Bitcoin puzzle transaction ~32 BTC prize to who solves it

[WanderingPhilospher]

completey unrelated to this topic and basically a noob question. but i read somewhere here on this thread that "directory.io has 904625697166532776746648320380374280100293470930272690489102837043110636675 pages thats almost all the private keys". if generating all the private keys combinations takes more then the age of universe with current technology, how come they generated them all.

Each page is only generated if visited...basically each page contains say 100 keys/addresses, so the website script knows if you visit page 900 to calculate the keys for 9000 through 9100.

thanks for the reply but why do they have this at the bottom of their page:"It took a lot of computing power to generate this database. Donations welcome: 1LBCPotwPzBvBcTtd7ADGzCWPXXsZE19j6". i thought all the address were already generated.

Negative, or Bitcoin would be at zero...now they may have empty rows setup in a database to fill in as each page is visited but I bet they haven't generated 1 percent of all keys possible.

[1714079130]

1714079130

[1714079130]

1714079130

[1714079130]

1714079130

[WanderingPhilospher]

completey unrelated to this topic and basically a noob question. but i read somewhere here on this thread that "directory.io has 904625697166532776746648320380374280100293470930272690489102837043110636675 pages thats almost all the private keys". if generating all the private keys combinations takes more then the age of universe with current technology, how come they generated them all.

Each page is only generated if visited...basically each page contains say 100 keys/addresses, so the website script knows if you visit page 900 to calculate the keys for 9000 through 9100.

I have another noob question for you about the website privatekeys.pw. Why they can provide a Search Private Key queries but not Search Address? If their website able to Search Private Key, why can't they do Search Address for their pages?

probably because their pages are in key sequential order, so if you type in a private key the system knows what web page number it should fall on versus someone searching for an address in which case they would really have to generate private keys to pubkeys...all the way to base58 address. Just more proof they haven't generated all addresses/been through all private keys

[nomadsena]

completey unrelated to this topic and basically a noob question. but i read somewhere here on this thread that "directory.io has 904625697166532776746648320380374280100293470930272690489102837043110636675 pages thats almost all the private keys". if generating all the private keys combinations takes more then the age of universe with current technology, how come they generated them all.

Each page is only generated if visited...basically each page contains say 100 keys/addresses, so the website script knows if you visit page 900 to calculate the keys for 9000 through 9100.

I have another noob question for you about the website privatekeys.pw. Why they can provide a Search Private Key queries but not Search Address? If their website able to Search Private Key, why can't they do Search Address for their pages?

on privatekeys.pw. new private keys are generated on the go using your system resources and the balances are verified. its not some kind of database

[nioctib.co]

completey unrelated to this topic and basically a noob question. but i read somewhere here on this thread that "directory.io has 904625697166532776746648320380374280100293470930272690489102837043110636675 pages thats almost all the private keys". if generating all the private keys combinations takes more then the age of universe with current technology, how come they generated them all.

Each page is only generated if visited...basically each page contains say 100 keys/addresses, so the website script knows if you visit page 900 to calculate the keys for 9000 through 9100.

I have another noob question for you about the website privatekeys.pw. Why they can provide a Search Private Key queries but not Search Address? If their website able to Search Private Key, why can't they do Search Address for their pages?

on privatekeys.pw. new private keys are generated on the go using your system resources and the balances are verified. its not some kind of database

I guess I know what you mean. We can solve address through Private Key on the go since it's not a database.

[nioctib.co]

completey unrelated to this topic and basically a noob question. but i read somewhere here on this thread that "directory.io has 904625697166532776746648320380374280100293470930272690489102837043110636675 pages thats almost all the private keys". if generating all the private keys combinations takes more then the age of universe with current technology, how come they generated them all.

Each page is only generated if visited...basically each page contains say 100 keys/addresses, so the website script knows if you visit page 900 to calculate the keys for 9000 through 9100.

I have another noob question for you about the website privatekeys.pw. Why they can provide a Search Private Key queries but not Search Address? If their website able to Search Private Key, why can't they do Search Address for their pages?

probably because their pages are in key sequential order, so if you type in a private key the system knows what web page number it should fall on versus someone searching for an address in which case they would really have to generate private keys to pubkeys...all the way to base58 address. Just more proof they haven't generated all addresses/been through all private keys

Yup, got it now because I think nomadsena answer make sense now. It generate on the go instead of a database search.

