Closed winnery2022 closed 9 months ago
It’s a little unclear what you wanted to say by this, since some of the numbers are closed.
It’s a little unclear what you wanted to say by this, since some of the numbers are closed.
I know Alberto will understand this. I can’t put everything out there for everyone to see because there’s just one movement missing for the full formula. Lust number of keys is not xxx
It’s a little unclear what you wanted to say by this, since some of the numbers are closed.
I know Alberto will understand this. I can’t put everything out there for everyone to see because there’s just one movement missing for the full formula. Lust number of keys is not xxx
Just explain what you want to convey to Alberto? What's the point? Do you use only keymath, or also keysubtracter?
I use two programs. My total file weighs a lot, about 400,000 mb. An intersection with half a key is possible here.
I use two programs. My total file weighs a lot, about 400,000 mb. An intersection with half a key is possible here.
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
It’s a little unclear what you wanted to say by this, since some of the numbers are closed.
I know Alberto will understand this. I can’t put everything out there for everyone to see because there’s just one movement missing for the full formula. Lust number of keys is not xxx
to crack all the keys, there is also only one step missing... determine the even key or not. )))
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
How do you imagine it? After all, we cannot check the result of division, whether it is correct or not.
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
these words are followed by a ban))
I can no longer answer here. I'll wait for Alberto's answer.
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
How do you imagine it? After all, we cannot check the result of division, whether it is correct or not.
Это был очередной мамкин хакер, который знает проверку чёткости, но он лёгких путей не ищет )))
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
How do you imagine it? After all, we cannot check the result of division, whether it is correct or not.
Это был очередной мамкин хакер, который знает проверку чёткости, но он лёгких путей не ищет )))
Show me the easy way, maybe you hacked 120 and 125?
Realy?
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
How do you imagine it? After all, we cannot check the result of division, whether it is correct or not.
C:\Users\win>C:\Users\win\Desktop\ecctools\keymath.exe 035dee726b0fe623224fa1f4d79af66a9a1829213ec675875c619266a5f184e7d1 / 2 Result: 0307814838a8456397f0f7523300e07418da15484b169467bd74e7d1e5947e8981
03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852 / 2 0337374e00a32eaf009e9946035c0e69085627b60a844637d2b958dd83bcfa4383
C:\Users\win>C:\Users\win\Desktop\ecctools\keymath.exe 0307814838a8456397f0f7523300e07418da15484b169467bd74e7d1e5947e8981 - 0xx000000000000000000000000000000xxxxd0cf2d4b7ee5eaddxxxxxxxxxxxxxxxxxxx Result: 0xxxxxxdb096b7d5fb22bf9786db97cc48a55930545265d9737c40a41e5f3b982
I can do it too: n-n*1/2= target )))
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
How do you imagine it? After all, we cannot check the result of division, whether it is correct or not.
Это был очередной мамкин хакер, который знает проверку чёткости, но он лёгких путей не ищет )))
Show me the easy way, maybe you hacked 120 and 125?
They were revealed using a large number of video cards. They wrote about this on Bittalk. Probably the next key will be revealed by the same miner.
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
How do you imagine it? After all, we cannot check the result of division, whether it is correct or not.
Это был очередной мамкин хакер, который знает проверку чёткости, но он лёгких путей не ищет )))
I made sure that you know perfectly well how to check this. So I'll delete this.
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
How do you imagine it? After all, we cannot check the result of division, whether it is correct or not.
Это был очередной мамкин хакер, который знает проверку чёткости, но он лёгких путей не ищет )))
Show me the easy way, maybe you hacked 120 and 125?
They were revealed using a large number of video cards. They wrote about this on Bittalk. Probably the next key will be revealed by the same miner.
I know about it. The person is just talking off topic. You asked a question, I answered. I don’t think it’s worth discussing this with you since I don’t get any reasonable answers. It seems you know more than me.
if you knew how to check the parity, then you wouldn't have any questions at all, because parity check = complete hacking of all keys that have a public key and you don't need to invent anything else. if you know the parity of the key, then it will take less than 1 second to crack one key.
You probably don't know about divisions. Firstly, the question is not about even and odd numbers. Secondly, even if you practice honesty, it will not give you anything. You will have to divide it 90 times and that is more than 10,000 million. If you miss once, you lose.
if you knew how to check the parity, then you wouldn't have any questions at all, because parity check = complete hacking of all keys that have a public key and you don't need to invent anything else. if you know the parity of the key, then it will take less than 1 second to crack one key. 034107489c19ef940da1accec6279675921d5eb289662e628bf698bf8090b03183b even 130 bit 0386e7c62ffab6f175f30396bbb7cb462060b27f5dc58ccf9e51f39f0989b4fb28 even 130 bit
Prove that you are not your mother's hacker.
You probably don't know about divisions. Firstly, the question is not about even and odd numbers. Secondly, even if you practice honesty, it will not give you anything. You will have to divide it 90 times and that is more than 10,000 million. If you miss once, you lose.
