Bitcoin puzzle transaction ~32 BTC prize to who solves it

[kTimesG]

Hi, didn't you notice? Someone else posted after you. They deleted their message, and unfortunately, I didn't have time to copy it to check the logic behind the reverse sha256 conversion. Maybe you saved it? The person who posted it said they didn't have much time to finish this theory.

Do you have the smallest understanding of what a hash function even is? Hash functions do not revert back to anything, as there's an infinite amount of inputs that hash back to a given hash (or none at all, at the same time - impossible to prove until someone brings up a full on complete 256-bit sized rainbow table).

99% of the posts in this thread deserve to be deleted, as they're plain out idiotic non-sense.

[brainless]

Here. I shall give you a glimpse of hope that these keys are designed to be solved with both logic and bruting . Including a starter formula for #70. Also:  The second half of each puzzle key is simply the first half inverted and mirrored but with the digits on the inverted side to summed if it doesnt exceed 10 in the process . so, you will notice for example if 9 is present on the left half it will almost always followed by or preceded by 81, 18, 27,72 , 45,54 as its inverted counterpart on the right half. If you can think of it as the left half of the key is the
right half compressed , sqeezed so that any two numbers next to each other will add and become as one new number representing them both (as long as it doesnt exceed 10 )That is frustrating but if you stare angrily enough at these problems, solutions will surface.  This will have to be studied for needed extra precision when the puzzle keys are larger. As of right now, a mere 21 digit key like for puzzle 70 is quicky bruteforced in minutes, that is if you could just correctly calculate the first half of it, of course.
 Perhaps, even, with a formula like this:
 2^70^.0071×47.5027
 
 ....which happens to calculate to 97.043621153516297713.

Hi, didn't you notice? Someone else posted after you. They deleted their message, and unfortunately, I didn't have time to copy it to check the logic behind the reverse sha256 conversion. Maybe you saved it? The person who posted it said they didn't have much time to finish this theory.

Are u looking below message
Copied...
 the order is not random. then there must be a structure that is embedded in the order of pkeys that can potentially be used  to unlock the big-ol Ripemd(Sha256(secp256k1(Key))  topology, Ripemd has 160 fixed point of 2^32 evaluation points. secp256k1 is dynamic. logically I would have embedded the structure in the earliest steps of secp256k1 one example would be, if the first step is implemented as below

code:
x1=str2sym('0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798');
y1=str2sym('0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8');
G(:,1)= [2*y1;p]
G(:,2)= [1;0]
G(:,3)= [0;1]
i=1;
while  (G(i,1)>2) % GCD
i=i+1;
G(i+1,1) = G(i-1,1) - G(i,1)*fix(G(i-1,1)/G(i,1));
G(i+1,2) = G(i-1,2) - G(i,2)*fix(G(i-1,1)/G(i,1));
G(i+1,3) = G(i-1,3) - G(i,3)*fix(G(i-1,1)/G(i,1));
end

x2=mod(mod(3*x1*x1*G(1:i,2),p).^2-x1-x1,p);
y2=mod(mod(3*x1*x1*G(1:i,2),p).*(x1-G1)-y1,p);

it can be embedded in the G's as mod(G(2),2^70 -1 ) or a combination of the Gs ..1,2 and 3. but the values in the Keys are dense so I am guessing it could be a more involved function of Gs

Another approach that I made a huge progress with involves implementing sha256 in complex .. then there is a map of  real(W) -> Ripemd and  imag(W)-> secp256k1. W can be computed in reverse from the sha256 hash. I am able to find the map to Ripemd but the other leg has become a bit tricky . I am throwing the towels at this point but I will give it a go again when I get the time. I have a few other promising approach's but I don't have luxury of time. But I know

[BlackAKAAngel]

For those who are writing that they have solved some puzzles, I have news for you: I have 10 bots waiting for you, and this time even Mara doesn’t help!
Good luck to all the solvers;
I love this puzzle!

[kTimesG]

For those who are writing that they have solved some puzzles, I have news for you: I have 10 bots waiting for you, and this time even Mara doesn’t help!
Good luck to all the solvers;
I love this puzzle!

