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 understand! But either way, all tips should be tested;) I use bitcrack, with this tip, I can create starting points.
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29-35 is the amount of 1´s in a binary number.
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This script generates a binary value according to the tips of MrFreeDragon
he puts from 29 to 35 "1´s" in binary
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bounty0z, there was no some pattern. All the keys were random. However, i made some analysis of all the private keys for the "opened" wallets, and can say that more likely the key has the following property: Quantity on bit "1" in the key is 45%-55% from the total bit quantityThis means that for the wallet #64 the total quantity of "1"s in the key is from 29 to 35 with the leading first bit "1" (first bit is always "1" for all keys in the puzzle). To increase your chances youl should take 29-35 "1"s, randomly mix them with the remaining "0"s and you will receieve the key. The key consists from let's say 10 "1"s bit and 54 "0"s bit is not the key for #64, as here only 15% are "1"s and the majority are "0"s. How did I receieve this? I had a look at all the private keys in BIN format, and found that the average quantity of "1" in all the keys is 51%, the total range is 40-60%, but the most likely is 45-55%. Please have a look at all the known private keys in BIN formmat and the calculation of "1"s and "0"s in them. No. K%rng 1's 0's %1's Private Key in BIN
01 100,00% 1 0 100% 1
02 100,00% 2 0 100% 11
03 100,00% 3 0 100% 111
04 0,00% 1 3 25% 1000
05 40,00% 3 2 60% 10101
06 58,06% 3 3 50% 110001
07 20,63% 3 4 43% 1001100
08 76,38% 3 5 38% 11100000
09 83,14% 6 3 67% 111010011
10 0,59% 2 8 20% 1000000010
11 12,90% 4 7 36% 10010000011
12 31,07% 8 4 67% 101001111011
13 27,37% 4 9 31% 1010001100000
14 28,73% 5 9 36% 10100100110000
15 63,99% 9 6 60% 110100011110011
16 57,20% 8 8 50% 1100100100110110
17 46,22% 11 6 65% 10111011001001111
18 51,57% 6 12 33% 110000100000001101
19 36,39% 12 7 63% 1010111010010011111
20 64,66% 10 10 50% 11010010110001010101
21 72,78% 11 10 52% 110111010010100110100
22 43,41% 12 10 55% 1011011110010000001111
23 33,49% 12 11 52% 10101010110111001010010
24 72,00% 9 15 38% 110111000010101000000100
25 97,80% 17 8 68% 1111110100101111011100101
26 62,54% 11 15 42% 11010000000011001001101110
27 66,82% 14 13 52% 110101011000011100001110101
28 69,60% 14 14 50% 1101100100010110110011101000
29 49,28% 16 13 55% 10111111000100101010100011110
30 92,44% 16 14 53% 111101100101001100110101100100
31 95,80% 21 10 68% 1111101010011111110011101000111
32 44,05% 15 17 47% 10111000011000101010011000101110
33 66,18% 16 17 48% 110101001011011001010100011011000
34 64,53% 16 18 47% 1101001010011001011001000100011101
35 17,07% 15 20 43% 10010101110110100100001000101110000
36 23,36% 17 19 47% 100111011110100000100000101001111100
37 45,89% 22 15 59% 1011101010111011101010110101010010011
38 6,94% 17 21 45% 10001000111000001011111010110011010000
39 17,77% 20 19 51% 100101101011111100000110000001111101001
40 82,56% 22 18 55% 1110100110101110010010010011001111010110
41 32,63% 21 20 51% 10101001110000110100110101100110001011011
42 31,67% 19 23 45% 101010001000100001110001011000110110001111
43 68,48% 24 19 56% 1101011110100111011001001111100010110010001
44 75,13% 22 22 50% 11100000001010110011010110100011010110001111
45 13,67% 19 26 42% 100100010111111001010000101000011110000000101
46 46,11% 19 27 41% 1011101100000110000011100010001101010101000100
