Bitcoin puzzle transaction ~32 BTC prize to who solves it

[SRG02289]

649a3dd0486c96e70b       1PWo3JeB9jdvp56gzG8navJjUMFQxxb997   
649aa0a809fe13a924        1PWo3JeB9jgLYkm8cEj6yywPn3kk41BFaY 
649a103b78b3e74856       1PWo3JeB9jRLz2NjTHsZ8uTBnqcjnu66Ak 
649ade9933e47de861       1PWo3JeB9jSVntJs1FpYM2iTNAoADNv2YD 
649a737f258abb8058        1PWo3JeB9jeFuotNeYDrWkEhqyZ359abB2 
649af9e3471ed5c49b        1PWo3JeB9jgFbdXaZG1h8ng9hpYzg3CAPw 
649a05d73d20978e62       1PWo3JeB9jT926gmLys26TBwr6pwStgwLb


Donation my

bc1qurlzuk6v7pruzemjfyjdttjxymhckcml9sw0w8


thanx

ha-ha

Code:

0x649a05d73d20978e62	1PWo3JeB9jT926gmLys26TBwr6pwStgwLb	F6F5431D25BBEE131726CA1F86672474B6E4A3D5	1855772920885219397218	0302751DAA96290A083BDDCA8CD23AC319B8D95116F40C2A9AA8CF122022C0033C	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBinmvSXpixVnKso3SZd
0x649a0ad2d9876795c4	1PWo3JeB9jS1ttWtFXgTCy2hnCYcW1dD6V	F6F5431D25BBED9B9A8F0E621784AC4752C869AA	1855774323434284619204	034F9223C37A4581C73C5F634221AC1D16453DFEF5EA3719930EBBBFAE3C846AA2	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBinsNm4f2xk6MCfAEAi
0x649a103b78b3e74856	1PWo3JeB9jRLz2NjTHsZ8uTBnqcjnu66Ak	F6F5431D25BBED54360606F002AB373B8AC542C8	1855775845842023827542	0312E89B2E156E19CC496873335BCF6D80FB7328C6E2A227F9E40019B563F76D1D	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBinyJ7SyJ1aHWsGeWtv
0x649A133876C6F063BE	1PWo3JeB9j6ffmBQZ4tTuWxaByxBvPKVbB	F6F5431D25BBE59099EE9B35FB4A89AC2576371A	1855776686960148505534	02A10ECB207E3E5335F44AB18A4F17CC90D55DBC975092AF85A2E978A97617E35F	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBio2ZozDD88oDkSiZt8
0x649a137ea2f2eaff0b	1PWo3JeB9j2EU5JxsW28UfCTp5ZFVjNk3b	F6F5431D25BBE3B8B38C1F296849159383301AA7	1855776764115678854923	03139D1F9A29A3B1115D76A9EB92B7A5964A93FB3A4E46894FDFBF004598EA4981	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBio2sDFH9PGjudKZJty
0x649a15cfc7095be9d9	1PWo3JeB9ju3Yq8s97QRtqE3U3p2YfdxxD	F6F5431D25BBF8D7DAA1C6A4B6CEDDC92A1366D3	1855777416281069447641	02EAE5E2B0781507F32419246878C4DE3E5595D1B094CCD310691EB9ADB3EEAAE2	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBio5QJ87uHAZRbdU6pC
0x649a205b29228381cc	1PWo3JeB9jMaH65f34qnG3DbtatLXrrwBn	F6F5431D25BBEBC2F0E6939FA87374E4A5E14B66	1855780384284281635276	02574E65ACAAF97C225FF819319ED0C3FCBABDA823EEA39663AE6C6AA59F62F764	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBioGwg87UGRobFd1ZB2
0x649a2f071642db50f8	1PWo3JeB9j3RLp4NNvVodUbn1movwwhSg7	F6F5431D25BBE43711B7EAE31900ECFBAC43EA7F	1855784513968893808888	03272F4AC555500472266F44CEC55E1D6B62DB8AD065E72C36CD1A44E75D9E43F1	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBioZ13uEvPTv5eBa9LS
0x649a3dd0486c96e70b	1PWo3JeB9jdvp56gzG8navJjUMFQxxb997	F6F5431D25BBF28F431ED96532EBAFA65BF5931E	1855788675835853465355	02D93E9A0DC609B3AD8C3E080B1DD10C425C037C53CB8313A2031C1CDCAECBD907	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBioqBgemYg4VSEkEFo5
0x649a737f258abb8058	1PWo3JeB9jeFuotNeYDrWkEhqyZ359abB2	F6F5431D25BBF2B24DCCBCDF87F2ABFEFFB4EEA8	1855803786274335850584	02ED65C797BFB5C842A9584ACC5F3713F86B5E3FDCC75ECB0B41E16E60B714DF72	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBipqwa1JY7VeYq3zfDL
0x649aa0a809fe13a924	1PWo3JeB9jgLYkm8cEj6yywPn3kk41BFaY	F6F5431D25BBF38FA4743F51AF14BCD67463B7D0	1855816497609940642084	02B1CDB21A56CD4B92840C59534420B75751B3AFBE664FA2DE37F824D862071CA2	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBiqhNP4PNciudJAoe6j
0x649ade9933e47de861	1PWo3JeB9jSVntJs1FpYM2iTNAoADNv2YD	F6F5431D25BBEDCEC969FFE21AD9F3D3648B48D3	1855833932745781667937	0353D262FA353A54E493CAA7E3C4104CB2824EAE9BB1C08BF513B6E02D479FD802	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBirsAZRi1dmwLPdfAD8
0x649af9e3471ed5c49b	1PWo3JeB9jgFbdXaZG1h8ng9hpYzg3CAPw	F6F5431D25BBF3868F42E4B88C52FCF50D1773EE	1855841614016596526235	0200D5D2F9393034EDAA730B7665EA0127DFCCF3F796D46CA4C05CB7DD4C5C1457	KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rBisP2vxweiiYwjdEHkj