Kewl. Why would you need ten bots when one is enough, so they can fight each other or what?

[Bram24732]

For those who are writing that they have solved some puzzles, I have news for you: I have 10 bots waiting for you, and this time even Mara doesn’t help!
Good luck to all the solvers;
I love this puzzle!

Kewl. Why would you need ten bots when one is enough, so they can fight each other or what?

If you spread them accros the world there’s a case to be made about latency and propagation times I guess ?

[BlackAKAAngel]

For those who are writing that they have solved some puzzles, I have news for you: I have 10 bots waiting for you, and this time even Mara doesn’t help!
Good luck to all the solvers;
I love this puzzle!

Kewl. Why would you need ten bots when one is enough, so they can fight each other or what?

When you are good at what you do, you should know why

[TheMissingNTLDR]

... and this time even Mara doesn’t help!

howd'ya know this?

[JakobianMatrix]

Here. I shall give you a glimpse of hope that these keys are designed to be solved with both logic and bruting . Including a starter formula for #70. Also:  The second half of each puzzle key is simply the first half inverted and mirrored but with the digits on the inverted side to summed if it doesnt exceed 10 in the process . so, you will notice for example if 9 is present on the left half it will almost always followed by or preceded by 81, 18, 27,72 , 45,54 as its inverted counterpart on the right half. If you can think of it as the left half of the key is the
right half compressed , sqeezed so that any two numbers next to each other will add and become as one new number representing them both (as long as it doesnt exceed 10 )That is frustrating but if you stare angrily enough at these problems, solutions will surface.  This will have to be studied for needed extra precision when the puzzle keys are larger. As of right now, a mere 21 digit key like for puzzle 70 is quicky bruteforced in minutes, that is if you could just correctly calculate the first half of it, of course.
 Perhaps, even, with a formula like this:
 2^70^.0071×47.5027
 
 ....which happens to calculate to 97.043621153516297713.

Hi, didn't you notice? Someone else posted after you. They deleted their message, and unfortunately, I didn't have time to copy it to check the logic behind the reverse sha256 conversion. Maybe you saved it? The person who posted it said they didn't have much time to finish this theory.

I wrote that message but I deleted it because it is an involved math which doesn't make much sense without detail explanation. but if you are  genuinely interested I could share more.

[k2laci]

I recently came across this BTC Puzzle challenge and have already read quite a lot about it. What I would like to know is: apart from the official scanning, can scanners outside the pool manage to avoid scanning a range that has already been checked by someone else?

[Cricktor]

... I have 10 bots waiting for you, and this time even Mara doesn’t help!

Yadda, yadda...

Care to explain how a transaction submitted to a non-public mempool as Mara's Slipstream service is attackable for the public (or you mastermind specifically)?

Oh, wait, you're such a genius that you convinced Mara to collude with you. I knew it!

Maybe you can convince me and others how superior you are, which I doubt you can. But hey, surprise me us!

And if you can't or are afraid of, I'd suggest to STFU, you don't add any value to this thread with your blabber. Who cares how many dudes have bots running, when it's clear as pure water that nobody sane should broadcast solutions for puzzles like #71, #72 and so on to public mempools.

[Tony8989]

Look at this madness going on here. I'm going crazy too but at least I don't argue about shit I don't have.

We got people from Nigeria and New Delhi saying they'll only talk when the puzzle creator shows up in the thread. Then you got new people who just discovered this puzzle two days ago trying to guess the keys like it's some lottery ticket. Then you got the same people repeating the same discussions over and over.

Maybe there's a shortcut, maybe there isn't. This puzzle started in 2015 - if there was some easy way to solve it back then, you think after 10 years nobody would have found it?

So guys, stop arguing about something you don't even have yet. And if by some chance you actually get a key, that's your business to protect it. You can use Mara, Luigi, or Mario - whatever works for you.

[frozenen]

Look at this madness going on here. I'm going crazy too but at least I don't argue about shit I don't have.