47 70,06% 24 23 51% 11011001101011000010000101101010011110010111010
48 35,86% 31 17 65% 101011011110011011010111110011100011101110011011
49 45,35% 25 24 51% 1011101000001011101101011000000010101111101001101
50 8,56% 24 26 48% 10001010111101010000111100001011101001001101010100
51 82,86% 19 32 37% 111010100000111000010100001101000000000100111010100
52 87,25% 30 22 58% 1110111110101110000101100100110010111001111000111100
53 50,18% 24 29 45% 11000000001111000100011100100011111100011001001101100
54 10,74% 32 22 59% 100011011011111011011011010101101011010001111101000011
55 66,79% 31 24 56% 1101010101111100001111110011011011001111110000100010100
56 22,73% 34 22 61% 10011101000110001011011000111010110001001111111111011111
57 91,85% 29 28 51% 111101011001001011100100100000111100101011101011000011100
58 38,76% 25 33 43% 1011000110011101011011100001010010000110001001101000100001
59 82,17% 33 26 56% 11101001001011011001011101110000111110010101011010001001111
60 96,89% 32 28 53% 111111000000011110100001100000100101001101100111101110111110
61 23,67% 29 32 48% 1001111001001011010100011011101000010111101100100100100000110
62 69,49% 34 28 55% 11011000111101010101000001111010110110000100011010101111101110
63 95,01% 36 27 57% 111110011001110010111101111110110101100110011110110100000001000
65 65,71% 29 36 45% 11010100000111000101100010011010100000101101100100110100001100111
70 64,40% 34 36 49% 1101001001101110000100101101100100001100011010011011000100111011110001
75 19,32% 33 42 44% 100110001011100111000010001010001101000011010100001001100110110111000000111
80 82,89% 37 43 46% 11101010000110100101110001100110110111001100000100011011010110101101000110000000
85 9,03% 39 46 46% 1000101110010000011000100111100000001100011010101000110111000110011101011101110101000
90 40,23% 43 47 48% 101100111000000000101110110010000100110110101001000100010111000111000111101000010110111111
95 28,87% 48 47 51% 10100100111101001111001001010110001100000111100011111110110010010100000111010001011000111110100
100 36,98% 53 47 53% 1010111101010101111111000101100111000011001101011100100011101100011001111110110100100100100000100110
AVG 53% 51% Average for wallets > #15
Keep in mind that the fist 15 keys are not representative due to small length, so the average is calcualted for wallets greater #15. Additional analytics for information purposes (i did not find any patterns here): K%rng is the actual private key range point (in %%). It means where the actual private key was found in the context of total range (0% is the 1st point of the range, 100% is the last point of the range). K%rng for wallet #n is (PrivKey(n) - 2^(n-1)) / 2^(n-1) * 100% Cool! I made a php script to generate these results b848af95c31c7000 - 33 b479223fc70f9800 - 33 a51b778c1dfa9800 - 35 cb5da3803496e000 - 31 8565165d73359800 - 31 ad9aa9f1d9c8d800 - 32 eae9d06a0dda3800 - 31 fc8699e06b942800 - 33 a87bcad3cafd3000 - 35 8b1d03767b88a800 - 30 fee40c87ff868000 - 31 9220d796b49bf000 - 32 9cc62606451f3000 - 29 b434b1ce9248f800 - 30 df1f7444b94e1000 - 34 fce5d65a4292f800 - 33 b642d85cb2fc2000 - 32 ac41754b49cb7800 - 32 bab931fe60882800 - 29 b7460cbcef883800 - 33 974fec034a650000 - 33 c6e327436c5e2000 - 31 e296a30a15cb6000 - 32 e9a6e8e8e14aa800 - 35 e369e4cce3a34800 - 35 886a4d6f9e9da000 - 35 da38c5fbc6962000 - 34 935afac54cff3000 - 35 f888618436e3f000 - 31 f3765981ce08f000 - 35 f6e9901d68c6a000 - 33 acb846ffc364a800 - 34 d0c4871bc5f30800 - 31 808784ef692fa000 - 33 8e4f264b62afe800 - 35 c7c80bdf8662d800 - 35 f9e2c298a10a7800 - 31 b97fee8bcc381000 - 35 ecb66958f8647800 - 34 ab4c5897a1462800 - 30 8e7c489527c5b000 - 33 88bc8a4e511b9000 - 29 81b2dc63cd700000 - 29 dfad627e5a748000 - 33 ddac4c0b7c2a5800 - 33 bc11ee44dc75a000 - 31 f46cc5146107a800 - 32 9f6aceb734112000 - 34 cf729a5e068cb800 - 33 ae12f3f0b5c44800 - 34 dd04d9ccba00a800 - 29 e9c3b64ad352c000 - 34 ba1c7dbb13ab0800 - 34 c4b25d1e22e95000 - 32 b5a90a950dfd7800 - 33 ea260d7b0d521800 - 34 ecbb5827bd50a800 - 33 89de30a6ba1d4000 - 34 9bed13cbcb5c7000 - 35 dacfbbe538855800 - 35 fcd919c8fe255800 - 33 a17db402f748d800 - 33 code: <?php
