Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 153. (Read 215539 times)

Thanks for the info bro. I use this calculator, it works great! And how do we subtract half of n from the point X and Y coordinates?, or do we need to get the coordinates x, y from half of n? (if there is, then what are the coordinates of half of n). Or was the subtraction just an example to show so that I would understand?.

OK, tomorrow I'll try to test this key))

(p2wsh) =bc1qjtgt74dxanh4pcmwnfs9usskcg8n33cxxhp0kcnaa82qg6ye26eqrjqvzm
p2wpkh) = bc1q5fyj9pfq2x5sqt4lfjry54dvkadmfnm4sdlecz

ETH= 0x10E2d3289af14a510c88e5107B11d69c47Dd5683
Tron = TBWVYCE2x7fsszXZ16iXDEmfr8QvrAEuU2



Knowing the public key is enough to generate addresses.

Hi Guys, I'm back, and i found the "p2sh-segwit" of 130 # puzzle address "1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua"


"p2sh-segwit": "3J9ANKe3NQgDvtS33JX1v4CB1tyDZAYgrS"

maybe there's some clue or something to work with p2sh ?

because i see the pattern of the puzzle, if someone not spend the fund, the P2SH not generated on privatekeys.pw

what would the scanner get for wasting time and resources if the keys needed don't fall within these ranges?

Download and run this java calculator, it's very slow and heavy but it's good for manual calculations which is why you can only do manual calc with it.
https://github.com/MrMaxweII/Secp256k1-Calculator

Select scalar on both sides, and place private keys in hex format, then select * , + , - , ÷ , and see for yourself what would be the result.

What you should consider when dividing an odd key, dividing any odd key by 2 will always give you .5 ( point 5, half ), anything on the left side of the point . Is your actuall result no matter if you are dividing by 2 or any other number, but for example we use 2, and anything on the right side of the point . Is a 2^255 + key, we don't want that, so we subtract that from our result to get the actual answer.

3 divided by 2 = 1.>5 <  this 5 here means half of n, so if we subtract it from 1.5, we will get 1, our actual result.

Now lets make it a bit difficult, let us divide 7 by 3 = 2.33333333333333333333333333333333333333333333333333333333333333333333333333333 3333333333333333333333333333333333333333333333333333333333333333333333333333333 3333333333333333333333333333333333333333333

Now to get the n/33333........... more 333333..... etc, no need to do any complicated calculation, we just divide 7 by 3 mod n to quickly get the result, then we subtract it from the result to get our key.

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Never mind all the above, I have something to twist your minds, take the following key and double, divide, do many other things with it to get really confused about how EC works. 😂

Introducing to you 2^256 of secp256k1


Try multiplying it by 2, 3, 4, 5 etc as well as dividing it, this little sucker is hiding it's half under the ground!

As previously mentioned, I have a total of 588 ranges, where is a private key of puzzle 66. Each range requires a 48-bit computation (I am referring to GPU computation). and I am seeking an individual who is capable of performing these computations within few minutes or at least an hour.

As for our collaboration approach, I am currently developing a script to scan all 588 ranges. Once the private key is obtained, the script will automatically perform the following actions: 2% of the amount will be deducted as fees, 59% will be sent to the individual performing the computations, and 39% will be sent to me. my bitcoin address and the computation performer's bitcoin address will be pre-entered into the script. The reason for allocating 59% to the computation performer is due to their resource usage, whether through GPU rental or electricity consumption. I intend to share the entire script with the individual to ensure transparency, while keeping the ranges known only to me. Once the script initiates, neither of us will have any access to it. But when scanning of one range completed, it will be saved along with the proof of work. Certainly, it will take me some time to perform all these processes.

So why not continue searching for that person in the meantime?

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(This range is an example of 588 ranges: 3b07a000000000000:3b07affffffffffff, and this address is for proof of work: 1GYpdGZBaqRDKkpFWbaRezrVeGy6g6UNfE, 1Ngdkik5LLGzmzYUWKiPYDQ6CRRGWCPo5L, 19XbiMqyqeaJLDSvHLzhJkxngDYsBPXTQo, 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so)

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The person who will perform the counting of this range needs to be scanned and to post here with proof of work, or they can personally DM me as well.  In this interaction, I will assess their operational speed in completing this 48-bit range. For scanning, they can use any software, but my suggestion is to use 'VanBitCrackenS v1.0' because here they can take advantage of multi-GPU. If the puzzle address falls within this range, then it will be yours. 👍

I don't really know python myself! bro for such simple tasks, you can use ChatGTP, he can write such tasks in code + with explanations!

hello everyone! You can explain what happens when multiplying by an odd key.

Hi, so  i found 2 address similar 11 digits of hash160 in extraction of bit 130 that means if you scan on this range and have the private key of those one of two addresses we will split the 13 BTC ,

1st

1AoNf67iZUmz1Ck9eefUfTzWQJMN5pgcPG
02113ba90a97c020ade0f3d8d0369981a723fe2bc4352815df22dd3eafae13c5a5
6b7e582a29a549cc60b591279a963f02eff02f99

pk:00000000000000000000000000000000000de4cfcadfc034c963dd053d719e88

address to search in :116 bit range
1AoNf67iZrwMSYPTDbk3Sh1yXJCARQbD7a
039d0a0241abe2411f64b4f6d29f2e1b6c837b26b6bdded577c3fc93574d3d735c
6b7e582a29a7601b79761f9f153c300c3d988231
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88


2nd
1PtStkm2bWKryHVduVjciPUxVx9UeDcCXG
02dc52ba09b16bc5cbd25aca7c82dd924f81cd31ecf29ecb264fa2cc45393728b9
fb0d9859584e68c24c1698eea4d05d2822fe4b70
pk: ad0f6ba584b355089cf6ce9cc9774

address to search in 116 bit range
1PtStkm2bLM7EK7g1rnTLBxu6aLouVuULV
03ef06cec3b3e35f68ba78618e5a5cf8663cc1a3b685dcfd197c1c0030530b1293
fb0d9859584d782df3fe652d2da5a21c30f137f9
range: 804cfcadfc034c963dd053d719e88:fffffcadfc034c963dd053d719e88




if the pk found of one of those address we will split the prize.
they have the same 11 digits of hash160 that means maybe 90 pourcent is in that range

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

See this site and you will get answer.

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

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

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.

Look forward to explanations! Good luck bro!

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Tell me how to make each pub write from a new line, and not all in one line?

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

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?

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

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.

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