[eddie13]

Further, quite entertaining reading, on this subject of the priv key “database”..
https://bitcointalk.org/index.php?topic=1527225.0

Also.. Don’t search for YOUR priv key there, because then the site owner has your priv key!!

[nioctib.co]

Further, quite entertaining reading, on this subject of the priv key “database”..
https://bitcointalk.org/index.php?topic=1527225.0

Also.. Don’t search for YOUR priv key there, because then the site owner has your priv key!!

Very good advice. I mean there's also no point for someone to search their own private key there other than out of curiosity.

[fxsniper]

Further, quite entertaining reading, on this subject of the priv key “database”..
https://bitcointalk.org/index.php?topic=1527225.0

Also.. Don’t search for YOUR priv key there, because then the site owner has your priv key!!

Very good advice. I mean there's also no point for someone to search their own private key there other than out of curiosity.

just read that thread LOL
I think easy to create web page that generate private key on the fly with out large database
Just put script generate bitcoin key inside web and then make input from URL page so when you put any keyboard to that will generate key as seed you input
may be google help to store that cache page and search found
ฺีBut, Accidentally by chance can possible meet with you bitcoin private key That's it happen random

[fxsniper]

use archive.org roll back to see page

it same many page use scripts generate bitcoin create webpage a lot people do like this

I think is fun. to do.

https://allbitcoinprivatekeys.com/

http://[Suspicious link removed]/
https://isidoroghezzi.bitbucket.io/directory-js/
https://allkeys.cash/
https://addresskeys.com/
https://188.166.21.28/bitcoin/1
https://braliman.com/bitcoin/1
https://lbc.cryptoguru.org/dio/
https://keys.lol/bitcoin/
https://privatekeys.pw/

some people do for make search engine optimization
some people do for make money by banner advertising (AdSense or other)

some people can do generate key and check buy user ip check and if found key have price will sent email to record at server they have price and tell web creator to know may be them will take it possible

just script  but real private key
technical people use script do like this smart to save storage

you can use puzzle address search here may be can found private key

[snaiperkg]

use archive.org roll back to see page

it same many page use scripts generate bitcoin create webpage a lot people do like this

I think is fun. to do.

https://allbitcoinprivatekeys.com/

http://[Suspicious link removed]/
https://isidoroghezzi.bitbucket.io/directory-js/
https://allkeys.cash/
https://addresskeys.com/
https://188.166.21.28/bitcoin/1
https://braliman.com/bitcoin/1
https://lbc.cryptoguru.org/dio/
https://keys.lol/bitcoin/
https://privatekeys.pw/

some people do for make search engine optimization
some people do for make money by banner advertising (AdSense or other)

some people can do generate key and check buy user ip check and if found key have price will sent email to record at server they have price and tell web creator to know may be them will take it possible

just script  but real private key
technical people use script do like this smart to save storage

you can use puzzle address search here may be can found private key

+ in your list

http://privatekeys.info with automatical searching.

[WanderingPhilospher]

want to secure your coin or what to learn about bitcoin mining. contact me

No, but could you tell us the exact location of puzzle #s 64 through say 119?

[fxsniper]

problem of puzzle address that not have public key. key it is still standalone. blockchain still connect to only that key can coming to get funds withdraw out.

one public key still make bitcoin like clean data not have mush to reference


on reason puzzle #64 still search because still on low bits level.

I think all other puzzle 9 address that have public key bay be easy to scan

I try puzzle #64 and #120

but for now all public key still on high bits level may be need to use method like bitcoin

brute force die when using on full 256 bits keys

kangaroo still good one but on high bits need to have some algorithm help to sharp focus target and hits straight

anyone can develop bitcrack mix with kangaroo strategy to hit puzzle #64

public key may be need formula like bitcoin vulnerability  to help calculate

use pure brute force only still long time used

[Andzhig]

Well, or so look...