We got people from Nigeria and New Delhi saying they'll only talk when the puzzle creator shows up in the thread. Then you got new people who just discovered this puzzle two days ago trying to guess the keys like it's some lottery ticket. Then you got the same people repeating the same discussions over and over.

Maybe there's a shortcut, maybe there isn't. This puzzle started in 2015 - if there was some easy way to solve it back then, you think after 10 years nobody would have found it?

So guys, stop arguing about something you don't even have yet. And if by some chance you actually get a key, that's your business to protect it. You can use Mara, Luigi, or Mario - whatever works for you.

The puzzle creator should never return to this forum and anyone with the vast resources and ability to find these puzzles would never get scammed! scammers are morons if they think they will get anything here.

[Gerbie]

Today is the historical day I finally discovered the secret behind this puzzle.

I will reply whenever the creator of this puzzle replies to the Threat .

I am not lying and whenever I get a replay I will prove it to everyone I don't want to ruined this beautiful puzzle I want to keep it going.

The method is simple I'm the same person as Bitcoin71.

I HAVE A SOLUTION The whole puzzle from 1 - 160 just need a password or a seed to solve all combinations, what i need is anyone who knows the seed or the passkey by analyzing the 32 bitcoin puzzle or by analyzing any available HINTS. I'm waiting for a replay on my post and whenever i get the correct passkey or someone discovers it I will share the half wit him/her.

Example when seed Bitcoin is used results will be :

1
3

....
15A8E63672C1B9F193
36E74118874EA56A80

You might not have understood how this puzzle was created... the result was never 1, 3, 5 or 2345, or even 36E74118874EA56A80. They were always full 256-bit keys, which were then reduced to the puzzle bit size.  Which means that for the first puzzle (1 bit) the first 255 bits were thrown away. Knowing the seed will give you all the keys, but since we don't know what was thrown away, we don't know the actual keys which were generated by the seed and therefore is't not that easy to reverse engineer the seed.
The same for all patterns many here are looking for: since the most parts of the keys were disregarded, you're looking for a pattern in the leftovers. Spoiler: there isn't one. If there would be a pattern, it would be in the full keys. But that's just my humble opinion, I invite you to prove me that I'm wrong.

[Niekko]

Let the DLP resolve itself, when the universe is ready to return the key. Do not force the solution, feel the Force flowing through the scalars, and one day, perhaps, it will find you.


May the Curve be with you.

[kTimesG]

Let the DLP resolve itself, when the universe is ready to return the key. Do not force the solution, feel the Force flowing through the scalars, and one day, perhaps, it will find you.

resolve itself = useful collision.
universe = baby points / DPs match.
return the key = solve the collision delta arithmetic.
Do not force = deploy 500 GPUs, setup telegram private messages, and go on vacation.
feel the Force = check up the stats once in a while
flowing = make sure the secure websockets are still working and all instances are healthy
through the scalars = minimize EC multiplications to increase throughput
one day = estimate roughly the standard deviation of solve steps, to get an idea of how long it will take
it will find you = you read about on bitcointalk on how the puzzle you're trying to solve was sweeped by some guy who never posted on the forums.

[ee1234ee]

Today is the historical day I finally discovered the secret behind this puzzle.

I will reply whenever the creator of this puzzle replies to the Threat .

I am not lying and whenever I get a replay I will prove it to everyone I don't want to ruined this beautiful puzzle I want to keep it going.

The method is simple I'm the same person as Bitcoin71.

I HAVE A SOLUTION The whole puzzle from 1 - 160 just need a password or a seed to solve all combinations, what I need is anyone who knows the seed or the passkey by analyzing the 32 bitcoin puzzle or by analyzing any available HINTS. I'm waiting for a replay on my post and whenever I get the correct passkey or someone discovers it I will share the half wit him/her.

