Bitcoin puzzle transaction ~32 BTC prize to who solves it

[digaran]

Do people really think Satoshi is reading all the posts here every day?
I don't know if anyone here is a giver, unfortunately we are here to take the loots.

Welcome to the jungle, and happy hunting.😉

Wow guys! I just discovered something, everytime I double a key I can get double the distance with my original base key!

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?  A good thing about mod function in EC, you could keep halving for eternity without reaching destination, as we are moving towards bitcoin halving, we should start key halving to solve some puzzles before the halving event! 😉

[1715548098]

1715548098

[1715548098]

1715548098

[elvis13]

Digaran @
You can't offer anything useful. An endless stream of insanity.!

[vjudeu]

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?

You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation.

Code:

SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7
           mask125=0000000000000000000000000000000020000000000000000000000000000000
(hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7
             key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814
             key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6
         (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773

See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits.

[batareyka]

digaran
If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers
I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve.
There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve.
There are special numbers on the curve that can be used to get exactly 1 digit less.
A third of the number you are looking for.
A quarter of the sought number.
You can get the mirror part of the number you are looking for, which can also be compared with the original.
And this is not even a 10th part of what is there.
You can get a lot.



But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.

I wish you success.

[digaran]

Digaran @
You can't offer anything useful. An endless stream of insanity.!

Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?

You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation.

Code:

SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7
           mask125=0000000000000000000000000000000020000000000000000000000000000000
(hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7
             key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814
             key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6
         (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773

See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits.

I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉

digaran
If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers
I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve.
There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve.
There are special numbers on the curve that can be used to get exactly 1 digit less.
A third of the number you are looking for.
A quarter of the sought number.
You can get the mirror part of the number you are looking for, which can also be compared with the original.
And this is not even a 10th part of what is there.
You can get a lot.



But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.

I wish you success.

Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle.

I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods.

Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣

[lordfrs]

Digaran @
You can't offer anything useful. An endless stream of insanity.!

Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?

You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation.

Code:

SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7
           mask125=0000000000000000000000000000000020000000000000000000000000000000
(hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7
             key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814
             key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6
         (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773

See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits.

I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉

digaran
If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers
I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve.
There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve.
There are special numbers on the curve that can be used to get exactly 1 digit less.
A third of the number you are looking for.
A quarter of the sought number.
You can get the mirror part of the number you are looking for, which can also be compared with the original.
And this is not even a 10th part of what is there.
You can get a lot.



But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.

I wish you success.

Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle.

I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods.

Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣

My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm .

I don't like working, my brain is almost numb from the blood pressure pills
i can't think

I'm sorry my English is bad, I'm writing from translate, but I hope you understand what I mean.

[digaran]

My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm .

If your algorithm is working for one key, it should work for all other keys, unless it's not working, if you can solve a known key and not any other key, you are doing things in a wrong way. Manage to find a key which you don't know it's private key, then you can claim success.

Hope you feel better.

[Gianluca95]

I think that there isn't any way to find a solution for this puzzle; the only thing to solve this is the strength of community. It requires many and many power of hash/sec to be solved and with maths you can't do many things.

It would be cool to use something with a speed of an ASIC, in this way you'll be able to discover keys from 66 to 76 in very easy way !

[Denis_Hitov]

Digaran @
You can't offer anything useful. An endless stream of insanity.!

Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?

You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation.

Code:

SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7
           mask125=0000000000000000000000000000000020000000000000000000000000000000
(hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7
             key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814
             key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6
         (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773

See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits.

I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉

digaran
If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers
I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve.
There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve.
There are special numbers on the curve that can be used to get exactly 1 digit less.
A third of the number you are looking for.
A quarter of the sought number.
You can get the mirror part of the number you are looking for, which can also be compared with the original.
And this is not even a 10th part of what is there.
You can get a lot.



But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.

I wish you success.

Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle.

I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods.

Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣

My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm .

I don't like working, my brain is almost numb from the blood pressure pills
i can't think

I'm sorry my English is bad, I'm writing from translate, but I hope you understand what I mean.

Hello.
Could you share your algorithm?
I wonder how you managed this: "Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds".

Sincerely,
be healthy.

[GR Sasa]

fake

[albert0bsd]

Hello.
Could you share your algorithm?
I wonder how you managed this: "Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds".

Sincerely,
be healthy.

Don't believe in that, it is fake.

fake

Agree.

Why from 115 to 28 bits? in that case why not up to 1 bit?

Guys please don't believe in that, in any case we only need to know one bit on the right side to break ECDSA

[lordfrs]

Hello.
Could you share your algorithm?
I wonder how you managed this: "Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds".

Sincerely,
be healthy.

Don't believe in that, it is fake.

fake

Agree.

Why from 115 to 28 bits? in that case why not up to 1 bit?

Guys please don't believe in that, in any case we only need to know one bit on the right side to break ECDSA

Private Key                                                                    Public key

1090246098153987172547740458951748   # puzzle 110   0309976ba5570966bf889196b7fdf5a0f9a1e9ab340556ec29f8bb60599616167d
545123049076993586273870229475874
272561524538496793136935114737937
136280762269248396568467557368968
68140381134624198284233778684484
34070190567312099142116889342242
17035095283656049571058444671121
8517547641828024785529222335560
4258773820914012392764611167780
2129386910457006196382305583890
1064693455228503098191152791945
532346727614251549095576395972
266173363807125774547788197986
133086681903562887273894098993
66543340951781443636947049496
33271670475890721818473524748
16635835237945360909236762374
8317917618972680454618381187
4158958809486340227309190593
2079479404743170113654595296
1039739702371585056827297648
519869851185792528413648824
259934925592896264206824412
129967462796448132103412206
64983731398224066051706103
32491865699112033025853051
16245932849556016512926525
8122966424778008256463262
4061483212389004128231631
2030741606194502064115815
1015370803097251032057907
507685401548625516028953
253842700774312758014476
126921350387156379007238
63460675193578189503619
31730337596789094751809
15865168798394547375904
7932584399197273687952
3966292199598636843976
1983146099799318421988
991573049899659210994
495786524949829605497
247893262474914802748
123946631237457401374
61973315618728700687
30986657809364350343
15493328904682175171
7746664452341087585
3873332226170543792
1936666113085271896
968333056542635948
484166528271317974
242083264135658987
121041632067829493
60520816033914746
30260408016957373
15130204008478686
7565102004239343
3782551002119671
1891275501059835
945637750529917
472818875264958
236409437632479
118204718816239
59102359408119
29551179704059
14775589852029
7387794926014
3693897463007
1846948731503
923474365751
461737182875
230868591437
115434295718
57717147859
28858573929
14429286964
7214643482
3607321741
1803660870
901830435
450915217
225457608
112728804
56364402
28182201
14091100
7045550
3522775
1761387
880693
440346
220173
110086
55043
27521
13760
6880
3440
1720
860
430
215
107
53
26
13
6
3
1  0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

[albert0bsd]

Private Key                                                                    Public key

1090246098153987172547740458951748   # puzzle 110   0309976ba5570966bf889196b7fdf5a0f9a1e9ab340556ec29f8bb60599616167d
...
1  0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

That is easy if you have the privatekey (X = X % 2 ? X/2 : (X-1)/2), but we can't do that if we don't have the privatekey, because there is no way that know if the publickey belongs to an even or odd privatekey.

But proof your self and show us.. here is a Publickey in the 110 bit range:

0301040c4e33b6ab297ba2a9c2858d15be5f6844febfc003ef7646796608f7f819

[GR Sasa]

1090246098153987172547740458951748   # puzzle 110   0309976ba5570966bf889196b7fdf5a0f9a1e9ab340556ec29f8bb60599616167d
1  0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

What is this mess? This doesn't prove anything XD

[albert0bsd]

What is this mess? This doesn't prove anything XD

Maybe he is trying to reduce the privatekey based only on the current information of the privatekey without do any calculation with the publickey.

