 Show Posts
|
|
Pages: [1] 2 3 »
|
Hi..Digaran..Meet you after a long time..How are you..Lol
I also have this suspicion about that user and other users, a lot of newbie accounts recently came up to this topic, all of them writing non-senses the probleme is not there, the probleme is we didint see digran reply from a long time, so you know the conclusion now xd we need mod here to check their ip and ban those users, |
|
|
|
|
im asking the admin to ban digran and his multiple accounts, this thread become useless with his useless ideas
@modeator, verify those replies you will find the same ip address
|
|
|
|
Some rocket science, some dice, a few weeks of calm and quiet, a few cups of coffee and finally an HD wallet to generate 256 keys and then masking each key with 0s to place each key in a certain bit range. Like : 9845224217be692ed430097d85a368999c01f3d14e5cf30c3b0ce72921e9d0cf Replace the strike through characters with 0s and good to go.
I managed to take a snapshot of the situation when #66 is solved. some mod ban this user please  his always reply with nothing to say about the puzzle, this is not a forum for memes, where are your new accounts ? we didint see your self reply to yourself anymore, stay away from this thread we are trying to solve a puzzle here, P66 is in 20-2F range not in 30-3F. |
|
|
|
So what is it you want? You explained in detail about 2+7=9, which is half of 18, but what are you getting at? Because you are looking for X, which is unknown, do you have any value related to X?
Never mind that, let me give you a hint: Let your X be 35, your end range be 50, now you can sub 35 -50 = 15, sub 15 from half of 50, so 15-25 = 10, then sub 10 from half of 25, so 10-12.5 = 2.5, now you should multiply by 4 to get 10, by 5 to get 12.5, by 6 to get 15 and so on. Someone here was talking about being good in math, this should be easy for mathematicians to figure out, after all we have the start and end range. bro we dont need your advices, you are the professor of maths and crypto here, please dont wright comments that dont have any meanings, for 1 year and you are typing nothing in this thread. except big talk and there is nothing, this topic become useless i swear, , if any mod here please close this thread. newbie now cant see all +200 pages on this thread thats why we will have lot of newbie asking for pk of their substracted public keys without knowloadge they must see all replys and this topic become useless. |
|
|
|
I think your comments/scripts confuse people more than helps them, IMO.
I thought everyone is a cryptography expert around here, or worked for more than 25 years on EC, if my methods confuses you is because you don't try working only with scalars to penetrate N the group order. If I explain everything step by step, where would be the fun in that? Have you shared anything you could discover by operating over scalars? Whatever method, equation you can use to get definitive results over scalars, applies to points as well. I have explained how to get meaningful results based on that before. To Alek on a few previous pages, regarding how to get target /1024 for sure. Digaran, you're like a toddler when it comes to ECC. Without the proper knowledge about ECC, group fields, etc , you will always be mesmerized by every result you're getting when you perform some arithmetical operations on the curve. ..if my methods confuses you is because you don't try working only with scalars to penetrate N the group order
That is so stupid in so many ways. The order is a prime number and you are trying to find a divisor that is different from 1 and the number itself....I'm speechless! Whatever method, equation you can use to get definitive results over scalars, applies to points as well.
