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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 80781 times)
brainless
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December 30, 2018, 11:50:51 PM
 #621

This Info Correct ?

this month one big account was pickedup by someone
https://www.blockchain.com/btc/address/1AhTjUMztCihiTyA4K6E3QEpobjWLwKhkR
66000 btc were picked on 4 dec 2018
and moved fast into next account
then split into pieces, maybe in further accounts and exchnages
https://www.blockchain.com/btc/address/1Ko8t47Mkvq8vyT2XgqLmdouVxM8tTryc
https://www.blockchain.com/btc/address/1Gjkd1hwrJxM9h5Sj1W5bfEN6km1qkVCg4
some 8000 btc stored inside
all time is first 18:53, and moved one account to othere at same time 18:56
66000 btc to split 661 btc each act, then split to 10 btc each other act Smiley
https://www.blockchain.com/btc/address/1MYv4C4hZ7hC5sbHrPkzvmNoozQgnHKeAU
one an other big act holder   moved some 35000  btc for safe an other acct for security Smiley
https://www.blockchain.com/btc/address/1M4NKCeps3jWRWnc8dpJg3zTf3FGYNScGy
https://www.blockchain.com/btc/address/1XB3AQu8V5eJHJrLa9UKqTLBGytbNGqLi
https://www.blockchain.com/btc/address/18saWjMCchDTGWunefuDJSWNdz4p6KtfBL
66000 > 661 each > 10 each ----> back to different address with 8000 btc balance
just want to know main 66000 btc picked up by key finder, or owner moving itself after long time ?
Correct. So? what is your point?
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racminer
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December 31, 2018, 12:15:41 AM
 #622

This Info Correct ?

this month one big account was pickedup by someone
https://www.blockchain.com/btc/address/1AhTjUMztCihiTyA4K6E3QEpobjWLwKhkR
66000 btc were picked on 4 dec 2018
and moved fast into next account
then split into pieces, maybe in further accounts and exchnages
https://www.blockchain.com/btc/address/1Ko8t47Mkvq8vyT2XgqLmdouVxM8tTryc
https://www.blockchain.com/btc/address/1Gjkd1hwrJxM9h5Sj1W5bfEN6km1qkVCg4
some 8000 btc stored inside
all time is first 18:53, and moved one account to othere at same time 18:56
66000 btc to split 661 btc each act, then split to 10 btc each other act Smiley
https://www.blockchain.com/btc/address/1MYv4C4hZ7hC5sbHrPkzvmNoozQgnHKeAU
one an other big act holder   moved some 35000  btc for safe an other acct for security Smiley
https://www.blockchain.com/btc/address/1M4NKCeps3jWRWnc8dpJg3zTf3FGYNScGy
https://www.blockchain.com/btc/address/1XB3AQu8V5eJHJrLa9UKqTLBGytbNGqLi
https://www.blockchain.com/btc/address/18saWjMCchDTGWunefuDJSWNdz4p6KtfBL
66000 > 661 each > 10 each ----> back to different address with 8000 btc balance
just want to know main 66000 btc picked up by key finder, or owner moving itself after long time ?
Correct. So? what is your point?

This account has been active since 2014 ... if you look at some transactions , for instance this one
https://www.blockchain.com/btc/tx/1d6580dcd979951bd600252b741c22a3ea8e605e43168f8452c68915c3ea2bf3

you can see that if you check the first unspent wallet such as https://www.blockchain.com/btc/address/3D2oetdNuZUqQHPJmcMDDHYoqkyNVsFk9r there are some 138 660 BTC !!!
and there are lots of them

looks normal to me  Smiley

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December 31, 2018, 12:53:02 PM
 #623

What has this to do with this Puzzle transaction we discuss in this thread?