for($i=0;$i<100;$i++){
$bytes = 1 . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) . rand(0,1) ;
if(substr_count($bytes, '1') >= 29 && substr_count($bytes, '1') <= 35){
echo dechex(bindec($bytes)) . " - " . substr_count($bytes, '1') .'<br>';
}
}
?> |
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Hi!! Is it easier to break an address with the public key? could you explain to me which method? I need public key for non find number .. 64, 66 etc  nobody help my? |
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1) ‘Why The Comb...’ is poor grammar (it should be ‘Why Does The Comb...’). I assume this is deliberate (OP appears to have native English), and that ONLY these letters are required.
2) I assume it’s an anagram of the question for that reason.I had not thought of that before, but it makes sense ... I'll post my thoughts so far here. They may or may not be useful - I hope so. It looks like blockladder has deleted posts he made in the past. A Google search shows up the odd quoted post for which the original has been deleted. It's hard to tell whether these are sarcastic or trolly. Either way, I've hit a dead end. Here's what I've assumed so far. Some of these assumptions could be false, of course. Thoughts/Assumptions 1) ‘Why The Comb...’ is poor grammar (it should be ‘Why Does The Comb...’). I assume this is deliberate (OP appears to have native English), and that ONLY these letters are required. 2) I assume it’s an anagram of the question for that reason. 3) The ‘.txt’ is difficult; it could be part of the answer, or could simply indicate that the answer is plaintext. Including ‘txt’ increases the proportion of consonants to a problematic level. 4) I assume that the 8 camel case words are to be hashed (SHA256) to create a 32-byte privkey. It is technically possible that the 8 words ARE the key. However, this would require that several letters were discarded (as the question contains too many), and there is no obvious indication of which ones. 5) ‘Natasha Otomoski’ is an anagram of ‘Satoshi Nakamoto’. The letters were clearly grouped to indicate this, so I assume ‘Satoshi Nakamoto’ is a part of the answer. This massively reduces possible permutations for 8 words. If it is not, and is simply deliberately misleading, it becomes vastly more difficult to figure out the answer. 6) There may be a second layer in the puzzle, whereby the answer references something in the question rather than just using its letters. ‘Why….?’ implies an answer ‘Because...’ (the right letters aren’t present though) or ‘To...’ I spent a while considering the joke, ‘What has teeth but cannot chew’. Answer: A comb. ‘ToChewThe21MbyteHashOfSatoshiNakamoto’ doesn’t hash to the given address. 7) The significance of ‘21’ (or ‘12’, or ‘1’ and ‘2’ separately) in the answer is unclear. The letters aren’t present for ‘21 Million bitcoins’, though they are for ‘21m Btc’.  A rail fence cipher doesn’t appear to produce anything useful. That's all from me. It felt I was getting close. Assuming there is a solution, I hope this helps someone find it.
