Private key missing 29 characters

[iconbtcx]

Hai guys,

            I have missing 29 characters of a wif private key with 488btc in it does anyone have idea about it how to reconstruct using permutation method and python i know a java script but it is good to find missing 8-10 characters but not for long characters, i know starting and ending but i dont know middle if any python script with gpu attribution added to it then its trying worth a shot.
Your views and suggestions please this is the address 1C8oHWB7htH139na4y8kG4w99MFrepseUv
Thank you

[1714114236]

1714114236

[1714114236]

1714114236

[1714114236]

1714114236

[1714114236]

1714114236

[pooya87]

A bitcoin compressed private key encoded as base58 has 52 characters, missing 29 of it means you are missing more than half of the private key which is 2¹²⁸ bits and is going to be impossible to brute force. Not only GPU but even with a specialized ASIC you won't be able to find this.

Could you tell us how you came by this private key?

[dansus021]

A bitcoin compressed private key encoded as base58 has 52 characters, missing 29 of it means you are missing more than half of the private key which is 2¹²⁸ bits and is going to be impossible to brute force. Not only GPU but even with a specialized ASIC you won't be able to find this.

Could you tell us how you came by this private key?

indeed, its like find one needle on earth. its impossible

is that really your money or you just find the first key and the end key

[NeuroticFish]

I'd add that if it's bought from somewhere/somebody, I wouldn't even be sure if the existing half of the key is actually good... (or from that address).

Indeed, OP, if it's your own key indeed, you need much more than that, you'll need searching deeper and for that I wish you good luck. If, on the other hand, you've acquired this from somewhere, just drop it, since you have, by far, not a chance.

[mabeyak]

I really believe this was bought from some scammer.
I remember years ago on some darknet markets some scammers were selling lists of "stolen" private keys which obviously were empty.
I guess they had to improve their tactics, and make it impossible to know for sure that it's a scam.

[DeathAngel]

Finding the remaining characters to this privkey is like finding ricking horse shit, it’s not going to happen. No currently existing tech can bruteforce a key with that many missing characters. Sorry OP but however you actually came across this privkey, it’s a waste of time.

[iconbtcx]

No i am not a scammer i am recently working on retrace back the position of private keys from a partial key
L39HAHFF1p6vxjBEe4YMxno
L39HAHFF1p6vxjBEe4YMxno
1Btud1pqADgGzgBCZzxzc2b1o1ytk1HYWC
i Used the bitaddress.org and slowly changed the hex position back and forth if the partial key for the above is true then the i exactly calculated what is the range the probable key
might be, to do with compurters it will take lot of years but with human logic i figured out that first 23 characters in key of course it took me five hours
i found the private key pattern starts with 1-9, then A-Z, a-z for every sequential position it moving forward
so i thought a python script with permutation is possible so we need 29 out of 61
so that its 61P29
P(n,r)=P(61,29)
=61!/(61−29)!
=1.929003152E+48
= 1929003152166198552081169486526708420444160000000
so its not needle on earth
and the address lies between the range below
B0BEA32711DBB7429BA37464FFBB4C8CDA2FFFFFFFFFFFFFFFFFFFFFFFFFFFFF -start
B0BEA32711DBB7429BA37464FFBB4C9090100000000000000000000000000000 -end