Example when seed Bitcoin is used results will be :

1
3

....
15A8E63672C1B9F193
36E74118874EA56A80

You might not have understood how this puzzle was created... the result was never 1, 3, 5 or 2345, or even 36E74118874EA56A80. They were always full 256-bit keys, which were then reduced to the puzzle bit size.  Which means that for the first puzzle (1 bit) the first 255 bits were thrown away. Knowing the seed will give you all the keys, but since we don't know what was thrown away, we don't know the actual keys which were generated by the seed and therefore is't not that easy to reverse engineer the seed.
The same for all patterns many here are looking for: since the most parts of the keys were disregarded, you're looking for a pattern in the leftovers. Spoiler: there isn't one. If there would be a pattern, it would be in the full keys. But that's just my humble opinion, I invite you to prove me that I'm wrong.

It is not about that these are just an outcome of a specific seed like [bitcoin] when used as a seed the results would be
1
3

....
15A8E63672C1B9F193
36E74118874EA56A80

But when using other seed word or passkey look at the results :

1
2
5
F
11
32
47
E0
1C8
295
762
B9A
1305
3853
47C9
FF92
1C87F
36656
65340
AFDDF
10C517
2522F4
6C48E2
8BF072
10F0EDD
2C5F137
42C84A9
DF89BA3
1D244D3E
37A253EC
4C077986
CD2BEAF7
12114260B
3F89B6A5D
74294BB3F
C93D7999A
1C1887DE30
397A860AF2
6133519AA9
ACD349B531
1E96E75982B
298BCE09B41
65F72C8AD44
84C0266042A
1C189538C314
27B4BB65B986
55C31C5B2A93
EF6BFF77B8F5
1D7D68A0C3CD7
24E8A3926FC0D
5B09C6344DCC2
FA0AEA11200B6
137654B7D8FBBD
2BA1DAB89048BE
6D3A3B8D46D3AA
DAE99BAC88ECB0
1786B7F1964B814
32BF3FC2D0995DD
70F386B94B16C2B
DD3D367959B57A3
1CCB06C5414FBFC9
2083926DA003D3C7
75F6DF2FD556FC40
91A835A66D01CF79
137A2BA73053FE2ED
2898FE481B2AD25E3
6EF39C26F2D67306D
E5B3136DA244FE69B
17D45A09B28DAE4DF8
349B2E6276D53B046A

Even if they are not the actual key's, They will give you a hint for where you can find the next one. I cannot prove anything to anyone until the puzzle 71 is found ! i will give you a hint , it will be found in this range 60000...6ffffff
so until that time when it will be discovered I'm not doing anything or I'm unable to prove it!

I'm NOT a scammer and I don't like to draw attention to myself I just like to help and get help!

Another scammer has emerged. I said the private key range must be between 70000 and 7FFFF. Do you believe it?
If you had the ability, you would have transferred Bitcoin long ago. Why waste time here

Today is the historical day I finally discovered the secret behind this puzzle.

I will reply whenever the creator of this puzzle replies to the Threat .

I am not lying and whenever I get a replay I will prove it to everyone I don't want to ruined this beautiful puzzle I want to keep it going.

The method is simple I'm the same person as Bitcoin71.

I HAVE A SOLUTION The whole puzzle from 1 - 160 just need a password or a seed to solve all combinations, what I need is anyone who knows the seed or the passkey by analyzing the 32 bitcoin puzzle or by analyzing any available HINTS. I'm waiting for a replay on my post and whenever I get the correct passkey or someone discovers it I will share the half wit him/her.

Example when seed Bitcoin is used results will be :

1
3

....
15A8E63672C1B9F193
36E74118874EA56A80

You might not have understood how this puzzle was created... the result was never 1, 3, 5 or 2345, or even 36E74118874EA56A80. They were always full 256-bit keys, which were then reduced to the puzzle bit size.  Which means that for the first puzzle (1 bit) the first 255 bits were thrown away. Knowing the seed will give you all the keys, but since we don't know what was thrown away, we don't know the actual keys which were generated by the seed and therefore is't not that easy to reverse engineer the seed.
The same for all patterns many here are looking for: since the most parts of the keys were disregarded, you're looking for a pattern in the leftovers. Spoiler: there isn't one. If there would be a pattern, it would be in the full keys. But that's just my humble opinion, I invite you to prove me that I'm wrong.