[VinIVaderr]

I was trying to figure out a good way to break the key search into ranges.
I decided to use columns, ie. 17 columns. And group them into 4.

group 1  2000 - 2fff
group 2  0000 - ffff
group 3  0000 - ffff
group 4  0000 - ffff
group 5  0123456789abcdef

output:    group 1 has 4096 lines
      group 2 has 65536 lines
      group 3 has 65536 lines
      group 4 has 65536 lines
      group 5 is 16 lines.
         
My idea is to create 2000 0000 0000 0000 0
But I'm still not sure how to bring them together in a string.

[Evillo]

I was trying to figure out a good way to break the key search into ranges.
I decided to use columns, ie. 17 columns. And group them into 4.

group 1  2000 - 2fff
group 2  0000 - ffff
group 3  0000 - ffff
group 4  0000 - ffff
group 5  0123456789abcdef

output:    group 1 has 4096 lines
      group 2 has 65536 lines
      group 3 has 65536 lines
      group 4 has 65536 lines
      group 5 is 16 lines.
         
My idea is to create 2000 0000 0000 0000 0
But I'm still not sure how to bring them together in a string.

On linux or WSL type: paste file1.txt file2.txt > fullrange.txt

Obviously change file names in your command according to your actual file names

[digaran]

Guys I need some help over here! So here is what I have so far.

I have subtracted this from #125.
0000000000000000000000000000000010******************************
The result is :
02c473281c8531d524c1a2b92b9192ffe3e8124448e47ec306f146f2f4f35269ee

Now if I add that result to this one :
0387a3951644f3bf9ee08135bcb9a00a31ed7b09e26ff7850b83713eb6f0582920

I get this key :
000000000000000000000000000000000c******************************

Ok, now I have the distance between the green key above and #125, which is :
000000000000000000000000000000000 4******************************

So why can't I figure the correct key while I have my double triangle, 3 known keys plus 3 unknown keys?

 

[NotATether]

Guys I need some help over here! So here is what I have so far.

I have subtracted this from #125.
0000000000000000000000000000000010******************************
The result is :
02c473281c8531d524c1a2b92b9192ffe3e8124448e47ec306f146f2f4f35269ee

Now if I add that result to this one :
0387a3951644f3bf9ee08135bcb9a00a31ed7b09e26ff7850b83713eb6f0582920

I get this key :
000000000000000000000000000000000c******************************

Ok, now I have the distance between the green key above and #125, which is :
000000000000000000000000000000000 4******************************

So why can't I figure the correct key while I have my double triangle, 3 known keys plus 3 unknown keys?

Are the result and green keys public keys or private keys? Because without context, they could be either one.

[kalos15btc]

Guys I need some help over here! So here is what I have so far.

I have subtracted this from #125.
0000000000000000000000000000000010******************************
The result is :
02c473281c8531d524c1a2b92b9192ffe3e8124448e47ec306f146f2f4f35269ee

Now if I add that result to this one :
0387a3951644f3bf9ee08135bcb9a00a31ed7b09e26ff7850b83713eb6f0582920

I get this key :
000000000000000000000000000000000c******************************

Ok, now I have the distance between the green key above and #125, which is :
000000000000000000000000000000000 4******************************

So why can't I figure the correct key while I have my double triangle, 3 known keys plus 3 unknown keys?

 

i told you multiple times, to test that on any other puzzle 100 or 115 you will figure out how it works,, get the private keys of substracted puzzle 110 and you will understand bro, then test it on 125,, when you substract told you to do this  range 0:1fffffffffffffffffffffffffffffff
you will have 125 substracted maybe to 100 or even 80 bit, but you need to know where exactly you can do that -f address