99450339347000738420840890196426183340851711984539219866681622085109234118673
B0BEA32FE0BA35D8FF0E8B6BE2DA5042C904C5340A234BECFFB07740F939B011 -upto 5
B0BEA327A0BA35D8FF0E8B6BE2DA5042C904C5340A234BECFFB07740F939B011- Upot7
B0BEA32714BA35D8FF0E8B6BE2DA5042C904C5340A234BECFFB07740F939B011-Upto 8
B0BEA327120035D8DF0E8B6BE2DA5042C904C5340A234BECFFB07740F939B011
79943888330950932036171869693561191471870036593116878810279630846882028826641
79943888330950032036171869693561191471870036593116878810279630846882028826641
B0BEA32711D905C6996CA072025FB5C99394207A551210D0BFB07740F939B011 --Upto 9
B0BEA32711DBB6C6996CA072025FB5C99394207A551210D0BFB07740F939B011--Upto 12
B0BEA32711DBB6FFF96CA072025FB5C99394207A551210D0BFB07740F939B011
B0BEA32711DBB73DFFFCA072025FB5C99394207A551210D0BFB07740F939B011--upto 13
B0BEA32711DBB7428BFCA072025FB5C99394207A551210D0BFB07740F939B011--upto 14
B0BEA32711DBB7429CBFA072025FB5C99394207A551210D0BFB07740F939B011
B0BEA32711DBB7429BA3F072025FB5C99394207A551210D0BFB07740F939B011--upto 15
B0BEA32711DBB7429BA37072025FB5C99394207A551210D0BFB07740F939B011 --upto 16
B0BEA32711DBB7429BA37472025FB5C99394207A551210D0BFB07740F939B011 --upto 17
B0BEA32711DBB7429BA37465095FB5C99394207A551210D0BFB07740F939B011 --upto 18
B0BEA32711DBB7429BA37465005FB5C99394207A551210D0BFB07740F939B011 --upto 19
B0BEA32711DBB7429BA37464FFBBB5C99394207A551210D0BFB07740F939B011--upto 20
B0BEA32711DBB7429BA37464FFBB3FC99394207A551210D0BFB07740F939B011 --upto 21
B0BEA32711DBB7429BA37464FFBB4BE99394207A551210D0BFB07740F939B011 --upto 22
B0BEA32711DBB7429BA37464FFBB4C900000207A551210D0BFB07740F939B011 --upto 23
B0BEA32711DBB7429BA37464FFBB4C8CE4C4D3D1FA8B4C56B6265500F939B011
B0BEA32711DBB7429BA37464FFBB4C8CDEFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
B0BEA32711DBB7429BA37464FFBB4C8CDA2FFFFFFFFFFFFFFFFFFFFFFFFFFFFF -start
B0BEA32711DBB7429BA37464FFBB4C9090100000000000000000000000000000 -end
79943888330935919077744049499279907419069159360589124460632026927327352979455 --start
79943888330935919077744049499279907420331759803085321425406530658711145283584 --end
No of Keys need to be checked :-1262600442496196964774503731383792304129
I Used

[odolvlobo]

No of Keys need to be checked :1262600442496196964774503731383792304129

If you could check 1x10²¹ keys per second (10 times the bitcoin network hash rate), it would only take 1.262x10¹⁸ seconds, or 40,000,000,000 years to check the entire range, which is about 3 times the age of the universe.

GO FOR IT!!!

[iconbtcx]

Why should i check entire range if i know upto 136bit please check start and end hex values mentioned in above post
B0BEA32711DBB7429BA37464FFBB4C8CDA2FFFFFFFFFFFFFFFFFFFFFFFFFFFFF -start
B0BEA32711DBB7429BA37464FFBB4C90901000000000000000000000000000 -end
8CDA2FFFFFFFFFFFFFFFFFFFFFFFFFFFFF
90901000000000000000000000000000
This is actual range
ignoring first part
No of Keys need to be checked :-1262600442496196964774503731383792304129 to be exact
this is the address i am looking for 1Btud1pqADgGzgBCZzxzc2b1o1ytk1HYWC

[kaggie]

No i am not a scammer i am recently working on retrace back the position of private keys from a partial key
L39HAHFF1p6vxjBEe4YMxno
L39HAHFF1p6vxjBEe4YMxno
1Btud1pqADgGzgBCZzxzc2b1o1ytk1HYWC
i Used the bitaddress.org and slowly changed the hex position back and forth if the partial key for the above is true then the i exactly calculated what is the range the probable key
might be, to do with compurters it will take lot of years but with human logic i figured out that first 23 characters in key of course it took me five hours
i found the private key pattern starts with 1-9, then A-Z, a-z for every sequential position it moving forward
so i thought a python script with permutation is possible so we need 29 out of 61
so that its 61P29
P(n,r)=P(61,29)
=61!/(61−29)!
=1.929003152E+48
= 1929003152166198552081169486526708420444160000000
so its not needle on earth
and the address lies between the range below
B0BEA32711DBB7429BA37464FFBB4C8CDA2FFFFFFFFFFFFFFFFFFFFFFFFFFFFF -start
B0BEA32711DBB7429BA37464FFBB4C9090100000000000000000000000000000 -end

99450339347000738420840890196426183340851711984539219866681622085109234118673

I have so many questions.