Another stupidity. For example you can multiply 2 scalars over the group field but you cannot do that with 2 points. It seems that posting whatever crosses your mind, it's your full time job! This wouldn't be a problem if you will present concrete examples from A to Z or if you will have a solid understanding about ECC. The problem is that you're lacking both. There are tons of materials about ECC and how it works - you just have to google it. The more you'll learn, the more you'll realize what a masterpiece the secp256k1 is. "Penetrate N the group order" by using a combination of subtractions and divisions is at least pointless! I already told him, alert him, this user digran is talking nothing in this thread, he talk whatever came in his mind, with chatgpt, give us codes that he dsnt know how it work or whats the algorithm, he do actually create codes but with ai, just small chat, hi ai give me multiply 2 of 2 points, then he give us the codes, 60 pages in this thread with this user have nothing related to the subject of this thread, 60 pages with zero result only big talks and nothing, Men . we are a lot of users who work in silent, if you have something that work please upload it in github, you dont have nothing keep your random search in vanity search or keyhunt please men let us think quite and understand other users, you are 24hours talking with maths you never understand ald algo, programing, even python if i came to you in real world and tell you to create anything stupid in python you will be lost, please STOP DIGRAN STOP POSTING ON THIS POSTE, Well that BBC is your answer, but it seems nobody has figured it out yet. Good news for our dragon. The answer to what? See, you throw out numbers and math but none of it makes sense, is new, or doesn't help in knowingly how to reduce a range. Elaborate what is shows/tells you? If you can. YES Wandering, thats what i mean i just saw your reply now and im editing my reply, yes he never help with something everything is false non use have no sense you are not a programmer DIGRAN so please dont enter here to have the talk of the professional by the way i know you are talking a lot, you have a probleme im real life bro you just have 4 stars in this forum that didint mean you are a pro or old in btc community, stop this madness bro
I just got this idea, where is our code genius @mcdouglasx, 😉 how about a script which subtracts G as many times as we want from target and divides
YES YOU ALWAYS HAVE AN IDEA THATS A BIG PROBLEME MEN we understand nothing from your maths and non sense codes you are doing nothing, you want to substract ? https://github.com/WanderingPhilosopher/Windows-KeySubtractoryes use keysubstractor this program is have all you can sub anypub key to whatever range you want, you cant substract pk to a know point because you already dont know the pk, there is no way here to do something to the puzzle you have tools just search please and never post anything here you are talking nothing here except this thread become big with your replys and non sense maths. im asking the MODERATOR to mute The user DIGARAN. |
|
|
|
By the way, and you know why their puzzle addresses were removed from the 161st?
My thought is that this is the limit of public addresses. That is the further search in the theory goes on a circle.
Because all the funds from 161 through 255 were redundant (Bitcoin addresses only use 160 bits) the author of the puzzle moved all the funds from the 160-255 ranged down into the 1-160 range. It is explained up thread. Is it? I thought it uses almost 256 bits (the last possible private key is FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140) OH, HIII, Good morning from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('Here Compressed Public Key')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))
def ters(Qx,Scalar):
ScalarBin = str(bin(Scalar))[2:]
le=len(ScalarBin)
for i in range (1,le+1):
if ScalarBin[le-i] == "0":
Qx=ice.point_multiplication(k,Qx)
else:
Qx=ice.point_addition(Qx,neg1)
Qx=ice.point_multiplication(k,Qx)
return ice.point_to_cpub(Qx)
for x in range(1,65536):
f= (ters(pub,x))
data= open(“halfpub.txt”,”a”)
data.write(f+”\n”)
data.close() Note this is where you decide how many bits should be reduced for x in range(1,65536): For example reducing 26 bits requires 67108864 to be generated, 1 of them is the correct 26 bit reduced key. Will you release for public to use? Yes this is my code, I can write bit reduction code in another way if you want. If you want, you can reduce bits by subtraction. I'm developing a new algorithm for downgrading from a known bit range to the bit range I want. When I complete it by giving only 1 correct pubkey, I will share it here. @lordfrs, I was thinking, why do we need to keep all the 65556 keys if we are reducing 16 bits? One other thing, if we multiply one of the keys from the end of the output file by 65556 we should see our target, correct? I haven't thought about on how to calculate, since we are subtracting one dividing by 2, one result if multiplied by 2^16 should reach the target or close to target, so why not discarding 90% of the generated keys and start multiplying the remaining 10% by 2^16? Could you add such operation to your script? I guess if we could operate on private keys instead of public keys using your script, we could reach some results, ( as a test of course ) To test with scalar only, I use this script : from sympy import mod_inverse
N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
def ters(scalar):
k = mod_inverse(2, N)
scalar_bin = bin(scalar)[2:]
result = 2
for i in range(len(scalar_bin)):
if scalar_bin[i] == '0':
result = (result * k) % N
else:
result = ((result * k) % N + N - 1) % N
return hex(result)[2:].zfill(64)
for x in range(1, 20):
f = ters(x)
print(f) It was a lucky shot at AI generated code and it worked.😅 Btw, the first "result" in the script above, represents your target, I couldn't risk the AI to mess one thing she has done right. So result = 2, replace 2 with your own scalar in decimal. bro please stop replying with those comments copy past from chatgpt you are doing nothing except replying and talk too much, you are doing nothing for real men, stop it bro,  thank you |
|
|
|
no the gpu find nothing only if i search with cpu can i find keys
-t 30 1000mk/s finds key after 10 sek
-gpu 10000mk/s find nothing after 1 hours
i thing the gpu engine is wrong or so
im search on rtx 4060
if i use bitcrack for searching i can search only on gpu and find keys
-gpu -t 0
you are using the wrong software if you are looking for the puzzle, thats a vanity search and there is no range search, , try vanbitcracken and put your range to search https://github.com/WanderingPhilosopher/VanBitCrackenRandom2 |
|
|
|
Hello, I'm new here and have a question about
bojlahg/VanitySearchPublicKey.