It is not about that these are just an outcome of a specific seed like [bitcoin] when used as a seed the results would be
1
3

....
15A8E63672C1B9F193
36E74118874EA56A80


Even if they are not the actual key's, They will give you a hint for where you can find the next one. I cannot prove anything to anyone until the puzzle 71 is found ! i will give you a hint , it will be found in this range 60000...6ffffff
so until that time when it will be discovered I'm not doing anything or I'm unable to prove it!

I'm NOT a scammer and I don't like to draw attention to myself I just like to help and get help!

Another scammer has emerged. I said the private key range must be between 70000 and 7FFFF. Do you believe it?
If you had the ability, you would have transferred Bitcoin long ago. Why waste time here

You can say that but look at the Number 70 on my prevkeys that i generated 349B2E6276D53B046A It is so close to the found one!

So based on that I have generated 71-72-73..... and approximate ranges are clear!

Perhaps you don't understand programming at all. This puzzle is generated using VC++'s random number function, and the seed may be a random CPU clock time. The author cannot use a fixed seed to generate random numbers, this is common sense.
So your idea of finding the author's original random number seed is inherently wrong.

[kTimesG]

Perhaps you don't understand programming at all. This puzzle is generated using VC++'s random number function, and the seed may be a random CPU clock time. The author cannot use a fixed seed to generate random numbers, this is common sense.
So your idea of finding the author's original random number seed is inherently wrong.

Assuming the creator's claim (masked consecutive keys of a deterministic wallet), the seed that generates the deterministic addresses is fixed.

However cracking the seed is less likely than breaking the entire Bitcoin blockchain itself.

Cryptographically-strong random number generators don't use CPU clock time, that's lame and it can be cracked faster than it takes you to blink. More likely, a strong entropy source is used, like dedicated hardware on your mainboard that mixes up PSU power drain, energy fluctuations across the chips (basically electric noise), heat sensors, PWM whispering, and everything similar you can possibly imagine or not. Everything also gets combined with software sources like network activity at the moment, I/O timings and stats, idle stats, mouse movements, etc.

Good luck to everyone trying to reverse engineering this to get some stupid seed. It's easier to just crack all the puzzles via brute-force then trying to recreate the RNG state.

[pbies]

It can be short python script with fixed seed to generate the pvks,
but we don't know the body of the script to tell what algo was used.

And even if we knew the seed we won't know the script to generate pvks.

[nebnebneb]

🧩 Bitcoin Puzzle Transaction — Full Structural Breakdown
After careful reconstruction and verification, I present the complete logic behind the 256 puzzles from the legendary Bitcoin transaction:

TXID: 08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15 Total distributed: 32.89600000 BTC Structure: Each puzzle receives 0.001 × n BTC, from puzzle #1 to #256.

Each private key is adapted to its exclusive semi-open range:

Puzzle 𝑛⇒Private key∈[2𝑛−1,2𝑛)
No overlaps. No ambiguity. Just pure structural elegance.
📜 Why this matters
This transaction is not just a technical artifact — it's a collective opera of cryptographic generosity. Each bifurcation, each satoshi, is part of a legacy. By documenting it with precision and clarity, we honor the spirit of Bitcoin itself.
.

🙏 Support the effort
If you appreciate this work and want to support further documentation, structural analysis, or puzzle-solving efforts, feel free to send a donation to:

BTC address: 16K8YpSTimnW3cF26NR5dZoja7Y1RhnCey

Every satoshi helps continue the narrative.

