Bitcoin puzzle transaction ~32 BTC prize to who solves it

[kalos15btc]

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-KeySubtractor
and 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..

[digaran]

Hello everyone What happens when dividing a scalar from a point (vector/scalar) on the secp256k1 calculator from Mr.maxwell. How can I find out what result it gives when using known points for an example, can you show an example?!.I still don't understand how division works here, I can't find the right result in any way.

If you know the private key of your point, just use scalar/scalar instead, result is the same.
Note the last digit of your private key in decimal, for example if you are dividing a key ending with 8 by 3, you need to separately divide 8 by 3 and subtract the result from your first result which is your key ending with 8 divided by 3, then you will have the correct result.

Also note that when you are subtracting, the greater key must be on the left slot, otherwise you'll get -n result starting with a lot of fs.  When you see that though you could either change the keys slot or just subtract the result from n to have your result.

Here is a trick, if you are dividing a key by 2 digits number, then the previous method won't work, e.g, dividing a key ending with 8 by 25 won't work if you divide "8" by 25 and subtract both results, instead you'd need to know the last 2 digits of your key, example : if your key ends in 58, then you'd need to divide "58" by 25 separately then do the previous steps. ( might not work with all numbers ).


Here is another trick, if you are dividing by 25, try dividing n by 7 and compare both results.

Most of what I said won't make sense now, but in the future you'd come back to check this post again.😉

Ps, if you want advanced tricks, ask the master of cryptography aka kalos. I hear he is about to break rmd160 hash function by finding a collision.

[tptkimikaze]

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-KeySubtractor
and 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..

Actually I saw your previous post quite some time ago and did try on the python script on that closed thread many months ago. After generating millions of public key, I found out it doesn't help much after trying an hour or so, immediately I gave up.  LoL. Maybe I know too little information about it. I think I have some idea now, not sure if I am correct about it. By subtracting known 130 public key to lower bits to find the lower bits private key, once gotten that private key you do an inverse of lower bit private key back to 130 bits. Am I right?

[albert0bsd]

Maybe I know too little information about it. I think I have some idea now, not sure if I am correct about it. By subtracting known 130 public key to lower bits to find the lower bits private key, once gotten that private key you do an inverse of lower bit private key back to 130 bits. Am I right?

Yes that is correct if you have  some case like this:

Code:

P130 - PA = X
P130 - PB = Y
P130 - PC = Z
...
P130 - PM = N

if you found the privatekey of any X, Y, Z or N you only need to apply the inverse process to get the Private key of P130, you only need to be aware of what are the values of PA,PB PC and PM

the same appy for other operations like  addition, multiplication and division

Regards!

[tptkimikaze]

Maybe I know too little information about it. I think I have some idea now, not sure if I am correct about it. By subtracting known 130 public key to lower bits to find the lower bits private key, once gotten that private key you do an inverse of lower bit private key back to 130 bits. Am I right?

Yes that is correct if you have  some case like this:

Code:

P130 - PA = X
P130 - PB = Y
P130 - PC = Z
...
P130 - PM = N

if you found the privatekey of any X, Y, Z or N you only need to apply the inverse process to get the Private key of P130, you only need to be aware of what are the values of PA,PB PC and PM

the same appy for other operations like  addition, multiplication and division

Regards!

Thanks. Maybe will try it later. More focusing on 66 now. If my idea and approach works and able to solve 66, then 67 and 68 should be also solvable in short period of time.

[kalos15btc]

@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  

[digaran]

@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.

[kalos15btc]

@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]

Here is the solution for partnering with muscles ( those with hardware to scan ranges ) , before scanning, the one asking for the scan should give all his information to an escrow and then the one doing the scan should provide the key he has found to the escrow so that the escrow does the final key extraction and divide the prize according to contract.

Person who asks for the scan provides the key which was subtracted from puzzle key and also the offset which the muscle is going to search for, escrow can verify the authenticity of the keys by doing the subtraction himself. 😉

[lordfrs]

@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

[citb0in]

@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? 

[kalos15btc]

@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 §.?

[mcdouglasx]

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.

because if you look for pk=1361129467683753853853498429727072845823
subtracting...

Code:

import secp256k1 as ice


target_public_key = "03e067911ebf6bacf87a8088ab9344c95843aed80b070eed09f9d947c98dfc0249"
target = ice.pub2upub(target_public_key)
num = 136114 # number of times.
sustract= 10000000000000000000000000000000000 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub)
for t in range (num+1):
    h= (res[t*65:t*65+65]).hex()
    hc= ice.to_cpub(h)
    data = open("data-base.txt","a")
    data.write(str(hc)+"\n")
    data.close()

one of the pub in the result will be:
pk=9467683753853853498429727072845823

and looking for this         1361129467683753853853498429727072845823
is not the same as this               9467683753853853498429727072845823

[digaran]

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.😉

[kalos15btc]

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]

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

Can you show one example where you could find a key based on rmd160 similarities? I'd like to learn this non existing equation where we can estimate EC points range by finding rmd160 similar starting characters.


 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.
 

[albert0bsd]

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

[kalos15btc]

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 ??

[citb0in]

can you try to be more specific, please? At best use a translator that will translate from your native language to english. I'm pretty sure this will be helpful for everyone. Oh, one more thing : try to give an -at least- one expressive example.

[mcdouglasx]

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 ??

ECC and rmd160 are different, you won't find any similarity there.

the matches you find are just that, random matches.