000000000000665d70945c5e4b05
Result of #65 * 3 - #64: 00000000000000000000000000000000000000000000000443e3d1d5ae38ad
Now, we'll subtract #65:

00000000000000000000000000000000000000000000000665d70945c5e4b05 - 1a838b13505b26867 = ffc19ec6aa168b9e04d4b2c69f53549e
00000000000000000000000000000000000000000000000443e3d1d5ae38ad - 1a838b13505b26867 = 28aa1181a66f8fd3ecf5deaae87d02e6
So, the results of subtracting #65 from the previous two calculations are:

HEX: ffc19ec6aa168b9e04d4b2c69f53549e
*---HEX: 28aa1181a66f8fd3ecf5deaae87d02e6----*
If you have any more questions or need further calculations, feel free to ask.



My

I don’t mean to offend anyone—this is just my analysis. Everyone is entitled to their own opinion, and this is simply mine. I have no intention of offending anyone.

I'm a mathematician. I deal with logic - a part of mathematics.


Formulating the Argument.

Steps for a logical analysis:
Formalize the claims: The main claims can be expressed as premises in logic:

Premise 1: After puzzle 120 was solved, the creator increased the prize money on 2023-04-16 to 1000 BTC (public stunt hypothesis).
Premise 2: Puzzle 125 was removed on 2023-07-09, likely as another stunt (further public manipulation hypothesis).
Premise 3: Puzzle 130 was moved after the solving of puzzle 66, likely to generate more attention.
Premise 4: It is practically impossible to solve puzzle 130 without at least 1000 GPUs, making it improbable for anyone to solve it legitimately.


Logical Conclusion:
Using modus tollens (if P implies Q, and Q is false, then P must be false):

If it is true that solving Puzzle 130 is only possible with extreme computational resources (Premise 4), and no one is known to possess those resources, then no one could have solved it without manipulation.
The formula could look like:

If (someone solves Puzzle 130) -> (they have extreme computational resources).
Not (anyone has extreme computational resources).
Therefore, Not (anyone solved Puzzle 130) -> it may have been manipulated.


This reasoning framework doesn't "prove" the manipulation but shows that, given the constraints, the likelihood of a public stunt increases if the technical requirements are realistically impossible for most individuals.

I believe there are some parties and individuals who possess both above average intelligence and have private means which they used in order to develop some cutting-edge applications of technology not tied to any large corporations. While these are not supercomputers, or quantum circuits in line with mainstream public knowledge.

I can say with positive proof and personal observations of such systems, they are impressive in their capabilities when compared to the average software running in any programming language on the standard operational systems. They are not a quantum computer in any sense, but their computational abilities are staggering.

I also have a B.A. in mathematics with a concentration in computer science. And some other education, I think if people still accept my cereal box diploma.

I deal with logistics, material science and the electronic engineering of emerging technologies.

I am horrible at writing code from scratch, so I tend to modify broken code snippets fixing the one-off errors or faulty logic to build my own software for my needs.

I guess that would make something of a lesser mathematician, perhaps on a good day?


I would humbly like to offer, we are in a time of profound revolution of conventional models of development and production. Anyone with half a brain, determination and some diy skills, can build things to rival billion dollar corporations.

These last two decades of technological advancements, coupled with the vast libraries available via the internet.

Humor me a moment, and fathom, what a being with 130 IQ or higher, insomnia and an unquenchable thirst of knowledge could accomplish?

[AndrewWeb]

Come on, let's get real. I myself computed that if I had 150k $ lying around I would have solved 130 in less than 90 days. Simple crude math - divide total ops by number of op/s, again by ops/GPU/s, multiply with price/h.

So no supercomputer required here, just $ to pay for the most optimized piece of software to run, to minimize costs.

But now 135 requires 5 times more money to solve, or for GPU prices to go down, or for performance per $ to go up, which WILL happen. What I do not understand, is what the deal is with the 3Emi script address, is it a group, is it some sort of smart contract, why was it not spent already, etc.

Yes, 3Emi.... transferred

Puzzle 120 1,20 BTC   27.03.2023
Puzzle 125 12,50 BTC  09.07.2023
Puzzle 130 13,00 BTC  24.09.2024

to himself without spending anything and now has 26,70 BTC He probably waits for 1 BTC = 100.000 USD ?

But why is he hiding the privat keys to Puzzle 120, 125 and 130 ? we got all other solved privat keys so far

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

[citb0in]

...
 I will have to speak with you, that looks like a modified version and I can only speculate you fell upon the growth factor of 18 to 20. Since the puzzle is already solved and the factor only helps find #66 I don't mind sharing the math behind my similar conclusion. I can not upload the image\screenshot of the conversation with my custom, privately hosted model. I can share the text within a small section of it.

So let me disclose this first and foremost. I have a custom model trained and fine-tuned by myself for personal research and assistance. This is not any publicly avaliable chat model. I am lazy and use it for rapid calculations and assistance in holding large hex values and decimal values in a memory pool to save on computations and energy load.
------------------------------------------------------------------------------------------------------------------------------------------------

Model said:
GO GET SOME LIFE
---snip---

I believe there are some parties and individuals who possess both above average intelligence and have private means which they used in order to develop some cutting-edge applications of technology not tied to any large corporations. While these are not supercomputers, or quantum circuits in line with mainstream public knowledge.

I can say with positive proof and personal observations of such systems, they are impressive in their capabilities when compared to the average software running in any programming language on the standard operational systems. They are not a quantum computer in any sense, but their computational abilities are staggering.

I also have a B.A. in mathematics with a concentration in computer science. And some other education, I think if people still accept my cereal box diploma.

I deal with logistics, material science and the electronic engineering of emerging technologies.

I am horrible at writing code from scratch, so I tend to modify broken code snippets fixing the one-off errors or faulty logic to build my own software for my needs.

I guess that would make something of a lesser mathematician, perhaps on a good day?


I would humbly like to offer, we are in a time of profound revolution of conventional models of development and production. Anyone with half a brain, determination and some diy skills, can build things to rival billion dollar corporations.

These last two decades of technological advancements, coupled with the vast libraries available via the internet.

Humor me a moment, and fathom, what a being with 130 IQ or higher, insomnia and an unquenchable thirst of knowledge could accomplish?

Digaran, stop your consecutive posts and don't be a fool. Go get some education before talking crap and sharing your monologue with AI and then go get some life. Please