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I already gave up on this I'm mining ethereum classic  I suppose that you have either cheap or free electricity I have a generator mounted on a waterfall here on my property free and clean energy |
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I already gave up on this I'm mining ethereum classic |
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that ".txt" at the end of the question should have some meaning
The English dictionary has 171,476 words.
finding 8 words in 171k and still in the right order with 32 characters is very unlikely anyone can solve.
I think there's more chance of winning the lottery than solving this puzzle.
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Right. Either the 8 words are the privkey (which is possible if OP did some brute forcing - I've done something similar myself, several years ago now, for a puzzle), or it's a hash.
Either way, it's finding the right 32 chars. I have been assuming that they are from the question itself, like others have been.
Why would that require brute forcing?  You "simply" pick 8 words that total up to 32 chars as your answer: WeCreateAnAnswerThatIsEightWords Then pick whatever method you like to use that as your private key: char->hex method... which yields private key: 5765437265617465416e416e737765725468617449734569676874576f726473 (18LpdmS7UPHRPsNmD384LRRawHjVXprJy1) or brainwallet (sha256) method... which yields private key: 8EAFEFFE8CEE0E1E43A53314FFEC31BE09B378A4E733A516BBBAEB177CC151FD (15idakESXkA4N5Uyjts2xTFupBLD9Pibz2) You can then send your X BTC prize to the address you just "created" The problem we have is that there is no definitive answer as to whether we should be doing "char->hex" method or "brainwallet" method. I guess the solution to that problem is to simply check both! I fully agree |
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I think you're mistaken, if you insert a camelcase phrase it generates the hash respecting uppercase and lowercase letters. https://brainwalletx.github.io/#generatorjust insert your guesses into the passphrase field that it will convert to an address.  I feel close. I have a solution that matches the right criteria, with the double meaning implied by the question. Just need to figure out how to process it. ablosulty wrong way- brainwallet words is a SMALL case begins! not a @CamelCase |
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I've tested thousands of possibilities, I've given up on this challenge Has anyone been able to verify the signature? G/cbms/K/DNzcRin5v2B03iXdbpdVoZbTebt7KG95j3FUqnJvcP9rDYcGpSV27RLspR7SlPjqma4h0tDAMwovIo= I could not, it always gives invalid signature I'm pretty sure it's troll. Rooting for you guys, I hope its for real and someone here solves it.  Annoyingly, I've hit a dead end. I'll give it a while longer, then post my thoughts here in case they're of use to anyone else. |
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from what I understood, the 8 words in camelcase form 32 characters and then sha256 (8words) forms the private key I don't think it's a troll - I'll give it a little bit more time and then, if I can't solve it, drop my thoughts here. When you have enough letters to play with, you can make it fit (almost) anything. But the answer I've got so nearly fits the criteria, and reflects the principle hidden between the lines of the question very neatly (I can't say more without giving it away). But I cannot figure out how it's slimmed down to 32 letters. I am missing something. Thank you for the tool. |
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I made a javascript script to test multiple combinations with 8 words that add 32 characters and did not succeed, several people tried it, this tool is very good for testing https://github.com/dan-v/bruteforce-bitcoin-brainwalletI came to the conclusion that this is some troll, the tips are very bad, it is easier to find out other puzzles type this: https://bitcointalk.org/index.php?topic=1306983.640sorry for my bad english https://brainwalletx.github.io/#generatorjust insert your guesses into the passphrase field that it will convert to an address. I feel close. I have a solution that matches the right criteria, with the double meaning implied by the question. Just need to figure out how to process it. Thanks - BitAddress does the same if you enter a non-valid private key. I've been using that. It's not clear from the question whether the solution to the riddle is the private key itself, or whether it points to the private key. I don't think it will make too much difference either way but would help - knowing if the answer is 32 letters or not would be useful. |
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https://brainwalletx.github.io/#generatorjust insert your guesses into the passphrase field that it will convert to an address. I feel close. I have a solution that matches the right criteria, with the double meaning implied by the question. Just need to figure out how to process it.