 
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December 31, 2018, 01:32:59 PM
 #624

Oooops !
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December 31, 2018, 02:10:19 PM
 #625

 no,I do not understand all this, If I'm smart enough to do this, it would be nice to do it. Because the gift is very large 32 bitcoin it is a very big gift, we can buy anything by having 32 bitcoin. but i cant o this ..
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December 31, 2018, 04:00:54 PM
 #626

we can’t solve this puzzle because the supercomputer invented it and only he can solve it
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December 31, 2018, 04:07:20 PM
 #627

no,I do not understand all this, If I'm smart enough to do this, it would be nice to do it. Because the gift is very large 32 bitcoin it is a very big gift, we can buy anything by having 32 bitcoin. but i cant o this ..

You can not buy health or health for any currency. Such little wisdom at the end of the year: P

Hi Zielar
What brute force strategy do you adopt?
we could PM if you want.
A Very Healthy, Wealthy and Happy New Year to All

At the moment I am studying the operation of this script because I was interested:

Quote
import random
from bit import *
from PyRandLib import *
rand = FastRand63()
random.seed(rand())

c1 = str (random.choice("1"))
b1  = "01111111111111111111111111111111111111111111111111111111"
b2  = "00111111111111111111111111111111111111111111111111111111"
b3  = "00011111111111111111111111111111111111111111111111111111"
b4  = "00001111111111111111111111111111111111111111111111111111"
b5  = "00000111111111111111111111111111111111111111111111111111"
b6  = "00000011111111111111111111111111111111111111111111111111"
b7  = "00000001111111111111111111111111111111111111111111111111"
b8  = "00000000111111111111111111111111111111111111111111111111"
b9  = "00000000011111111111111111111111111111111111111111111111"
b10 = "00000000001111111111111111111111111111111111111111111111"
b11 = "00000000000111111111111111111111111111111111111111111111"
b12 = "00000000000011111111111111111111111111111111111111111111"
b13 = "00000000000001111111111111111111111111111111111111111111"
b14 = "00000000000000111111111111111111111111111111111111111111"
b15 = "00000000000000011111111111111111111111111111111111111111"
b16 = "00000000000000001111111111111111111111111111111111111111"
b17 = "00000000000000000111111111111111111111111111111111111111"
b18 = "00000000000000000011111111111111111111111111111111111111"
b19 = "00000000000000000001111111111111111111111111111111111111"
b20 = "00000000000000000000111111111111111111111111111111111111"
b21 = "00000000000000000000011111111111111111111111111111111111"
b22 = "00000000000000000000001111111111111111111111111111111111"
b23 = "00000000000000000000000111111111111111111111111111111111"
b24 = "00000000000000000000000011111111111111111111111111111111"
b25 = "00000000000000000000000001111111111111111111111111111111"
b26 = "00000000000000000000000000111111111111111111111111111111"
b27 = "00000000000000000000000000011111111111111111111111111111"
b28 = "00000000000000000000000000001111111111111111111111111111"
b29 = "00000000000000000000000000000111111111111111111111111111"
b30 = "00000000000000000000000000000011111111111111111111111111"
b31 = "00000000000000000000000000000001111111111111111111111111"
b32 = "00000000000000000000000000000000111111111111111111111111"
b33 = "00000000000000000000000000000000011111111111111111111111"
b34 = "00000000000000000000000000000000001111111111111111111111"
b35 = "00000000000000000000000000000000000111111111111111111111"
b36 = "00000000000000000000000000000000000011111111111111111111"
b37 = "00000000000000000000000000000000000001111111111111111111"
b38 = "00000000000000000000000000000000000000111111111111111111"
b39 = "00000000000000000000000000000000000000011111111111111111"
b40 = "00000000000000000000000000000000000000001111111111111111"
b41 = "00000000000000000000000000000000000000000111111111111111"
b42 = "00000000000000000000000000000000000000000011111111111111"
b43 = "00000000000000000000000000000000000000000001111111111111"
b44 = "00000000000000000000000000000000000000000000111111111111"
b45 = "00000000000000000000000000000000000000000000011111111111"
b46 = "00000000000000000000000000000000000000000000001111111111"
b47 = "00000000000000000000000000000000000000000000000111111111"
b48 = "00000000000000000000000000000000000000000000000011111111"
b49 = "00000000000000000000000000000000000000000000000001111111"
b50 = "00000000000000000000000000000000000000000000000000111111"
b51 = "00000000000000000000000000000000000000000000000000011111"
b52 = "00000000000000000000000000000000000000000000000000001111"
b53 = "00000000000000000000000000000000000000000000000000000111"
b54 = "00000000000000000000000000000000000000000000000000000011"
b55 = "00000000000000000000000000000000000000000000000000000001"
#xx = "00000000000000000000000000000000000000000000000000000001"