Donation my 13sChartsJcYYxpMM4FPUArfp7StLiXBQC

thanx

[L_u_x]

Stop the BS. Any developer who actually understands the math of the Kangaroo algorithm knows this is false. You don't need to recompute the whole walk; you only need a Pre-step Proof.

<some math BS...~

Why a "Scanner" fails here:
A sequential scanner (VanitySearch style) finds a point with 28 zeros by pure luck or brute force. It cannot produce a predecessor P(n-1) that satisfies the deterministic jump table. Finding such a predecessor without actually walking the path is as hard as solving the ECDLP itself.

Thanks for proving that you are indeed as retarded as the LLM that generates the crap you're posting.

Finding some whatever predecessor is so simple that it can be done on paper. You only need to find some valid point with a satisfactory X. However, it doesn't prove that such predecessor was part of a longer walk. You'll simply say that's where the walk started. What the fuck is wrong with you, comparing this to ECDLP hardness?

You should indeed get banned, not for the selloff, but for the BS.

You are stuck in a binary "all-or-nothing" verification trap.Probabilistic Auditing is how trust is handled in massive datasets.

Let's end the debate with a Merkle Tree Audit Protocol:

1. Commitment: I am publishing the Merkle Root Hash of my entire database. This locks every single byte. Once the Root is posted, I cannot swap a single "fake" point into the set.
2. Challenge: Pick any random segment index (out of 1000 chunks).
3. Reveal: I will provide the raw segment data plus the Merkle Path.

The Final Proof of Work:
If the audited segment proves Internal Path Connectivity (deterministic Kangaroo chains where P(n) + JumpTable = P(n+1)) and correctly hashes back to the Root, the validity of the entire net is statistically confirmed.

Why you can't fake this:
To cheat this protocol, I would have to synthesize a statistically consistent, connected path structure across  Merkle-committed dataset. To fake that without actually running the Kangaroo algorithm is computationally more expensive than solving Puzzle #135 itself.