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Hello, I just registered here. I`m trying to solve this puzzle from a week now and I tried everything that came to my mind. Tried to search for answen for: Why Satoshi Nakamoto made bitcoin amount to be 21 million - no eight english words answer Why combs have 21 teeth - no eight english words answer What is the meaning of the number 21 - it is the sum of all first 6 natural numbers - matematics, Fibonacci, spritual, what so ever involves the number 21, 21 century stuff, I tried and tried.... and no luck... I don`t think that even Satoshi Nakamoto(or Natasha Otomoski) can answer that question. I wish you people luck to crack that puzzle! This puzzle I suppose requires some technical workaround in cryptography. I've tried hashing the question into a sha256 and minus the transaction hash no eight english words there. This riddle or puzzle is extremely difficult to solve, but I think it is only a matter of months anyone would solve this. knowledge of cryptography is only a secondary requirement and it will only be needed in the last step, if at all. the first step is to understand the puzzle question and have a starting point instead of making guesses and/or brute forcing random words to get something out of it. at this point i am convinced that this either was a trolling attempt considering overall activity of OP or if it is an actual puzzle, it was very poorly designed which makes it impossible to solve. total waste of time. it is clear that it is a troll. |
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You, stubborn boyz, don't get simple thing: bounty is 100% compressed, so you need only -c switch.
For abandoned wallets you 100% need both -u and -c switches.
That's 2x drop.
bloom filter with thousands of addressed also cannot be free. Here you get another ~50% drop.
So, 1 address vs thousands is really around "3X difference".
Ok We "stubborn boyz" have shown you numbers not expectations. ./clBitCrack -c -d 1 -b 72 -t 256 -p 2048 15VF3MsCzjHmFQ3wK3SMrTEBTmFY8zhJnU [2019-02-27.13:22:58] [Info] Compression: compressed Ellesmere 2304/6473MB | 1 target 107.29 MKey/s (1,321,205,760 total) [00:00:10] ./clBitCrack -u -d 1 -b 72 -t 256 -p 2048 15VF3MsCzjHmFQ3wK3SMrTEBTmFY8zhJnU [2019-02-27.13:22:00] [Info] Compression: uncompressed Ellesmere 2304/6473MB | 1 target 85.06 MKey/s (1,283,457,024 total) [00:00:12] ./clBitCrack -c -u -d 1 -b 72 -t 256 -p 2048 15VF3MsCzjHmFQ3wK3SMrTEBTmFY8zhJnU [2019-02-27.13:20:27] [Info] Compression: both Ellesmere 2304/6485MB | 1 target 64.78 MKey/s (868,220,928 total) [00:00:11] ./clBitCrack -c -d 1 -b 72 -t 256 -p 2048 -o res.txt -i btc.txt [2019-02-26.03:06:54] [Info] Compression: compressed [2019-02-26.03:08:58] [Info] 2,749,473 addresses loaded (52.4MB) Ellesmere 2368/6450MB | 2749473 targets 83.47 MKey/s (4,420,641,226,752 total) [14:41:53]^C ./clBitCrack -u -d 1 -b 72 -t 256 -p 2048 -o res.txt -i btc.txt [2019-02-27.13:34:01] [Info] Compression: uncompressed [2019-02-27.13:34:40] [Info] 2,749,473 addresses loaded (52.4MB) Ellesmere 2368/6473MB | 2749473 targets 64.28 MKey/s (2,793,406,464 total) [00:00:39] ./clBitCrack -c -u -d 1 -b 72 -t 256 -p 2048 -o res.txt -i btc.txt [2019-02-26.23:52:15] [Info] Compression: both [2019-02-26.23:54:19] [Info] 2,749,473 addresses loaded (52.4MB) Ellesmere 2368/6481MB | 2749473 targets 46.05 MKey/s (2,213,925,617,664 total) [13:21:11] In Short: Ubuntu 16.04, Rx480 8Gb 1 target compressed 107 Mkey/s 1 target uncompressed 85 Mkey/s 1 target both 65 Mkey/s 2.7M targets compressed 83 Mkey/s 2.7M targets uncompressed 64 MKey/s 2.7M targets both 46 Mkey/s Show us your setting (OS, gpu type) and your numbers (Mkeys/s with 1 or many targets, with -c or -c -u) otherwise, no need to comment. Thanks. One more thing, about -b -t -p (to Zielar) Default setting: ~/BitCrack/bin$ ./clBitCrack -d 1 15VF3MsCzjHmFQ3wK3SMrTEBTmFY8zhJnU Ellesmere 16/6471MB | 1 target 48.35 MKey/s (1,997,537,280 total) [00:00:39] My setting: :~/BitCrack/bin$ ./clBitCrack -d 1 -b 72 -t 256 -p 2048 15VF3MsCzjHmFQ3wK3SMrTEBTmFY8zhJnU Ellesmere 2304/6491MB | 1 target 107.70 MKey/s (868,220,928 total) [00:00:06] Nice!!! |
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Thank you for you very competent comments.