You just wrote that it's easy to determine parity))))))
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
ВОТ ЭТОТ ПОСТ
You just wrote that it's easy to determine parity))))))
I proved it. If you don't understand, sorry. But you have not proven how to hack an even 130 bit wallet in 1 second.
Maybe so, but we don’t know the parity of the key and can’t check it. Even if you are sure that the target key is even and can be divided by 2, it is still not possible to solve it in a short time.
An even key is very easy to check.
ВОТ ЭТОТ ПОСТ
I don't have time to discuss this with you. I'll wait for Alberto's answer.
You just wrote that it's easy to determine parity))))))
I proved it. If you don't understand, sorry. But you have not proven how to hack an even 130 bit wallet in 1 second.
the thing is, it's impossible. You haven't proven anything, absolutely
the thing is, it's impossible. You haven't proven anything, absolutely
I repeat once again, you don’t have to have a conversation with me. She's not talking about anything. This can be checked very easily. I gave you 2 even keys. Prove to everyone here your words on how to hack in 1 second. Don't change the subject. You won't get any more information.
the thing is, it's impossible. You haven't proven anything, absolutely
I repeat once again, you don’t have to have a conversation with me. She's not talking about anything. This can be checked very easily. I gave you 2 even keys. Prove to everyone here your words on how to hack in 1 second. Don't change the subject. You won't get any more information.
Dude, you didn't get the point at all..
the thing is, it's impossible. You haven't proven anything, absolutely
I repeat once again, you don’t have to have a conversation with me. She's not talking about anything. This can be checked very easily. I gave you 2 even keys. Prove to everyone here your words on how to hack in 1 second. Don't change the subject. You won't get any more information.
Dude, you didn't get the point at all..
I just understood the point. Let's be more specific. The question here is not even or uneven numbers. You're probably unlikely to understand what's written at the top. You can keep even and odd numbers in parallel, but it will take a lot of time. This topic is clearly not worth discussing with you. I don't see the answer in 1 second with an even key. And if you meant that knowing each next subtracted key, then express your thoughts correctly (and once again, they can be done in parallel), but this will take a lot of time manually. If you manually proofread with such thoughts, you are unlikely to achieve success.
We can hit some whith this?
@winnery2022 can you be more specific with this? i don't know what are you asking.
the thing is, it's impossible. You haven't proven anything, absolutely
I repeat once again, you don’t have to have a conversation with me. She's not talking about anything. This can be checked very easily. I gave you 2 even keys. Prove to everyone here your words on how to hack in 1 second. Don't change the subject. You won't get any more information.
Dude, you didn't get the point at all..
I just understood the point. Let's be more specific. The question here is not even or uneven numbers. You're probably unlikely to understand what's written at the top. You can keep even and odd numbers in parallel, but it will take a lot of time. This topic is clearly not worth discussing with you. I don't see the answer in 1 second with an even key. And if you meant that knowing each next subtracted key, then express your thoughts correctly (and once again, they can be done in parallel), but this will take a lot of time manually. If you manually proofread with such thoughts, you are unlikely to achieve success.
You didn't understand the question from sergy about parity. To paraphrase: we do not know and cannot verify the parity of the key if its private key is unknown. If we could check the honesty, we would divide the honest key by 2, check the parity again and divide it by 2 again, and so about 90 times and then we would find the key by a simple search. Try to determine the parity of these public keys even or not the private key from them:
And never think that you know something more than everyone else, for example, I never publish my promotions. but I work in a completely different direction, more fundamental.
the thing is, it's impossible. You haven't proven anything, absolutely
I repeat once again, you don’t have to have a conversation with me. She's not talking about anything. This can be checked very easily. I gave you 2 even keys. Prove to everyone here your words on how to hack in 1 second. Don't change the subject. You won't get any more information.
Dude, you didn't get the point at all..
I just understood the point. Let's be more specific. The question here is not even or uneven numbers. You're probably unlikely to understand what's written at the top. You can keep even and odd numbers in parallel, but it will take a lot of time. This topic is clearly not worth discussing with you. I don't see the answer in 1 second with an even key. And if you meant that knowing each next subtracted key, then express your thoughts correctly (and once again, they can be done in parallel), but this will take a lot of time manually. If you manually proofread with such thoughts, you are unlikely to achieve success.
You didn't understand the question from sergy about parity. To paraphrase: we do not know and cannot verify the parity of the key if its private key is unknown. If we could check the honesty, we would divide the honest key by 2, check the parity again and divide it by 2 again, and so about 90 times and then we would find the key by a simple search. Try to determine the parity of these public keys even or not the private key from them:
- 02b30ef81221219e64d429f30e783a125e0988c8cf3ab9437a45b727618adb7385
- 03e227af2a43bf8eb8e50ca80e1ef920f06613a1450f07d1e3e6841c8c19557ba0
- 03bf68a3a688f57bdb8a0e2ce19b7cecd69354de3edad7a834c36bbed191c96ed6
And never think that you know something more than everyone else, for example, I never publish my promotions. but I work in a completely different direction, more fundamental.
Now you have expressed yourself in more detail. Are these 130 bit keys? We can divide every 2 next keys, then 4, and so on, but the file will be very large and difficult to do manually. I talked about this earlier. We can also conduct them in parallel. I think we can close the topic on division at this point.