[0xastraeus]

L_u_x... You have multiple people telling you to stop with the snake oil BS. I would advise you to do so...

You're already in violation of atleast 3 unofficial(official) rules of the forum (https://bitcointalk.org/index.php?topic=703657.0):
1. 32. Posting multiple posts in a row (excluding bumps and reserved posts by the thread starter) is not allowed.
2. 22. Advertising (this includes mining pools, gambling services, exchanges, shops, etc.) in others' threads is no longer allowed, including, but not limited to, in altcoin announcement threads. [8]
3. 13. Bumps, "updates" are limited to once per 24 hours per thread. Bumping multiple threads at the same time is allowed if it's not annoying. [2][e]

To everyone posting projects:
post your project in the development area first then post that forum link here if you want more feedback.

To the prefix posters:
Use the code block function

Code:

Like so

to post the prefixes so you don't flood the whole page. Or even better create a new forum for it.

You are stuck in a binary "all-or-nothing" verification trap.Probabilistic Auditing is how trust is handled in massive datasets.

[ik316m]

 I think RC will hit 135 soon in a month or two. It is been more than a year and half he solved 130. What you think?

[ldabasim]

Didn't he say its not economically sound so he's not working on it or was that another member? 2^67.5 ops, takes around 600 years to have a decent chance with 5000 Mkeys/s (A single good gpu). But it could be around 1300 years too if you're unlucky. So a thousand gpus for a few months to a year. 200 watts each, $200k+ per month in electricity alone. So doubt anyone is doing that. And if they aren't we can't make a prediction like that, its down to luck.
Edit:actually 20k in electricity so maybe. But you'd get about $150k per month if you rent them instead, still kinda profitable for 135 in the end with a bit of luck.

[holy_ship]

still kinda profitable for 135 in the end with a bit of luck.

what if puzzle creator didn't put 135-160 to supposed range? haha

[Cricktor]

,,,

And why should've the puzzle creator done that? Makes no sense at all.

[kTimesG]

Didn't he say its not economically sound so he's not working on it or was that another member?

135 is around 10 times cheaper to solve than 71, with a double-sized reward that's still greater than the estimated costs. 71 is a losing bet even before a single key is tested (unless BTC goes to 300k instead of dumping to new daily bottoms). It doesn't take a diploma in nuclear fusion to understand this.

While RC is probably gonna solve 135 any day now with his "pure SASS turbo kernel", everyone's posting vanity prefixes (e.g. wasting electricity indeed). GG.

[crytoestudo]

He already thought too much, let us just think one, hahaha

Didn't he say its not economically sound so he's not working on it or was that another member?

135 is around 10 times cheaper to solve than 71, with a double-sized reward that's still greater than the estimated costs. 71 is a losing bet even before a single key is tested (unless BTC goes to 300k instead of dumping to new daily bottoms). It doesn't take a diploma in nuclear fusion to understand this.

While RC is probably gonna solve 135 any day now with his "pure SASS turbo kernel", everyone's posting vanity prefixes (e.g. wasting electricity indeed). GG.

[kind_user]

How do you got this B9j? You have a lot of power to scan random or is over the time? In how much time you got this?

[bibilgin]

How do you got this B9j? You have a lot of power to scan random or is over the time? In how much time you got this?

He never actually obtained the hex of the wallets. He used what was shared in my Telegram group to solicit donations here.

Therefore, I advise him to stop begging. jonematt

[stillhere@2009]

Tired of these fuckin puzzles one day I will come again with a gpu and with a little hope

[jonematt]

How do you got this B9j? You have a lot of power to scan random or is over the time? In how much time you got this?

He never actually obtained the hex of the wallets. He used what was shared in my Telegram group to solicit donations here.

Therefore, I advise him to stop begging. jonematt

I'm not begging. I have an RTX 5090 graphics card. I've compiled some solutions, most of which are my own. I've deleted all the solutions and won't share them with anyone.

GPU: GPU #0 NVIDIA GeForce RTX 5090 (170x0 cores) Grid(1360x128)

[kind_user]

This is not a solution. If you want o share you are welcome, but asking for donations...

