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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 187201 times)
Feron
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January 11, 2023, 02:02:18 PM
Last edit: January 11, 2023, 05:08:50 PM by Feron
 #2001



It is slow compared to ICE secp256k1. Twice the speed...However, I am currently creating a script that will be Full ICE - to search for lost parts of the private key. About 2 million per second per CPU thread.. Grin




I would love to see your 2 mil/sec as I am stuck on 500k/s with ice. Would be interested to see which route you went.

Edit: 2 mil+... CPU dependent. But that is on a public key hunt. So for those who want to try their luck.... It's set for 120 bit key...I guess if you link enough machines with enough CPU's you may hit in a few hundred years.

Code:
import secp256k1 as ice
from timeit import default_timer as timer
import random
import multiprocessing

cores = 1
perRound = 1000000
bitRange = 120
target_Key = '02ceb6cbbcdbdf5ef7150682150f4ce2c6f4807b349827dcdbdd1f2efa885a2630'
    
def new_Number():
    doneSet = set()
    rotate = True
    with open("120-Nums.txt", "r+") as textfile:  
        for checking in textfile: doneSet.add(str(checking.replace("\n","").replace(",", "")))
        while rotate != False:
            number = round(random.randint(2**(bitRange-1),2**bitRange)/perRound)
            if str(number) not in doneSet:
                rotate = False
                textfile.write(str(number) + "\n")
            if str(number) in doneSet:
                print('Collision')
    textfile.close()
    return number

def seek(r,cores):
    startStart = timer()
    dog = False
    while dog != True:
        number = new_Number()
        p = (ice.scalar_multiplication(number*perRound))
        start = timer()
        check = str((ice.point_sequential_increment(perRound, p)).hex())
        if target_Key[5:25] in check:
            print('Success: ' + str(number))
            print('Test Run Time Length: ' + str((timer()-startStart))[:6])
            with open("results.txt", "a") as completed:
                completed.write(str(p) + '   ' + str(number))
            completed.close()
            dog = True
        else:
            print('Core: ' + str(r) + '   Key/s: ' + str(perRound/(timer()-start))[:10]  + '   Time: ' + str(timer()-start)[:6] + '   RandInt: ' + str(number))

if __name__ == '__main__':
    jobs = []
    for r in range(cores):
        p = multiprocessing.Process(target=seek, args=(r,cores))
        jobs.append(p)
        p.start()
I tried your code with fastrand but the speed is similar
Code:
from timeit import default_timer as timer
import secp256k1 as ice
import fastrand
#import random
perRound = 1000000
target_Key = '031a864bae3922f351f1b57cfdd827c25b7e093cb9c88a72c1cd893d9f90f44ece'
#donset = set()
#number = round(fastrand.pcg32bounded(134217727)/perRound)
dog = False
while dog != True:
 number = round(fastrand.pcg32bounded(134217727)/perRound)
 #if str(dog) in str(dog):
  #ke = ice.privatekey_loop_h160(0,0,True,number).hex()
 public = (ice.scalar_multiplication(number*perRound))
 checkk = (ice.point_sequential_increment(perRound,public)).hex()
 start = timer()
 if target_Key[2:66] in checkk:
 #if "0c7aaf6caa7e5424b63d317f0f8f1f9fa40d5560" in (ke):
  f=open("result.txt","a")
  f.write(str(target_Key[2:66])+"-start number for key "+str(number)+"\n")
  #f.write(str(ke)+"-"+str(number)+"\n")
  f.close()
  #dog = True
 else:
  print('Key/s: ' + str(perRound/(timer()-start))[:7]  + '   Time: ' + str(timer()-start)[:3] + '   start number for key: ' + str(number))

27 puzzle address is collision test
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January 11, 2023, 03:36:14 PM
 #2002

https://github.com/Jamerson1000/btc-puzzle  Wink
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January 11, 2023, 08:31:52 PM
 #2003