if I search there I find only keys with use cpu
-gpu is displayed but does not find any keys
for example:
VanitySearchPK.exe -t 0 -gpu -sk 2000000000000000000000000000000000 -p 633cbe Find no keys but 5000mk/s (0keys) 1 minute
VanitySearchPK.exe -t 1 -gpu -sk 2000000000000000000000000000000000 -p 633cbe find keys with 5600mk/s very bad (25keys) 1 minute
VanitySearchPK.exe -t 32 -sk 200000000000000000000000000000000 -p 633cbe find keys good but 1000mk/s (200keys)1 minute
I believe that the program recognizes the GPU
but the gpu performance does nothing
can any help for this
I can't fix this but I would be happy if someone who can do it would contact me.
it does actually, 8 core cpu searching 10 million key per seconde, you are searching with 5000 million key per seconde thats your gpu performence, so dont worry gpu search is working |
|
|
|
Looks like those who is searching for #66 are at 354d range, even myself also search at the same range. If my research correct, #66 range should be in between 354df - 358ae. Even this range will take ages to scan. LoL.
How did you come to this conclusion? By breaking down bits and some bits combination. Maybe just some wild guessing and not sure if I am correct because I spots seems like some obvious pattern. I hope someone can find it on this range thought even if I don't get it. If the results were in this range, then I can use the same methods to proceed to 67. Unfortunately, the most lowest I can go is only on current range. I guess it boost some morale maybe I am right when I saw zahid8888 post PK start at 354d and have similarity on #66 Hash160. thats what i said exactly, im substracting from 130 lower bit range and i get those address that start with same 14 digits of hash160 that means that address in that range of 90 or even 80 bit range So any update on the progress of finding this key?
First offset = 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710
Half of above =
0291001b0dc6e5a2628cb4698eb00a6fb7dbd276dc2b214795f2fe52e61243aa9b
Half of 130?
0337374e00a32eaf009e9946035c0e69085627b60a844637d2b958dd83bcfa4383
The following is the subtracted key from #130
Second offset =
03d99bb89e8db75d20b882f13f8086fb39221858fa211de0346c926a93ae259b3a
Half of above?
03a3dc00bf5f7e7eec691569c7f67a15d3cdbb3a9994c9a5ec1430cffdb622cf9f
Now subtract half of first offset from half of #130 to get half of second offset.
Second offset is known, we need to work on first offset's half, use -1 divide by 2 script to reduce 18 bits from it, you'll have millions of new offsets and one of them is the target, now divide the #130 range by 2, subtract 18 bits from it and use the new range as your search range, input those millions offset keys and search the range.
Don't just try blind searching.😉
its not blind searching 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 hash160 3aaccf438388c4aeb0b433c7b778f25cb6ab244c hash160 3aaccf438388c31f14410f489939d3d5eac88f19 pk: 48ea48b7a25627365cff38d 13 same hash160 digits, and its two substraction of 130 its not one substraction, 13 digits means its 80 pourcent in that range like puzzle 66 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 + 2786b52d106d22524ed9cf8a87d2 031ed6283a43d439eace1ee2815118cb6f16f475be60fa5ccc8598372d8c5f1995 # target 031ed6283a43d439eace1ee2815118cb6f16f475be60fa5ccc8598372d8c5f1995 # + 3......................................... 03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852 # target |
|
|
|
I'm bored, nothing interesting is happening around these woods no more. Nobody has new and exciting ideas no more.
Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?
And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.
I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?
Good luck to all.😴
i already find two puzzle with my methode of hash160 lowering the range of searching,and still im going to 130 135 my methode is working its just lowering and find the range in subed pub key to lower bit range 2 times yes when you find same 14 digits of hash160 than means that adress is in that range, but the search is not for one adress, i get a lot of addresses im trying , and no im not going to share the pk of any address i work hard for this, i advise you to try on 120 my methode, and get your bitcoin cash prize, ps: ill leave 50 satoshi on 130 for you, thank you |
|
|
|
its not blind searching
No its not blind search it actually a dummy search @kalos15btc There is no relationship if two publickey keys share a partial RMD hash even if you found an address with 159 bits exactly equals to another address, there is not going to be a relationshipt between the publickeys I have this rmd160 as my address
000000009f012a539f3df386f0bddbb874af55ec
I should get to work and find this address 1111111111111111111114oLvT2
With rmd160 of
0000000000000000000000000000000000000000 ? Lol I'm done.
This time I agree with you @digaran you did not understand, im searching for substracted public keys, im searching for their range where exactly,, 110bits 112 118 109 102 bits,, then like that public key 13 same hash160 digits, than means its in that range , so i scan the range if i found it there i can brake the 130,, thats how i think tell me if im wrong exemple 10 substracted of 130 i found same 10 first lettre of uncompressed adress i have the pk, so i search in that range of pk the address similar in substracted if i found it i do + and find the target,,, am i wrong ?? |
|
|
|
So any update on the progress of finding this key?
First offset = 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710
Half of above =
0291001b0dc6e5a2628cb4698eb00a6fb7dbd276dc2b214795f2fe52e61243aa9b
Half of 130?
0337374e00a32eaf009e9946035c0e69085627b60a844637d2b958dd83bcfa4383
The following is the subtracted key from #130
Second offset =
03d99bb89e8db75d20b882f13f8086fb39221858fa211de0346c926a93ae259b3a
Half of above?
03a3dc00bf5f7e7eec691569c7f67a15d3cdbb3a9994c9a5ec1430cffdb622cf9f
Now subtract half of first offset from half of #130 to get half of second offset.
Second offset is known, we need to work on first offset's half, use -1 divide by 2 script to reduce 18 bits from it, you'll have millions of new offsets and one of them is the target, now divide the #130 range by 2, subtract 18 bits from it and use the new range as your search range, input those millions offset keys and search the range.
Don't just try blind searching.😉
its not blind searching 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 hash160 3aaccf438388c4aeb0b433c7b778f25cb6ab244c hash160 3aaccf438388c31f14410f489939d3d5eac88f19 pk: 48ea48b7a25627365cff38d 13 same hash160 digits, and its two substraction of 130 its not one substraction, 13 digits means its 80 pourcent in that range like puzzle 66 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 + 2786b52d106d22524ed9cf8a87d2 031ed6283a43d439eace1ee2815118cb6f16f475be60fa5ccc8598372d8c5f1995 # target 031ed6283a43d439eace1ee2815118cb6f16f475be60fa5ccc8598372d8c5f1995 # + 3......................................... 03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852 # target |
|
|
|
@digaran indeed,, im about to breake puzzle 130 ---------------------------------------------------------------------------------------------------- Need someone to scan 89bit to 90 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 # + range 10005df8cf4d0f152b26859:49fd5df8cf4d0f152b26859 its easy range anyone have gpu can scan this in 15 min, we try this why not maybe im not wrong of course we split 13 btc if im correct and we found that pk of this adress,, yes @digaran you can call me master of crypto  89-90 bits 20000000000000000000000:3ffffffffffffffffffffff ok, I'm through, range scan done. Nothing found. Whats next?  where is the proof of the work §.? |
|
|
|