We can hit some whith this?
@winnery2022 can you be more specific with this? i don't know what are you asking.
I meant that different numbers from the main x2 division intersect relatively, this can be seen in my file. Is it possible to improve this process?
We can hit some whith this?
@winnery2022 can you be more specific with this? i don't know what are you asking.
I meant that different numbers from the main x2 division intersect relatively, this can be seen in my file. Is it possible to improve this process?
Invented the bicycle.) This is the natural and well-known behavior of numbers when dividing modulo))))
the thing is, it's impossible. You haven't proven anything, absolutely
I repeat once again, you don’t have to have a conversation with me. She's not talking about anything. This can be checked very easily. I gave you 2 even keys. Prove to everyone here your words on how to hack in 1 second. Don't change the subject. You won't get any more information.
Dude, you didn't get the point at all..
I just understood the point. Let's be more specific. The question here is not even or uneven numbers. You're probably unlikely to understand what's written at the top. You can keep even and odd numbers in parallel, but it will take a lot of time. This topic is clearly not worth discussing with you. I don't see the answer in 1 second with an even key. And if you meant that knowing each next subtracted key, then express your thoughts correctly (and once again, they can be done in parallel), but this will take a lot of time manually. If you manually proofread with such thoughts, you are unlikely to achieve success.
You didn't understand the question from sergy about parity. To paraphrase: we do not know and cannot verify the parity of the key if its private key is unknown. If we could check the honesty, we would divide the honest key by 2, check the parity again and divide it by 2 again, and so about 90 times and then we would find the key by a simple search. Try to determine the parity of these public keys even or not the private key from them:
- 02b30ef81221219e64d429f30e783a125e0988c8cf3ab9437a45b727618adb7385
- 03e227af2a43bf8eb8e50ca80e1ef920f06613a1450f07d1e3e6841c8c19557ba0
- 03bf68a3a688f57bdb8a0e2ce19b7cecd69354de3edad7a834c36bbed191c96ed6
And never think that you know something more than everyone else, for example, I never publish my promotions. but I work in a completely different direction, more fundamental.
Now you have expressed yourself in more detail. Are these 130 bit keys? We can divide every 2 next keys, then 4, and so on, but the file will be very large and difficult to do manually. I talked about this earlier. We can also conduct them in parallel. I think we can close the topic on division at this point.
-1 /2 -1 /2 ??? Wow, this is a discovery))
I got it, forgot the subject... bro....
We can hit some whith this?
@winnery2022 can you be more specific with this? i don't know what are you asking.
I meant that different numbers from the main x2 division intersect relatively, this can be seen in my file. Is it possible to improve this process?
Invented the bicycle.) This is the natural and well-known behavior of numbers when dividing modulo))))
the thing is, it's impossible. You haven't proven anything, absolutely
I repeat once again, you don’t have to have a conversation with me. She's not talking about anything. This can be checked very easily. I gave you 2 even keys. Prove to everyone here your words on how to hack in 1 second. Don't change the subject. You won't get any more information.
Dude, you didn't get the point at all..
I just understood the point. Let's be more specific. The question here is not even or uneven numbers. You're probably unlikely to understand what's written at the top. You can keep even and odd numbers in parallel, but it will take a lot of time. This topic is clearly not worth discussing with you. I don't see the answer in 1 second with an even key. And if you meant that knowing each next subtracted key, then express your thoughts correctly (and once again, they can be done in parallel), but this will take a lot of time manually. If you manually proofread with such thoughts, you are unlikely to achieve success.
You didn't understand the question from sergy about parity. To paraphrase: we do not know and cannot verify the parity of the key if its private key is unknown. If we could check the honesty, we would divide the honest key by 2, check the parity again and divide it by 2 again, and so about 90 times and then we would find the key by a simple search. Try to determine the parity of these public keys even or not the private key from them:
- 02b30ef81221219e64d429f30e783a125e0988c8cf3ab9437a45b727618adb7385
- 03e227af2a43bf8eb8e50ca80e1ef920f06613a1450f07d1e3e6841c8c19557ba0
- 03bf68a3a688f57bdb8a0e2ce19b7cecd69354de3edad7a834c36bbed191c96ed6
And never think that you know something more than everyone else, for example, I never publish my promotions. but I work in a completely different direction, more fundamental.
Now you have expressed yourself in more detail. Are these 130 bit keys? We can divide every 2 next keys, then 4, and so on, but the file will be very large and difficult to do manually. I talked about this earlier. We can also conduct them in parallel. I think we can close the topic on division at this point.
-1 /2 -1 /2 ??? Wow, this is a discovery))
I got it, forgot the subject... bro....
Сool Bro maiby you Satoshi D
I meant that different numbers from the main x2 division intersect relatively, this can be seen in my file. Is it possible to improve this process?
No, that is just some properties of the curve, not big deal.
Please don't use the issues section for this kind of post
Hi alberto.
We can hit some whith this?
C:\Users\win>C:\Users\win\Desktop\ecctools\keymath.exe