Code:

import hashlib, hmac, base58
from ecdsa import SECP256k1, SigningKey
from decimal import Decimal, getcontext

getcontext().prec = 50  # Precisión alta para BTC

# === Funciones auxiliares ===
def mnemonic_to_seed(mnemonic: str, passphrase: str = "") -> bytes:
    salt = "mnemonic" + passphrase
    return hashlib.pbkdf2_hmac("sha512", mnemonic.encode(), salt.encode(), 2048)

def hmac_sha512(key: bytes, data: bytes) -> bytes:
    return hmac.new(key, data, hashlib.sha512).digest()

def CKD_priv(parent_key: bytes, parent_chain: bytes, index: int) -> tuple:
    hardened = index >= 0x80000000
    if hardened:
        data = b'\x00' + parent_key + index.to_bytes(4, 'big')
    else:
        pubkey = SigningKey.from_string(parent_key, curve=SECP256k1).verifying_key.to_string("compressed")
        data = pubkey + index.to_bytes(4, 'big')
    I = hmac_sha512(parent_chain, data)
    child_key_int = (int.from_bytes(I[:32], 'big') + int.from_bytes(parent_key, 'big')) % SECP256k1.order
    return child_key_int.to_bytes(32, 'big'), I[32:]

def derive_path(seed: bytes, path: str) -> bytes:
    key, chain = hmac_sha512(b"Bitcoin seed", seed)[:32], hmac_sha512(b"Bitcoin seed", seed)[32:]
    for level in path.split("/"):
        if level == "m":
            continue
        index = int(level.replace("'", ""))
        if "'" in level:
            index += 0x80000000
        key, chain = CKD_priv(key, chain, index)
    return key

def pubkey_to_p2pkh(privkey: bytes) -> str:
    pubkey = SigningKey.from_string(privkey, curve=SECP256k1).verifying_key.to_string("compressed")
    sha256 = hashlib.sha256(pubkey).digest()
    ripe160 = hashlib.new('ripemd160', sha256).digest()
    payload = b'\x00' + ripe160
    checksum = hashlib.sha256(hashlib.sha256(payload).digest()).digest()[:4]
    return base58.b58encode(payload + checksum).decode()

# === Mnemonico base ===
mnemonic = "puzzle puzzle puzzle"
seed = mnemonic_to_seed(mnemonic)

# === Dirección origen confirmada
origin_address = "1Czoy8xtddvcGrEhUUCZDQ9QqdRfKh697F"
print(f"📦 Dirección origen: {origin_address} (32 BTC)")

# === Importes por puzzle (lineales, 0.001 × n BTC)
btc_per_puzzle = [Decimal("0.001") * Decimal(n) for n in range(1, 257)]

# === Transacción estructural con verificación
print("\n🧾 Transacción estructural:")
print("Puzzle,Privada Original,Direccion Original,Privada Adaptada,Direccion adaptada,BTC asignado")
for i in range(256):
    path = f"m/44'/0'/0'/0/{i}"
    privkey = derive_path(seed, path)
    priv_original_int = int.from_bytes(privkey, 'big')
    address_original = pubkey_to_p2pkh(privkey)

    # Adaptación al rango exclusivo [2ⁿ⁻¹, 2ⁿ)
    lower = 1 << i
    upper = 1 << (i + 1)
    range_size = upper - lower
    adapted = lower + (priv_original_int % range_size)
    address_adapted = pubkey_to_p2pkh(adapted.to_bytes(32, 'big'))

    # Validación estructural
    if not (lower <= adapted < upper):
        print(f"❌ Puzzle #{i+1} fuera de rango")
        continue

    btc_str = f"{btc_per_puzzle[i]:.11f}"
    print(f"#{i+1:03}:,{hex(priv_original_int)},{address_original},{hex(adapted)},{address_adapted},{btc_str}")

==================================================================
BTC address: 16K8YpSTimnW3cF26NR5dZoja7Y1RhnCey

[Cricktor]

What is it today, mental hospital free roaming vacation day?

~crap snipped~

You idiot deserve to be banned for trying to impost the presumably creator of the Bitcoin puzzle by using a very similar user name. And don't try to deny this wasn't deliberate. I wonder how shabby your life must be to do something that lame.

~more crap snipped~

OK, so you're already a moron who doesn't realize that posting code without [code]...[/code] tags might alter your code which happened at the end of your code crap. Find the missing piece...

OK, so which AI tool was used which you deliberately omit to declare?

Gosh, you fools deserve to be immediately banned.