Topic: Private key missing 29 characters - page 2. (Read 861 times)

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!!!

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

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.

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.

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.

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

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?

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