while True:
    spisok = [b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13,b14,b15,b16,b18,b19,b20,b21,b22,b23,b24,b25,b26,b27,b28,b30,b31,b32,b33,b34,b35,b36,b37,b38,b39,b40,b41,b42,b43,b44,b45,b46,b47,b48,b49,b50,b51,b52,b53,b54,b55]
    for element in (spisok):
        for spisok in range(1000):
            s = element
            d = ''.join(random.sample(s,len(s)))
            bina = (c1+d)
            b = int(c1+d,2)
            key = Key.from_int(b)
            addr = key.address
            if addr == "15c9mPGLku1HuW9LRtBf4jcHVpBUt8txKz":
                print ("found!!!",b,addr)
                s1 = str(b)
                s2 = addr
                f=open(u"C:/a.txt","a")
                f.write(s1)
                f.write(s2)      
                f.close()
                pass
            else:
                print (s,bina,b,addr)
    pass

I wonder how to change it so that he does not pick randomly only from the chosen threshold he flew up one by one. Does anyone have knowledge of how it could be remade?

BY USING OUR OPPORTUNITY, WE WOULD LIKE YOU TO SAY ALL OF YOUR SUCCESS AND JOY IN THE NEW YEAR!
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December 31, 2018, 06:08:29 PM
Last edit: December 31, 2018, 06:56:46 PM by Andzhig
 #628

there it is necessary little correct.

bina = (c1+d)
b = int(bina,2)


Quote
flew up one by one.
The meaning of this particular example is that it uses a fixed set, 01111111111111111111111111111111111111111111111111111111 and mixes it adding 1 in front so that it doesn’t appear there 0. So if you take 00000000000000000000000000000111111111111111111111111111 +1 = 100000000000000000000000000000111111111111111111111111111 72057594172145663 (can wrap and go in the opposite direction 111111111111111111111111111100000000000000000000000000000 144115187538984960) it will take as much time to find a number so that it satisfies the requests, in other words, it will be easy to go through step by step as without dancing with a tambourine.

Quote
i = 72057594037927936
while i <= 144115188075855872:
    ff = bin(i)[2:]
    ed = ff.count("0")
    if ed >= 10:
        if ed <= 35:
            print(i,ff,ed)
    i = i+1

try to change 10 and 35 to 30-31 and detect the time after which he will give you the result.

Quote
from bit import *
import random

while 1 == 1:
    a = 390000000000000000
    while a <= 554000000000000000:
        bina = bin(a)[2:]
        zeros = bina.count("0")
        if zeros >= 30:     # "00000000000000000000000000000011111111111111111111111111"
            if zeros <= 30: # "00000000000000000000000000000011111111111111111111111111"
                key = Key.from_int(a)
                addr = key.address
                if addr == "1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt":
                    print("found!!!",a,addr)
                    s1 = str(a)
                    s2 = addr
                    f=open(u"C:/a.txt","a")
                    f.write(s1)
                    f.write(s2)      
                    f.close()
                    break
                else:
                    print (a,bina,addr,zeros) #pass
        a = a +1
    pass

runs fast but still need gpu, the gain in speed is that he doesn’t compare unnecessary address spaces (using a bitcoin slowdown library), but nonetheless he goes through them.

Although, if we find crypto guru who will help us calculate to organize the program, we will be very grateful. for example if 2^256 this from 57896044618658097711785492504343953926634992332820282019728792003956564819968 to 115792089237316195423570985008687907853269984665640564039457584007913129639936 step by step and if we take numbers with only 127 zeros in the binary how much will it be in the end result, 2^?