120 scan delete print for run 10% speed increase
Code:
from timeit import default_timer as timer
import secp256k1 as ice
import random
while True:
 stackk = 1000000
 target = "02ceb6cbbcdbdf5ef7150682150f4ce2c6f4807b349827dcdbdd1f2efa885a2630"
 number = round(random.getrandbits(120)/stackk)
 public = (ice.scalar_multiplication(number*stackk))
 startt = timer()
 checkk = (ice.point_sequential_increment(stackk,public)).hex()
 if target[2:66] in checkk:
  f=open("results.txt","a")
  f.write(str(target[2:66])+"-start number for key "+str(number)+"\n")
  f.close()
 print('Key/s: ' + str(stackk/(timer()-startt))[:7]  + ' Time:' + str(timer()-startt)[:3]+' start number for key:'+str(number))
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January 13, 2023, 03:20:03 PM
 #2004


Can i ask what did you mean by "bitcrack doesn't work when workspace is larger than 999.999" ?

Cool Story Bro.
BTC: 1EviLLo1Y5VeNn2Lajv9tdZTkUuVgePVYN
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January 16, 2023, 08:50:52 PM
 #2005

@Andzhig And if we increase one more character of address 16jY7qLJn & 'x' then most binaries are started from '111'

few examples -

16jY7qLJnxLQQRYPX5BLuCtcBs6tvXz8BE   1110000000100110101001101101010100100011010011001000100000110110000   7013536A91A6441B0
16jY7qLJnX9uchnyf26t3QJnsUf78Xdikb   1110010000101000111010000001111110010000001011001101111011100000   E428E81F902CDEE0
16jY7qLJnX9eX8j612s8fnbn6uzR48xjua   1110100000001101111010110011001110101001011001111010000010001111   E80DEB33A967A08F
16jY7qLJnx2EZZumnYFke3GutCrRnHKs1M   111010110100110101001101101010111010101000110011101011001010110000   3AD3536AEA8CEB2B0
16jY7qLJnx2ixrxCnTLSraerkgyB3YYAiT   1110110111111001110011010110000000110101011011011100110000011001   EDF9CD60356DCC19
16jY7qLJnxHBp3dqwV2kzYq1LucfZzgxsH   1110111010111001101010110011001101001101111100100111011100001101   EEB9AB334DF2770D
16jY7qLJnX2cZXJ78wV1ef42e7cLAZJ1Vn   1111111000101000011001011100011011011011111111101100001110000011   FE2865C6DBFEC383
16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN   1111011100000101000111110010011110110000100100010001001011010100 F7051F27B09112D4


Could this also be some logic?

can you shed more light on this issue

1BGvwggxfCaHGykKrVXX7fk8GYaLQpeixA
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January 17, 2023, 05:01:30 PM
 #2006


I was checking exatly the same the other day, hi everyone im new into this puzzle, and im obsessed with it. Spend hours and hours looking for a pattern, and there is no pattern as the "owner" mentioned it but i was looking for the already found keys and i think i found something.

If someone is brute forcing this, you need to start from eb851eb851eb8000 to ffffffffffffffff, thats why this is not still be found, the hex is almost at the end of the range. Im bruteforcing this from feb851eb851eb800 to ffffffffffffffff at 25 MKey/s (yes is a little slow but is honest work Tongue ) and i will be checking from the back after finish a range.

I hope this helps anyone, and if i do please share something as i will share to 2 users of this forum (if i find the key) who gives me the idea Smiley


Regards from AR and sorry about my english.

based on what you say to start from that point of the range ?

so i was right, and i think where to start from for puzzle 66, i think i found a pattern. Ive also got a 1080 at a good price so im searching at 240 MKeys/s
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January 17, 2023, 05:53:57 PM
 #2007



I was checking exatly the same the other day, hi everyone im new into this puzzle, and im obsessed with it. Spend hours and hours looking for a pattern, and there is no pattern as the "owner" mentioned it but i was looking for the already found keys and i think i found something.