@digaran indeed,, im about to breake puzzle 130 ---------------------------------------------------------------------------------------------------- Need someone to scan 89bit to 90 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 # + range 10005df8cf4d0f152b26859:49fd5df8cf4d0f152b26859 its easy range anyone have gpu can scan this in 15 min, we try this why not maybe im not wrong of course we split 13 btc if im correct and we found that pk of this adress,, yes @digaran you can call me master of crypto So you want others to search for you, if the key was found you split the prize that easy? Son, what happens if they say they have found the key and you go rent some GPUs to get it yourself and keep the prize? You see why nobody would search that range for you, unless they are willingly trying to help you. Also there is no explanation about why you think that key is in that specific range. Should we be waiting for 587 other such ranges from you asking people to scan them for you? Lol. Here is my 50satoshis, before attempting to search for a key, make sure it's in range. you have concluded this ? and you judge me ? no im not sure but at least im trying to solve it instead of doing nothing like you; you just reply like you know everything and you dont do nothing;; im helping here if i find the 130 i will leave tips for everyone even you  so please dont judge like this ps/ im the first one here in this thread who shared the gpu rent website; if im searching and want for my self i didint post the link;; i posted then 120 andd 125 cracked;; i have no probleme and im happy who will crack this puzzle; |
|
|
|
@digaran indeed,, im about to breake puzzle 130 ---------------------------------------------------------------------------------------------------- Need someone to scan 89bit to 90 03982a5a42895a5cfe4b9b98e49f9389ebf9b3bf91c2289f1c5db3d944f46ec710 # + range 10005df8cf4d0f152b26859:49fd5df8cf4d0f152b26859 its easy range anyone have gpu can scan this in 15 min, we try this why not maybe im not wrong of course we split 13 btc if im correct and we found that pk of this adress,, yes @digaran you can call me master of crypto |
|
|
|
I have no idea why so many are so obsessed about this adding and subtracting public key thingy. Maybe I know nothing much about it and makes me think that way. This adding/subtracting method in my opinion are totally different from Pollard Kangaroo idea or BSGS. From my understanding, by adding/subtracting public key, you can generate a bunch of valid public key, nothing wrong about it. But if 1 private key matching only 1 public key(of course it does), then where is the difference between adding/subtracting and private key bruteforcing? IMHO, it doesn't eliminate any possible private key combination at all. Please correct me if I am wrong.
try this https://github.com/WanderingPhilosopher/Windows-KeySubtractorand you will understand with substraction, we can minimise the search range from 130bit to lower range, you could fild a private key of generated substracted public key in lower range exemple 100bit or even lower so that will be easy to crack 130 bit.. |
|
|
|
PS: @digaran, you are awesome , 3 months ago im teaching you how to substract from public keys
Modified the quote. WP taught me subtraction/addition, I learned division from garlonicon's posts. Please don't take credit for things you haven't done. Thanks for kind words.! @Op, in the output file I have several #targets, is that what we are looking for? How do I derive the private key now?
@Op, in the output file I have several #targets, is that what we are looking for? How do I derive the private key now?
You should only have 1 target in output file unless you ran the program more than once with same output file. The target key is the original public key you used in the settings. To derive the private key you need to search whatever range/bits of your original public key used to generate all of the offset keys in your output file. What now? I just double clicked on the bat file a few times, each time it finished in less than a second so I changed things and ran it again until I saw the output file is full. Now you say #target is my input, but why is it in my output file and why did the tool add + to them instead of subtracting? Because I selected -s only. I have 3 #targets, one of them is derived from 64 hex string and 2 of them are from 31 char hex string, should I search the one leading to the target? And why does it select the additions like this : eeeeeeeeeeeeee434343434343, did you just hit the keyboard for add/sub selection or is it doing it by itself? I got what you said about how it works, I have a problem with the results I have got, for example:
dc629ccec8a0c3c53e5a0707021a0053a325720e3321f5fb6db3f9339eac23f7 #+ 184ffe30916bcf90fdba43677100000 and I used -s, which means I have to actually add, but add what to what and how?