If you think it's recoverable from this, and worth $300 million, why would you post that much of the key online?
If it's yours, then do you know where you stored the rest of the key?

If it's not your keys, then how do you know that it's even part of real key? It could be literally anything or even nothing.

[BitMaxz]

Can you tell us the history of your private key and how the 29 characters have gone missing?

There might be another way to recover them than brute-forcing them. If you can tell us the exact history maybe we can find some way to recover them but if you just randomly generated the 23 character then it's impossible to find the exact private key of that wallet.

Are you sure this is yours I agree with kaggie post above why did you post the partial private key here?
It's risky I would suggest you to remove it if this is actually yours(Watch out for Big eyes out there).

[haidil]

No one can return it, even the support of the party you use will not be responsible because the private key is a keyword that is not the same as the others either based on the arrangement of numbers, sentences, and everything is left to the user, either in the form of Files etc. So when you lose all that, it means you can't find it again, if you forget to save it, if it's not stored properly then the business is done.

I've experienced it all.

[odolvlobo]

Why should i check entire range if i know upto 136bit please check start and end hex values mentioned in above post
B0BEA32711DBB7429BA37464FFBB4C8CDA2FFFFFFFFFFFFFFFFFFFFFFFFFFFFF -start
B0BEA32711DBB7429BA37464FFBB4C90901000000000000000000000000000 -end
8CDA2FFFFFFFFFFFFFFFFFFFFFFFFFFFFF
90901000000000000000000000000000
This is actual range
ignoring first part
No of Keys need to be checked :-1262600442496196964774503731383792304129 to be exact
this is the address i am looking for 1Btud1pqADgGzgBCZzxzc2b1o1ytk1HYWC

As you wrote, there are potentially 1262600442496196964774503731383792304129 keys to check. That is the number I used in the calculation to determine the time in my previous post.

However, there might be a way to theoretically optimize it down to to a much smaller number of keys (see Pollard's kangaroo ECDLP solver), which would give you an answer quickly (again assuming you could check 1x10²¹ keys per second), but it would require around a huge amount of storage (potentially more than exists on the planet).

Edit: Actually, I don't know what the optimized numbers would be, but that's not really important here.

[bigvito19]

Why should i check entire range if i know upto 136bit please check start and end hex values mentioned in above post
B0BEA32711DBB7429BA37464FFBB4C8CDA2FFFFFFFFFFFFFFFFFFFFFFFFFFFFF -start
B0BEA32711DBB7429BA37464FFBB4C90901000000000000000000000000000 -end
8CDA2FFFFFFFFFFFFFFFFFFFFFFFFFFFFF
90901000000000000000000000000000
This is actual range
ignoring first part
No of Keys need to be checked :-1262600442496196964774503731383792304129 to be exact
this is the address i am looking for 1Btud1pqADgGzgBCZzxzc2b1o1ytk1HYWC

As you wrote, there are potentially 1262600442496196964774503731383792304129 keys to check. That is the number I used in the calculation to determine the time in my previous post.

However, there might be a way to theoretically optimize it down to 35533089402642672055 keys (see Pollard's kangaroo ECDLP solver). That would give you an answer very quickly, but it would require around 1,137,058,861 TB of memory, which is about 4 times the world's total storage capacity.

Neither one has an outgoing transaction, you will have to have the public to use that and using that will still take a long time to search.