If someone is brute forcing this, you need to start from eb851eb851eb8000 to ffffffffffffffff, thats why this is not still be found, the hex is almost at the end of the range. Im bruteforcing this from feb851eb851eb800 to ffffffffffffffff at 25 MKey/s (yes is a little slow but is honest work Tongue ) and i will be checking from the back after finish a range.

I hope this helps anyone, and if i do please share something as i will share to 2 users of this forum (if i find the key) who gives me the idea Smiley


Regards from AR and sorry about my english.

based on what you say to start from that point of the range ?

so i was right, and i think where to start from for puzzle 66, i think i found a pattern. Ive also got a 1080 at a good price so im searching at 240 MKeys/s
Lol...okay, so where to start for puzzle 66? 20000000000000000?
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January 17, 2023, 06:08:16 PM
 #2008

it is completely pointless to "guess" or "estimate" a range since the author of the puzzle has already made a clear statement about it --> there IS NO PATTERN  Cool
However good luck to everybody!

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WanderingPhilospher
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January 17, 2023, 07:06:45 PM
 #2009

Quote
well you saying this stupid thing i know you didnt read my message at all, puzzle 64 start from 8000000000000000 to ffffffffffffffff i say start from eb851eb851eb8000 so yea, i save a lot of time from 8000000000000000 to Eb and the hex started at f7, so go F yourself we are not here making funny or stupid commenting things, so F you and i keep this pattern to myself, Good luck Smiley
Lol...I merely said it because you had a 1 in 8 chance of guessing correctly and you claim you have a good start for #66 but do not share, why? Maybe because you will be wrong? I have trillions of keys checked in each sub 48 bit range; so if you know the first 1 to 5 starting characters for #66, tell us, and I will tell you if it's been checked already.
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January 17, 2023, 07:13:25 PM
 #2010

yes, there is, if you really think automated random created things are really random, then think twice.

Ok, I will ... hold on a second ... finished.

Now, please tell me what the weather in South Florida will be on 2024/Feb/29 at 02:19 PM.

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WanderingPhilospher
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January 17, 2023, 08:14:42 PM
 #2011

yes, there is, if you really think automated random created things are really random, then think twice.

Ok, I will ... hold on a second ... finished.

Now, please tell me what the weather in South Florida will be on 2024/Feb/29 at 02:19 PM.
Warmer than what it will be where I am lol...
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January 18, 2023, 03:23:35 PM
 #2012

Hi everybody,
maybe someone already noticed.. not sure
PUZZLE 66.....
Looking at address "https://privatekeys.pw/address/bitcoin/13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so" I just realized that there is a DOGE coin address  with A BALANCE that has the SAME private key !!!
Does this means that the owner of DOGE address could theoretically take the Bitcoin in puzzle#66 ??
.. in addition...
I am really suprised of the probability that this can happen.... same key that have a Bitcoin and DOGE addresses both with balance ? it could be the puzzle owner DOGE account ? or maybe that key is derived from some simple schema ??
what do you think ?
I tried to search in google for that DOGE address but no relevant info found.
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January 18, 2023, 04:53:53 PM
Merited by nc50lc (1)
 #2013

The hash rmd160 is the same for almost all chains, yes you can derive the doge address from it the only change is the prefix byte.

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January 19, 2023, 07:47:35 AM
 #2014

-snip-
I am really suprised of the probability that this can happen.... same key that have a Bitcoin and DOGE addresses both with balance ? it could be the puzzle owner DOGE account ? or maybe that key is derived from some simple schema ??
what do you think ?
If you see any funded address, that doesn't mean that someone knows its private key.
That Dogecoin address is one example, someone just derived the dogecoin address from p66's address then sent dogecoins to it.

In your link, see that "Public Key Hash (Hash 160)" below the address?
That's basically the address when decoded with base58 without the network bytes and checksum.
Simply add any altcoin's network bytes, etc. and encode it with base58check to get the equivalent altcoin address.

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January 19, 2023, 08:36:02 AM
Last edit: January 19, 2023, 01:48:32 PM by citb0in
Merited by nc50lc (1), saxydev (1)
 #2015

Here's an example for python. The program takes the hash160 value as the first argument of the command line call. It then uses this value to generate both the Bitcoin and Dogecoin addresses by using the same value with different version bytes and Base58Check encoding rules.