Edit, one other thing, how does this adding and subtracting exactly work? Is it done using integers or hex and why do I get different results when I use decimal values to add and subtract? Could you explain this please?
./keysubtracter -p 02ceb6cbbcdbdf5ef7150682150f4ce2c6f4807b349827dcdbdd1f2efa88 5a2630-n 100 -b 120 Note the bolded part above, no space therefore it starts to produce invalid keys. 03f1d41da8acf0506f3bf7140b5629dd33a5cf546133479530cd8065e335a97666 # - 13292279957849158729038070602803446
02000000000000000000000000000000000000000000000000000000000000000 # + 13292279957849158729038070602803446
02b70ae2dcb442548570313f652b91ca093a3acac3a2441cb64614e8195505b6b8 # - 26584559915698317458076141205606892
0367dabeef20a6a8b7b5555b162cc8c8489e3134dcec624fe028573c34dbbf20f6 # + 26584559915698317458076141205606892
02a1d21298779f888cd8169f9ed59e4383219cdadbdb342ba886034ef9013b89be # - 39876839873547476187114211808410338
02ae015703cbaee9570dc648d7bce78ac7cb438630e09d69eef4f1208655e1027d # + 39876839873547476187114211808410338 Note one of the public keys above is the 0 public key aka k(0) public key, it shows up if you subtract your k from your p, basically point at infinity. Now our actual target's k is 13292279957849158729038070602803446 because we subtracted it from our target. Showing examples for newbies to understand, not that a few posts above I wasn't a total noob, now I act as if I know things, go figure.😅 work on any puzzle public key puzzle 100 or 105, you will understand how it work, and put the range like this Keysubstracter -s ------- 0:fffffffffffffffffffffffff -o puzzle100.txt do only -s dont add -a, you will have all public key from 0 to range 100 put this range always 0 to : end range and when you create the file.txt, just go and do the maths with https://www.boxentriq.com/code-breaking/big-number-calculatorjust do this privatekey of the puzzle 100 - number that you get,, in any line ,exemple af55fc59c335c8ec67ed24826 + ..... you will get the privatekey of the public key that you generate with keysubs,,,, do that with more that 20 lines randomly,,, you wil have smaller and bigger ranges pk,,, you will understand how you will search for puzzle 120 or 125 by this methode, because your work range with rangepuzzle120:rangepuzzle120 and -a and -s you only generate public keys between 800000000000000000000000000000 and ffffffffffffffffffffffffffffff, but the methode above and i wanna you to try it you will get all addresses txt file from 0 to puzzle,,, you can find any address in smaller range 10 or in 66 like 13zb1hQ did you get it ? you will now know how to search and work with this programme. its easy, you can substract 125 puzzle to 80 and 100 but you will have millions in this txt file, you can do bsgs on that file but its slow because it devide your speed to millions that you generated, so you can generate with addresses and search manually, and there is a lot of tricks how can you search , but sir your methode is totally wrong, in one case,, you generate and search with bsgs in that range already, but its the same this if you search for 1 public key 125 puzzle or a file contain 100 million public key substracted from 125 and speed is devided to 100 million, same thing already,, but you must work again on that program to learn how to use it its simple and try my methode you will get it simple, just do an exemple or old puzzle you will find how it works 4 months ago, you didint knew nothing about substraction,,, when i did reply last week about a range did you saw what you said to me ? 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 ---SNIP--- what makes you think that a partial string of one hash has a reference to another? This phase of learning, I call it kindergarten phase, I have also passed this era, I'm in preschool now. Lol. They need to learn their lessons through experience if verbal education is not effective. 😉 so please bro let us think about the puzzle and coloborate to crack it, and dont ever reply to me as the other member said that you are really act like you are genious ok? |
|
|
|
i have found same 14 digits of hash160 of substraction puzzle 130 1QKZKb6DjAxKkbRycPgC8n9Fs7pPfCZeTv # + 3............................ Hash160: ffcd5c308ebf83ad7ba414b068c69e6192d5a1ab address that i found, maybe its in that range: 1QKZKb6DjAxS4kGh9RohKCnJisM9geDp5o Hash160: ffcd5c308ebf83b910d0cd8b853df079c5b07452 pk: 33faa087a4ea5c40e696be9a1c1e8389 but the range is a little big,,,, but for 100 pourcent sure that 130 puzzle is between 30 and 3F dont waste your time searching in 2 so we can say 130 puzzle is located in 330000000000000000000000000000000:3ffffffffffffffffffffffffffffffff range you can search from there , im working on that if i found smaller range same 14 digits or more, that will be amazing to search there .. Some elliptic curve magic ahead!
