Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 245. (Read 192015 times)

The key is between 36028797018963968 and 72057594037927936
2^55 - 2^56

What is "358 formula"?
keys 1-55 are not rounded to decimal hundreds at all.

The key is from no 36054468590243300 to 71207575465730500
.
*And when simplified there are a few:
36054468590243300
36354922495162000
36655376400080700
36955830304999400
37256284209918100
37556738114836800
37857192019755400
38157645924674100
38458099829592800
38758553734511500
39059007639430200
39359461544348900
39659915449267600
39960369354186300
40260823259105000
40561277164023700
40861731068942400
41162184973861100
41462638878779800
41763092783698500
42063546688617200
42364000593535800
42664454498454500
42964908403373200
43265362308291900
43565816213210600
43866270118129300
44166724023048000
44467177927966700
44767631832885400
45068085737804100
45368539642722800
45668993547641500
45969447452560200
46269901357478900
46570355262397600
46870809167316300
47171263072235000
47471716977153700
47772170882072400
48072624786991000
48373078691909700
48673532596828400
48973986501747100
49274440406665800
49574894311584500
49875348216503200
50175802121421900
50476256026340600
50776709931259300
51077163836178000
51377617741096700
51678071646015400
51978525550934100
52278979455852800
52579433360771400
52879887265690100
53180341170608800
53480795075527500
53781248980446200
54081702885364900
54382156790283600
54682610695202300
54983064600121000
55283518505039700
55583972409958400
55884426314877100
56184880219795800
56485334124714500
56785788029633200
57086241934551900
57386695839470600
57687149744389200
57987603649307900
58288057554226600
58588511459145300
58888965364064000
59189419268982700
59489873173901400
59790327078820100
60090780983738800
60391234888657500
60691688793576200
60992142698494900
61292596603413600
61593050508332300
61893504413251000
62193958318169700
62494412223088400
62794866128007000
63095320032925700
63395773937844400
63696227842763100
63996681747681800
64297135652600500
64597589557519200
64898043462437900
65198497367356600
65498951272275300
65799405177194000
66099859082112700
66400312987031400
66700766891950100
67001220796868800
67301674701787500
67602128606706200
67902582511624800
68203036416543500
68503490321462200
68803944226380900
69104398131299600
69404852036218300
69705305941137000
70005759846055700
70306213750974400
70606667655893100
70907121560811800
71207575465730500

*BOLD= when simplified by 358 formula (although I do not like this formula because from the beginning did not form a pattern)

I have confirmed this, and can show from wallet 1 to 55

Does it really matter, PM with unknown persons or here, in public?

Arulbero is good (best here?) at math, but probably is a bit busy.

 I am newbi in bitcins, but i pay attention to journey puzzle 32 (112) btc.

I found a little pattern, but i dont know what i do to process it.

Can someone discuss with me? pm

Nobody bothered.


You are not authorized to guarantee that  unless you're our dear Allah.

What is wrong in what I said?

Also, there are no known collisions for both SHA-256 and RIPEMD-160 hash functions. Not a single one. Since there is close to 0 chance of finding collision for these hashes, there is even endlessly smaller chance that potential collision would be on BTC private key and not on some other series of bytes. No, you will not find 50.000 BTC wallet by searching for the solution of this puzzle, I guarantee you that.

Sooo wrong  There is no direct relation between 32BTC puzzle and "real wallets". While searching for modest .5BTC puzzle you can discover 50.000BTC wallet due to collision.

That's what attracts so many mommies's-crypto-kiddies to LBC

One problem is the probability of such discovery.... 

Miners have very fast hashing hardware.

This puzzle is about quickly calculating public keys from private keys - not hashing.

So all that fancy mining hardware cannot be applied to this puzzle.

If anyone has a "better way" to do this to the point that they can claim the remaining Bitcoins in this puzzle then they could claim anyone and everyone's Bitcoins, Bitcoin is broken, Bitcoin becomes worthless, they can then claim all of nothing.

Well, this is actually challenging for those who are accustomed to deciphering secret codes. I'm sure that only an intelligent person can do this right, organize the puzzle correctly to produce something meaningful. But, unfortunately I am not a part of the solution. I do not have much time to think about the challenge. Maybe the hackers who are experts in breaking the bitcoin wallet can do that, ha ha ha. That’s a bit of my opinion.  Maybe, I will offer this challenge to my friend who is very good at computer software. Good luck and I will share these results with you all.

After 4 years (maybe even earlier) when the reward for bitcoin dropped Instead of mining bitcoin will go in search (something similar LBC with Chinese capacities) of lost millions of bitcoins then this puzzle will not last long. However a clever mathematician with a calculator would have found a way to puzzle now not necessarily accurate mathematics but all sorts of paradox.

There is nothing to "think of anything other". The point of the puzzle is not to be cracked, to prove that it can not be crakced. When private keys even from this limited range of numbers are uncrackable, the ones from full range are absolutely untouchable. Bitcoin is safe. That's what anonymous author of the puzzle is telling us, and he backed his claim with all these BTC. If he is wrong, someone would pick the money from this puzzle. Fortunately no one could do it, and no one will ever be able to do it. As long as this puzzle is not cracked - we have nothing to worry about.

It would have been so much bitcoin for me i would not be sitting here.

And how this process looks, you can also generate a random and add "maskin". Something he wrote 000 0002...


If we can not think of anything other than scans of addresses, then it's like diving into the lake and farting there to stun and catch fish. To find 56 by scanning you need 100 1080ti cards and it will take a month or more. For the next step you need already 200 1080ti with the same time either increase the search time.

All the private keys for each range were created by generating a 256 bit secure random number and then masking the result down to the desired key length.  You could easily create a similar puzzle if you had enough Bitcoins to do it.

Maybe somehow the word (passphrase) tear out of the private key. Or there only one way, you can select 64 characters hexadecimal system and this will be a private key, you can get the same address through the word but through the address we can not see this word. As the author of a puzzle in general he scattered keys in space If he did not select the address manually. Or he has a hardware wallet (deterministic). How can we repeat the principle of his puzzle just like that scattered at the same addresses. Or was it an isolated case in the history of the universe? 

WOw that is really tiresome puzzle i would recommend you to take some photos and pictures on how you came up with that given so that we can fully understand on how are we going to crack that puzzle. I know that would lead to an account holder with huge money on it. Well let me help cracking that one but i need more data regarding the puzzle.

1. There is no true randomness in computers.
2. Can you count? 30 keys from 55 were in second half.

But OK, I've counted myself just now and was expecting a bit more than 30/55 chances

I do not understand all this, If I'm smart enough to do this, it would be nice to do it. Would the author be a miner? This also raises the question whether someone else rather than the author did cash out

Unless and until they don't.  The selection of key is totally random within the range with all values equally likely.  Seeing or creating "trends" in random data is a very human thing to do but unfortunately for your theory the private keys for each range were created by having a computer create a 256 bit secure random number and then masking the result down to the desired key length.

If you search the top half of the range for any key there is exactly a 50% chance of the key being in the top half of the range.

Not exactly. He just mentioned that values are random. But randomness can be very different.
If you look at the table with keys 1-55 you'll see that keys tend to sit in second part of range.

Theory is explained here (in short):
https://lbc.cryptoguru.org/about

and here (a lot of reading):
https://bitcointalk.org/index.php?topic=1877935.260

Software is available in at least two options:

LBC
or
https://bitcointalksearch.org/topic/faster-and-autonomous-large-bitcoin-collider-fork-3102823