Code:
#/usr/bin/env python3
# ./hash160_to_address.py <hash160>
# by citb0in, 2023/Jan/19
import sys
import hashlib
import base58

# hash160
h = sys.argv[1]

# version byte (for Bitcoin "00", for Dogecoin it's "1E")
version_btc = b"\x00"
version_doge = b"\x1E"

# hash160 + version byte
h_btc = version_btc + bytes.fromhex(h)
h_doge = version_doge + bytes.fromhex(h)

# double-sha256
r_btc = hashlib.sha256(h_btc).digest()
r_doge = hashlib.sha256(h_doge).digest()
r_btc = hashlib.sha256(r_btc).digest()
r_doge = hashlib.sha256(r_doge).digest()

# first 4 bytes of sha256(sha256)
checksum_btc = r_btc[:4]
checksum_doge = r_doge[:4]

# hash160 + version byte + checksum
address_btc = h_btc + checksum_btc
address_doge = h_doge + checksum_doge

# base58 conversion to get the address
address_btc = base58.b58encode(address_btc)
address_doge = base58.b58encode(address_doge)

print(f"\n\nHash160 = {h}")
print(f"Bitcoin (BTC) address = {address_btc.decode()}")
print(f"Dogecoin (DOGE) address = {address_doge.decode()}")

Code:
$ python3 hash160_to_address.py 20d45a6a762535700ce9e0b216e31994335db8a5 
Quote
Hash160 = 20d45a6a762535700ce9e0b216e31994335db8a5
Bitcoin (BTC) address = 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so
Dogecoin (DOGE) address = D88gYxMEoumtZSJAC9mwa2EWfm6g3uocon

Now for additional example take the hash160 of puzzle #67 which is 739437bb3dd6d1983e66629c5f08c70e52769371 and calculate the DOGE address:
Code:
$ python3 hash160_to_address.py 739437bb3dd6d1983e66629c5f08c70e52769371
Quote
Hash160 = 739437bb3dd6d1983e66629c5f08c70e52769371
Bitcoin (BTC) address = 1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9
Dogecoin (DOGE) address = DFgDofYSD4T6CwdVykSc2CLJ9s8A6Bqt97

If you or someone else send some coins to DFgDofYSD4T6CwdVykSc2CLJ9s8A6Bqt97 you will see the same effect.

EDIT:
You can find an extended version of this tool on my github repository. It will also show the native segwit (bech32) address of the hash. I have added also Litecoin as an additional example because I know that Litecoin supports bech32, too. I don't give a **** about shitcoins, it should just serve as programming example to see how this works.

Code:
$ python3 ./hash160_to_coin_addresses.py 20d45a6a762535700ce9e0b216e31994335db8a5
Quote
Hash160 = 20d45a6a762535700ce9e0b216e31994335db8a5
Bitcoin (BTC) address = 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so
Bitcoin (BTC) bech32 address = bc1qyr2956nky56hqr8fuzepdccejse4mw994lyftn
Dogecoin (DOGE) address = D88gYxMEoumtZSJAC9mwa2EWfm6g3uocon
Litecoin (LTC) address = LNDYGuiRbA7fHEoidhmgJH8fzqjeseATsU
Litecoin (LTC) bech32 address = ltc1qyr2956nky56hqr8fuzepdccejse4mw993r7dnr

.
.HUGE.
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jl33213
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January 20, 2023, 09:44:47 PM
 #2016

For half a year I use my cpu to scan for the #67 puzzle. I am 0.000039% scanned addresses of the range ;( A lifetime more to wait. I wonder how many supercomputers out there scan the bitcoin chain for these keys every minute. If you remember the pipeline hack, the FBI had already the privkey for the address used and it was not obtain during a hack or seize.
But Why the 67th puzzle ?.. why not the easier one? I mean the amount of private keys in 66 is unbelievably enormous already .. why go for the double of that gigantic range?
because someone is more likely to find 66 before 67 he doesnt wanna spend a shit ton of time on 66 and someone find it before him so he does 67 by time someone does find 66 he will be at head start( to most people anyway ) thats what i think at least
Evillo
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January 21, 2023, 12:40:00 AM
 #2017