1C8oHWB7htH139na4y8kG4w99MFrepseUv
1Btud1pqADgGzgBCZzxzc2b1o1ytk1HYWC

[TelolettOm]

Missing your private key characters means that you cannot access it. Moreover, if it is those many amounts of the characters. It is only about 2 characters, you may check them randomly one by one although you will need many hours maybe.
But 29? I think that it is very difficult to get your private key back or can access your wallet.

[pooya87]

Actually the smallest and biggest values for the half key posted above (which only has 23 characters by the way) are wrong, the correct values are the following:
3803496f04b8c0f3e8335eb24b07a4a15335d37bff03bafbb19640a287e0609a
3803496f04b8c0f3e8335eb24b07a4a16378ffd2a031ba2c431c7ca071115174
And the difference between the two values is 1.37E+51 which makes sense since 55% of the key is missing.

[iconbtcx]

nope brother
the private key starts with L3
the key ranges you posted private keys start with K
3803496f04b8c0f3e8335eb24b07a4a15335d37bff03bafbb19640a287e0609a
Ky6bKVgbJPFRiAPxvBo4ftiAMkKCBjAbWfVRWEevsHxgULTthBFB
i can able to recover last or even mid where missing seven to eight or upto 10 in matter of minutes thats not an issue
its more than 10 that whats the issue is i am expermienting on it anyway thanks for your suggestions brother

[Ebede]

A bitcoin compressed private key encoded as base58 has 52 characters, missing 29 of it means you are missing more than half of the private key which is 2¹²⁸ bits and is going to be impossible to brute force. Not only GPU but even with a specialized ASIC you won't be able to find this.

Could you tell us how you came by this private key?

This what i have been trying to know the root about bitcoin private key, bitcoin private key is not phrase code that when it's lost or misplaced the person dont have access to the wallet, is any way we can retrieve the key online if been misplaced, because i dont understand the word  GPU and ASIC, please i need more education for these.

[king_of_1]

Hai guys,

            I have missing 29 characters of a wif private key with 488btc in it does anyone have idea about it how to reconstruct using permutation method and python i know a java script but it is good to find missing 8-10 characters but not for long characters, i know starting and ending but i dont know middle if any python script with gpu attribution added to it then its trying worth a shot.
Your views and suggestions please this is the address 1C8oHWB7htH139na4y8kG4w99MFrepseUv
Thank you

let me tell you something broh..
if you are sure it is yours..then go for it. it deserves to be searched.. i know it is impossible, but you don't have any other choice.
BUT, if you got it from DARK WEB, for get it, it is waste of time.

[pooya87]

nope brother
the private key starts with L3
the key ranges you posted private keys start with K
3803496f04b8c0f3e8335eb24b07a4a15335d37bff03bafbb19640a287e0609a
Ky6bKVgbJPFRiAPxvBo4ftiAMkKCBjAbWfVRWEevsHxgULTthBFB

Sorry, I made two mistakes.
First is adding 1 less character (total 51 instead of 52). The correct values are:
Minimum

Code:

L39HAHFF1p6vxjBEe4YMxno11111111111111111111111111111
b0bea32711dbb7429ba37464ffbb4c8cda31ea17c6d85d063c0aa4d2c8d5e30f

Maximum

Code:

L39HAHFF1p6vxjBEe4YMxnozzzzzzzzzzzzzzzzzzzzzzzzzzzzz
b0bea32711dbb7429ba37464ffbb4c908969f5b84b442e0734743c599dec7459

Diff:
1253753004473247624297767682974128968010 = 1.25E+39 (which is close to your initial value).

Second is that the number I posted is all permutations of the base58 characters if they are placed in their missing positions (ie. 58²⁹) instead of using the method above since all missing characters are from one place at the end without anything in between.

i can able to recover last or even mid where missing seven to eight or upto 10 in matter of minutes thats not an issue
its more than 10 that whats the issue is i am expermienting on it anyway thanks for your suggestions brother

Up to 12 characters from the end can be recovered very easily since you'd essentially be checking less than a million keys. The problem is that the numbers grow very fast as the number of missing chars increase.