***

Happy holiday https://www.youtube.com/watch?v=VVOXHmvRRh8  Grin
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January 01, 2019, 06:22:01 PM
 #629

What has this to do with this Puzzle transaction we discuss in this thread?
Dear Sir
i apologies if my question is not related to thread, i like to explain why i ask these question comes in my mind related to this thread,
1. every one posting here about random search scripts, some post about brute force techniques, some trying to explain logics for find 32 BTC PUZZLE
2. above my posted link, when explorer, you will find some services used by website, who offer protection for btc coins, what technique used is split btc into pieces and transfers to new series generated accounts, then more splits into new series generated accounts, all generated accounts are same time in seconds, (mean no manual transaction one by one),
3 same in 32 BTC PUZZLE series of satochi transferd to newly generated accounts same time with sequence of each satoshi plus in series of bit space
4 in 59 and 60 bit space, peoples need to explorer by brute force or random etc equal to total hardware used for btc minning in Ph/s,
5 for this reason people must think an other ways to definatly in formula by math, by brainwallet, logics will help
6 for path and ref , i post links for other to think and find solutions deeply by explorer links
7 for deep leaning inside for solve PUZZLE, IN MY VIEW , above and this link could change thinking, and aproch to solve puzzle
8 https://www.blockchain.com/btc/tx/1d6580dcd979951bd600252b741c22a3ea8e605e43168f8452c68915c3ea2bf3
   RETURN PUSHDATA(60)[424553544d495845522e494f207c204d495820594f555220424954434f494e5320544f2053544159205341464520414e442050524f54454354454421]
   (decoded) BESTMIXER.IO | MIX YOUR BITCOINS TO STAY SAFE AND PROTECTED!
9 bestmixer.io workstyle happen in 32 BTC PUZZLE and recenet transaction of https://www.blockchain.com/btc/address/1Ko8t47Mkvq8vyT2XgqLmdouVxM8tTryc
10 IN MY VIEW, brainwallet char series by formula \ calculation of series by formula,  could generate address in all Bit Space, etc, and work arund this may fruitfull instead of brute force by series or random

Sir if you feel Good about my refrence of research related to Solve PUZZLE, encourage me with thumb up, in other just update me, i will remove my post, like to say apologies again

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January 01, 2019, 08:14:46 PM
Merited by hatshepsut93 (1)
 #630

What has this to do with this Puzzle transaction we discuss in this thread?
...
Sir if you feel Good about my refrence of research related to Solve PUZZLE, encourage me with thumb up, in other just update me, i will remove my post, like to say apologies again
...


There is no puzzle to solve, and you do not have to delete anything. Here is the post from the creator of this "puzzle", it may get you to see things clearly:

I am the creator.

...

A few words about the puzzle.  There is no pattern.  It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty).  It is simply a crude measuring instrument, of the cracking strength of the community.

See? Nothing to solve, nothing to crack, from the man who created this. The "puzzle" is just a measurement of the brute forcing abilities of everyone together,  making us sleep peacefully showing that Bitcoin can not be bruteforced. The more of these private keys get discovered by this joined brute-forcing effort, the more (double exactly) difficult it gets to find next private key, and it continues to grow exponentially. There is nothing more to it, human brain is useless with this "puzzle", and any effort to use your brain to accomplish finding private keys for these 32 BTC is just big waist of your time.

 
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January 02, 2019, 01:35:36 AM
 #631

It takes brain power to come up with ways to check keys fast. Why do you think the BitCrack program can test a billion keys per second on a single video card while the Large Bitcoin Collider could test millions on the same card?
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January 02, 2019, 09:09:14 AM
 #632

Because you can run Bitcrack with a lot of cards without cpu.
At collider you need cpu power for every card so when you want to run lot of cards you need big cpu power.
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January 05, 2019, 09:54:44 PM
 #633

Are we looking for specific wallets or just wallets in general that have a balance?
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January 05, 2019, 10:03:13 PM
 #634

Are we looking for specific wallets or just wallets in general that have a balance?

Neither, we are expecting that no private keys from a single transaction will ever being found faster then expected by current bruteforce difficulty. That's the best outcome - that everyone fails and Bitcoin is safe.