For half a year I use my cpu to scan for the #67 puzzle. I am 0.000039% scanned addresses of the range ;( A lifetime more to wait. I wonder how many supercomputers out there scan the bitcoin chain for these keys every minute. If you remember the pipeline hack, the FBI had already the privkey for the address used and it was not obtain during a hack or seize.
But Why the 67th puzzle ?.. why not the easier one? I mean the amount of private keys in 66 is unbelievably enormous already .. why go for the double of that gigantic range?
because someone is more likely to find 66 before 67 he doesnt wanna spend a shit ton of time on 66 and someone find it before him so he does 67 by time someone does find 66 he will be at head start( to most people anyway ) thats what i think at least

This analogy may seem smart if we're dealing with normal amounts like million or even billions. But when talking about tens of million trillions, that "head start" gets engulfed to almost nothing. The human mind can't grasp how extremely improbable it is to find ONE address among 73 million trillion ones, let alone double that digit. That being said, why are we still searching?
Because luck and coincidence do happen. But that's just that, you don't need to try outsmarting your odds in order to land on the key. The only factor you need is luck, and you're surely not doing yourself any more favours by doubling the odds against yourself, aka: puzz 67

Cool Story Bro.
BTC: 1EviLLo1Y5VeNn2Lajv9tdZTkUuVgePVYN
The_Prof
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January 21, 2023, 02:50:20 AM
 #2018

Well in essence, lets say he eventually reaches 0.005 percent when someone hit's 66 on a lucky hit in a year and a bit. if he has large saved segments of what he has searched he can go to tdd/ttd can't remember and say, here are the ranges checked and get his according percentage as was done with 64 when it started. You move from private to a team split. It's the very long game to play. Has it's pro's and cons. 

Look over there...
jl33213
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January 22, 2023, 08:45:14 PM
 #2019

For half a year I use my cpu to scan for the #67 puzzle. I am 0.000039% scanned addresses of the range ;( A lifetime more to wait. I wonder how many supercomputers out there scan the bitcoin chain for these keys every minute. If you remember the pipeline hack, the FBI had already the privkey for the address used and it was not obtain during a hack or seize.
But Why the 67th puzzle ?.. why not the easier one? I mean the amount of private keys in 66 is unbelievably enormous already .. why go for the double of that gigantic range?
because someone is more likely to find 66 before 67 he doesnt wanna spend a shit ton of time on 66 and someone find it before him so he does 67 by time someone does find 66 he will be at head start( to most people anyway ) thats what i think at least

This analogy may seem smart if we're dealing with normal amounts like million or even billions. But when talking about tens of million trillions, that "head start" gets engulfed to almost nothing. The human mind can't grasp how extremely improbable it is to find ONE address among 73 million trillion ones, let alone double that digit. That being said, why are we still searching?
Because luck and coincidence do happen. But that's just that, you don't need to try outsmarting your odds in order to land on the key. The only factor you need is luck, and you're surely not doing yourself any more favours by doubling the odds against yourself, aka: puzz 67

and im always tryna think of a way to explain that to people you just cant people dont understand cuz its that many numbers lol people who dont know much about btc that is. bitcoins security is literally based on how many numbers it is
jl33213
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January 23, 2023, 06:14:56 AM
 #2020