Welcome. I claim that an Rx480 does 105MHkey/s on 01 target and 85 MHkey/s on 2,749,473 targets (compressed only)... draw you own conclusions. This is no BS.
1. Most abandoned wallets are uncompressed. 2. If speed with one address is almost the same as with 2.7M, then you haven't found right values for -b A -t B -p C No other options. 8Gb Rx480 -b 72 -t 256 -p 2048 Most abandoned wallets are uncompressed .. you are certainly right. my rx480 4gb config: -c -b 72 -t 256 -p 2048 5,000,000 addresses = 85Mkey/s -c -u -b 72 -t 256 -p 2048 5,000,000 addresses = 42Mkey/s |
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hello guys, today this wallet appeared 1JMT8uyrnVAvkhiexbTL9dyyMwTqjKygU9
in my search
take a look at the transactions, can anyone explain what this is?
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Woah that https://www.blockchain.com/en/btc/address/1FeexV6bAHb8ybZjqQMjJrcCrHGW9sb6uF?offset=250&filter=6 address is filled with $300 million! I thought this whole riddle was about 32 BTC? Btw I am not into this cracking kinda stuff enough to participate into this. Just clicked through the posted addresses in this thread and found this HUGE wallet which a user above me posted. Insane. I guess if this key is crackable, the reward is so insanely high because it is nearly impossible to crack. I'd love to know the guy knowing the result to this riddle and get to know how to solve this lol. That'd be awesome. Like "Hey dude you heard about this riddle I made with Bitcoin? You aren't into BTC anyway, so here's how to solve it...." "Oh that sounds fun! No worries I won't participate in solving it. *Opens Bitcoin wallet and imports priv key*" By the way: Ever thought of where this huge stack of Coins might come from? Could the source be some illegal hacking stuff offering the coins for grabs via a riddle because they can't sell them anyway? (IF it is related to any riddle at all.) Could it be an exchange wallet not related to a riddle at all? Would be really interested in seeing some blockchain analysis here to get to know who owns/owned these coins and how they have been acquired. 2011 is a long long time ago. Or did Mark Karpeles forgot a paper wallet sitting in between his couch again? That one must be an exchange cold wallet. Transaction happened within the last 12 months also. I bet it's Coinbase. Or Bitfinex. Btw [...]e+12 not familiar with these kind of numbers that begin with "e", what does it mean? I'm stupid I know. I'm also not very good at math. Would be happy to have explained why you wrote those "e"-numbers behind the addresses. Thanks! hi!! Prism switches to scientific notation when the values are very larger or very small. For example: 2.3e-5, means 2.3 times ten to the minus five power, or 0.000023 4.5e6 means 4.5 times ten to the sixth power, or 4500000 which is the same as 4,500,000 This is a standard notation used by many computer programs including Excel. Entering a value in this form is not the same as entering the logarithm of a number. This is simply a shortcut way to enter very large values, or tiny fractions, without using logarithms Note that in other contexts, e = 2.71828183, the base of natural logarithms. But when used in displaying large or small numbers, e means "times ten to the power of...". |
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