 
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anythingtechpro
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January 05, 2019, 10:07:11 PM
 #635

Are we looking for specific wallets or just wallets in general that have a balance?

Neither, we are expecting that no private keys from a single transaction will ever being found faster then expected by current bruteforce difficulty. That's the best outcome - that everyone fails and Bitcoin is safe.

Is it true that someone found the private key for this wallet? https://www.blockchain.com/btc/address/1AhTjUMztCihiTyA4K6E3QEpobjWLwKhkR
racminer
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January 05, 2019, 10:25:32 PM
 #636

Are we looking for specific wallets or just wallets in general that have a balance?

Neither, we are expecting that no private keys from a single transaction will ever being found faster then expected by current bruteforce difficulty. That's the best outcome - that everyone fails and Bitcoin is safe.

Is it true that someone found the private key for this wallet? https://www.blockchain.com/btc/address/1AhTjUMztCihiTyA4K6E3QEpobjWLwKhkR

You may notice that many big wallets are receiving multiple very small deposits from https://www.blockchain.com/btc/address/1AK4LYE6PYwBmSYHQX3v2UsXXHTvCAsJeK  (Bitcoin buy or sell JUBTC.COM)
It seems to be a company which help people retrieve their private keys (they surely try to crack some for themselves Wink ).
I think that if you narrow space search ( for example if you know part of mnemonic words ) , you can retrieve private key from transaction scripts ...

Any thoughts on this?
Does it seem possible?
Lolo54
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January 06, 2019, 06:39:52 AM
 #637

arulbero Help!
what key can there be 1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf
2..............
3..............

Code:
Private key : 00000000000000000000000000000000000000000000000002c675b852189a21
Public key  : 11569442e870326ceec0de24eb5478c19e146ecd9d15e4666440f2f638875f42 524c08d882f868347f8b69d3330dc1913a159d8fb2b27864f197693a0eb39a23
 
PrKey WIF c.: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjv2kTamEYQT9BNC7o1
Address c.  : 8c2a6071f89c90c4dab5ab295d7729d1b54ea60f
Address c.  : 1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf

Bravo and thank you Arulbero you are impressive! If I have followed the topic well let’s say that from 59 to 65 addresses in just a few seconds you could get the private key?!
Could you get there keys for this address 1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt
1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt
racminer
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January 06, 2019, 03:19:19 PM
 #638

arulbero Help!
what key can there be 1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf
2..............
3..............
Code:
Private key : 00000000000000000000000000000000000000000000000002c675b852189a21
Public key  : 11569442e870326ceec0de24eb5478c19e146ecd9d15e4666440f2f638875f42 524c08d882f868347f8b69d3330dc1913a159d8fb2b27864f197693a0eb39a23
 
PrKey WIF c.: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjv2kTamEYQT9BNC7o1
Address c.  : 8c2a6071f89c90c4dab5ab295d7729d1b54ea60f
Address c.  : 1Dn8NF8qDyyfHMktmuoQLGyjWmZXgvosXf

Bravo and thank you Arulbero you are impressive! If I have followed the topic well let’s say that from 59 to 65 addresses in just a few seconds you could get the private key?!
Could you get there keys for this address 1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt
1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt

No  one can. Unless whoever finds  the pvkey  spends partially or totally the  0.59 BTC.

 arulbero's program needs as input a spending script. And it is fast because the  pvkey range is small.
umadbro
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January 10, 2019, 01:52:23 AM
 #639

I've modified bitcrack to work with hashtopolis (https://github.com/s3inlc/hashtopolis) in an attempt to better track and distribute keyspaces that i've already tried for TX 59. My current hashing power is around 6 GHS. If anyone is interested in running the client we could use this as a pool and combine our efforts. I realize the large bitcoin collider exists for this purpose but it is very slow in comparison.

If you're interested in joining, pm me and we can discuss details.
zielar
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January 10, 2019, 02:57:39 AM
 #640

Hello, how have you "changed" bitcrack? As you can see from my posts on the last pages - I would gladly come in as a part of some mutual help. How can I check your revised version and check it on my 8x 1400mkeys / s?
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