@Andzhig And if we increase one more character of address 16jY7qLJn & 'x' then most binaries are started from '111'

few examples -

16jY7qLJnxLQQRYPX5BLuCtcBs6tvXz8BE   1110000000100110101001101101010100100011010011001000100000110110000   7013536A91A6441B0
16jY7qLJnX9uchnyf26t3QJnsUf78Xdikb   1110010000101000111010000001111110010000001011001101111011100000   E428E81F902CDEE0
16jY7qLJnX9eX8j612s8fnbn6uzR48xjua   1110100000001101111010110011001110101001011001111010000010001111   E80DEB33A967A08F
16jY7qLJnx2EZZumnYFke3GutCrRnHKs1M   111010110100110101001101101010111010101000110011101011001010110000   3AD3536AEA8CEB2B0
16jY7qLJnx2ixrxCnTLSraerkgyB3YYAiT   1110110111111001110011010110000000110101011011011100110000011001   EDF9CD60356DCC19
16jY7qLJnxHBp3dqwV2kzYq1LucfZzgxsH   1110111010111001101010110011001101001101111100100111011100001101   EEB9AB334DF2770D
16jY7qLJnX2cZXJ78wV1ef42e7cLAZJ1Vn   1111111000101000011001011100011011011011111111101100001110000011   FE2865C6DBFEC383
16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN   1111011100000101000111110010011110110000100100010001001011010100 F7051F27B09112D4


Could this also be some logic?

can you shed more light on this issue

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5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEw6QwynKpQi          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rVkthFNsQ6i7
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5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kFD3cZqtqDcV          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7siAXycwkwRQg
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5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kFzbD1oFi4uP          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7wBBz2KJQdASx
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kH9V2Fos6zmj          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M82GSgY8p5EkUe
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5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB43UMHpcm3f8ag          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9NRuiZFAX6XciCX
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5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB6ikvy2duGEu2D          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9aFJuCJDo5F6Jm7
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB84u9Cbd5fwdR6          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9gCD9CBomewdcUD
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEBCKvkRRBUtpNdX          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9zzYEemjCVJ3vo9
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEBRhv6ue8TGcLBe          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgdB23bP5LsYN8Krv7
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEBqQAxxEAdgvLgN          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgdCpdGyxm7PWoNQdr
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEC51Sd31pLUWx3i          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgdDrgx63tbte4scLA
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEDg6uLuezrDySzX          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgdLwb42sQhwTBJDnG
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEHAPod6LNRWAZBf          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgdcLTenJYVVVTRzxN
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nELpQuZ1gQQYrZ1J          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgdtUGWxHzvDi28MfR
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEXxDuJBiUbiVA9e          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgejcjwprJ4MAYLw8e
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEiGwVv6mtxTe9ip          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgfXBMYdNVA4EjUMzg
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nFzFBD1zxALEsLqK          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgm9fa43QnU2CpULaK
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nHtVssx1RqUdSdVj          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjguYJTZsntce5oQQrh
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nL8eoQu4hLSpXwa4          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjh5SmqrUoK2mhwSYFV
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nNxiLZixsgdChZjz          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjhHvuTMSDchRp5hktc
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4ngeHA7b4zPaJ3ga1          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjidyYKE5NcJQZYvknA
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nu8VixoNGBCuaLCm          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjjb65nDHeqyjiBaJXv
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4pKxzyzRYGJZF2CUQ          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjqtiAvYTJzYEmqup7b
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4qGMGYDHfP5bxwCnj          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjv2kTamEYQT9BNC7o1
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4vCkbw1TKC5xefUVS          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYkHpsTBP19HvTFqiU6i  
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip55U8L18x1xXBdqu84          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYkzijLsc5qE43yZ5eLV
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip59tGc46u8joQDakRT          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYmLDHsih379uP9zbHSD
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip5obPQhDC8LRVrSnwc          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYpCemuaUp7NigjvtJug
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip78q4h8QmfB6MwMuF1          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYv5Z9J7hv7VYYN3XL3Y
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip9SbPnb1C74XPPH9cU          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZ6FxoaD5r1kYegmtbaT
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ipCnYRNeQuRFKarWVVs          KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZM21gaY8WN2CdwnTG57

for those still doing puzzle:) here is WIF 1-65 compressed and uncompressed hope this helps. if so tip me up bc1q7h54ddkhy7c2fxqvhg34jxejpq89e8k483632a

i give up now:(
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