[jonematt]

This is not a solution. If you want o share you are welcome, but asking for donations...

If someone finds a solution and I'm the reason for it, then there's no problem if they donate to me.
 https://i.postimg.cc/mDqkHJqw/556.png

[olnev]

While RC is probably gonna solve 135 any day now with his "pure SASS turbo kernel", everyone's posting vanity prefixes (e.g. wasting electricity indeed). GG.

Is it possible to ban idiots who post here vanity prefixes and other spam craps?

[Cricktor]

If someone finds a solution and I'm the reason for it, ...

Hm, how would someone else benefit from your or any other prefix samples? What exactly do they indicate toward the solution? That you searched exhaustively all of the range around respective published private keys? Why should anybody believe that?

I just can't wrap my head around it, seriously!

Is it possible to ban idiots who post here vanity prefixes and other spam craps?

Just use the Ignore link at everybody you think is a time waster here, don't look back and enjoy the silence.  

[rt07]

This is not a solution. If you want o share you are welcome, but asking for donations...

If someone finds a solution and I'm the reason for it, then there's no problem if they donate to me.
 https://i.postimg.cc/mDqkHJqw/556.png

It doesn't work. Bitcoin addresses are generated using irreversible hash functions (SHA256 + RIPEMD160). It's impossible to reduce the search range for a private key from the address prefix.
It's like trying to guess a number between 1 and a billion knowing only the last two digits of the result—it doesn't help.

For example your 1st prefix 649a3dd0486c96e70b       1PWo3JeB9jdvp56gzG8navJjUMFQxxb997
649a3dd0486c96e70b is just a hex string (18 characters)
A full Bitcoin private key: 64 hex characters (256 bits). You only have ~72 bits. You're missing 184 more bits.
Number of options remaining 2^184 ≈ 2.4 × 10^55(24 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000)
Even if it's really part of the key, in what order are they? You still have to search randomly. You can't find the rest sequential. And how many prefixes are there? Which one should you rely on? You're making -random search even more complicated.

As for donations, I think someone who finds key for any puzzle using someone else's code wouldn't donate even to the developer.
“Man is a creature for whom nothing is ever enough: once he has everything, he begins to want more.”

[kind_user]

Why everybody is stuck on BTC formula? Think out of the box!!!

For me the prefixes are helping a lot because i developed something that is giving me B9 in 8 of 10 ranges. The output is very acurate because not all the ranges contain B or B9. And depends the number of data i enter, as much as the result is more acurate.
I do not generate 44 bits and i use 40 bits instead for ranges.
My challenge was the distribution of keys and that's why i need so many keys to be more acurate. Now i do not use a LLM or AI but in future i will implement for faster search. I do not expect someone to believe what i say because everybody is stuck on BTC formula and the ireversible method.

Just to give everybody a hint: in a 48 bits you will not find two B9j, so you will stop searching there because you waste time.(this is from my database distribution and i hope i am not wrong). If anyone found two of B9j in a 48 bits range please tell me i am wrong, if not do not come with aberations....

[kTimesG]

Just to give everybody a hint: in a 48 bits you will not find two B9j, so you will stop searching there

So if I start searching 2**48 random different keys (which is no different than searching 2**48 sequential keys), I should stop searching after I found the B9j, and then what?

Unless you found a flaw in how sequential scalars, related somehow by the generator, somehow affect sequential public keys, which in turn somehow affects the SHA256 outputs, which in turn somehow affects the RIPEMD-160 hash, then this is bullshit. There is absolutely no relation between some human-readable sequence of some numbers and how the H160(kG) looks like.

By that time though, you'd have been broken three proven secure cryptographic primitives (secp256k1, SHA256, and RIPEMD-160). ALL OF THEM.

If you're in the club that thinks they are filtering some beach area by dismissing it because a grain of sand was hit, well, the next grain of sand in in Alpha Centauri, not near the (n - 1) grain of sand. The counting is irrelevant, only exists in our mind.