...
I just enjoy making a simple division difficult and twisted! 🤣, now chop chop start your brain's engine and do some calculation, large fractions could be solved by accounting for the above results, not telling you how.
Dive deep and let your brain solve it.😉
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
...
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.
...
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
you constantly act as if you are a genius who broke SECP256K1 but doesn't want to share his genius findings. On the other hand, you struggle in vain and miserably like many others to solve a puzzle. Find the mistake. PS: @digaran, dont reply to me i dont need your answer nether your guessing or anything of you, like what @citb0in tell, you are acting like you are awesome and know everything, 3 months ago im teaching you how to substract from public keys now you say to me old school and bla bla, please bro dont waste my time replying something have nothing to do with puzzle to just say im old school and you do new things, i dont need your answer really bro,,, @citb0in i totally aggree with you |
|
|
|
scaning with hash160 is 2 time faster ... i run 588 ranges and have this error C:\Users\Wafa>588_ranges.exe =========================== VanBitCrackenS1-Distributed-Puzzle-66-Address-Scanner ================================= Enter the last range you are scanning: Resumming your scan... Sorry.. This Range does not exist in the list. Press any key to exit. Traceback (most recent call last): File "588_ranges.py", line 1205, in <module> NameError: name 'exit' is not defined [11720] Failed to execute script '588_ranges' due to unhandled exception! C:\Users\Wafa> |
|
|
|
i found this address in substraction of 125 13zb1hQ664XDDhM2LkWLLa23cdbhHwya7c # 19.................... hash 160 : 20d45a6a51bc082aeb4d344be937b1bd4ea1f238 am i close if i search in the range 66 cuz the address looks like its in that range it starts with same 7 digits with 66 puzzle and same 8 first digits of hash160... 02c0643bd28d11d650b24dae3143b1d3578e6a3597a9395f43c25ead8ed2a9298f who can scan in kangaroo 66 bits in 5 min or less ? if you find this address we will split the 12 btc  , of course if it is in that range im not sure im asking you guys ? └─$ ./keyhunt -m bsgs -f tests/testpubkey.txt -k 1500 -S -t 10 -b 66 - Version 0.2.230519 Satoshi Quest (legacy), developed by AlbertoBSD
- K factor 1500
- Threads : 10
- Mode BSGS sequential
- Opening file tests/testpubkey.txt
- Added 1 points from file
- Bit Range 66
- -- from : 0x20000000000000000
- -- to : 0x40000000000000000
- N = 0xfffb4000000
- Bloom filter for 6291456000 elements : 21566.38 MB
- Bloom filter for 196608000 elements : 673.95 MB
- Bloom filter for 6144000 elements : 21.06 MB
- Allocating 93.00 MB for 6144000 bP Points
- processing 6291456000/6291456000 bP points : 100%
- Making checkums .. ... done
- Sorting 6144000 elements... Done!
- Writing bloom filter to file keyhunt_bsgs_4_6291456000.blm .... Done!
- Writing bloom filter to file keyhunt_bsgs_6_196608000.blm .... Done!
- Writing bP Table to file keyhunt_bsgs_2_6144000.tbl .. Done!
- Writing bloom filter to file keyhunt_bsgs_7_6144000.blm .... Done!
- Total 36893523135430656000 keys in 720 seconds: ~51 Pkeys/s (51241004354764800 keys/s)
End Your precious key is not in this public key range. Provide some more let's scan. We will hit the key soon and share some funds too. i have news for you bestie, not 7 digits this time 11 same digits in this post https://bitcointalk.org/index.php?topic=1306983.msg62736366#msg62736366 |
|
|
|
|