import time

a1= [(0, 3), (3, 0), (1, 2), (2, 1)]
a2= [(0, 11), (11, 0), (1, 10), (2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1)]
a3= [(0, 11), (11, 0), (1, 10), (2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1)]
a4= [(0, 14), (14, 0), (1, 13), (2, 12), (3, 11), (4, 10), (5, 9), (6, , (7, 7), (8, 6), (9, 5), (10, 4), (11, 3), (12, 2), (13, 1)]
a5= [(0, 4), (4, 0), (1, 3), (2, 2), (3, 1)]
a6= [(0, 2), (2, 0), (1, 1)]
a7= [(0, 10), (10, 0), (1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]
a8= [(0, 2), (2, 0), (1, 1)]
a9= [(0, 10), (10, 0), (1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]
a10=[(0, 10), (10, 0), (1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]

count = 0

for b1 in a1:
    for b2 in a2:
        for b3 in a3:
            for b4 in a4:
                for b5 in a5:
                    for b6 in a6:
                        for b7 in a7:
                            for b8 in a8:
                                for b9 in a9:
                                    for b10 in a10:
                                        count += 1
                                        d = b1+b2+b3+b4+b5+b6+b7+b8+b9+b10
                                        print(d) #517492800
    

print(count)
time.sleep(260.0)

[/size]

2×8×4×4×5×3×13×15×11×2       16473600      1135041350219496382
5×7×15×12×6×7×14×9×17×2     1133546400  1425787542618654982
12×8×10×7×6×5×15×4×6×2       145152000    3908372542507822062
17×12×4×18×13×7×8×15×13×8  16680867840 8993229949524469768
4×12×12×15×5×3×11×3×11×11  517492800    30568377312064202855 (3×11×11×14×4×2×10×2×10×10)

for pz 30568377312064202855 517 million 492,800...

all possible 18x18x18x18x18x18x18x18x18x18 = 3570467226624 steps.

roughly pz 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN between 9223372036854775807 18446744073709551615

previous less 9223372036854775807 pz 8993229949524469768
which is higher 36893488147419103231 pz 30568377312064202855

then it should start with [(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7)]

9223372036854775807

10...
11...
12...
13...
14...
15...
16...
17...

18446744073709551615

import random
from bit import Key
import time

list = ["16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN","13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so","1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9",
        "1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ","19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG","1DJh2eHFYQfACPmrvpyWc8MSTYKh7w9eRF",
        "1PWo3JeB9jrGwfHDNpdGK54CRas7fsVzXU","1JTK7s9YVYywfm5XUH7RNhHJH1LshCaRFR","12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4",
        "1FWGcVDK3JGzCC3WtkYetULPszMaK2Jksv","1DJh2eHFYQfACPmrvpyWc8MSTYKh7w9eRF","1Bxk4CQdqL9p22JEtDfdXMsng1XacifUtE",
        "15qF6X51huDjqTmF9BJgxXdt1xcj46Jmhb","1ARk8HWJMn8js8tQmGUJeQHjSE7KRkn2t8","15qsCm78whspNQFydGJQk5rexzxTQopnHZ",
        "13zYrYhhJxp6Ui1VV7pqa5WDhNWM45ARAC","14MdEb4eFcT3MVG5sPFG4jGLuHJSnt1Dk2","1CMq3SvFcVEcpLMuuH8PUcNiqsK1oicG2D",
        "1K3x5L6G57Y494fDqBfrojD28UJv4s5JcK","1PxH3K1Shdjb7gSEoTX7UPDZ6SH4qGPrvq","16AbnZjZZipwHMkYKBSfswGWKDmXHjEpSf",
        "19QciEHbGVNY4hrhfKXmcBBCrJSBZ6TaVt","1EzVHtmbN4fs4MiNk3ppEnKKhsmXYJ4s74","1AE8NzzgKE7Yhz7BWtAcAAxiFMbPo82NB5",
        "17Q7tuG2JwFFU9rXVj3uZqRtioH3mx2Jad","1K6xGMUbs6ZTXBnhw1pippqwK6wjBWtNpL","15ANYzzCp5BFHcCnVFzXqyibpzgPLWaD8b",
        "18ywPwj39nGjqBrQJSzZVq2izR12MDpDr8","1CaBVPrwUxbQYYswu32w7Mj4HR4maNoJSX","1JWnE6p6UN7ZJBN7TtcbNDoRcjFtuDWoNL",
        "1CKCVdbDJasYmhswB6HKZHEAnNaDpK7W4n","1PXv28YxmYMaB8zxrKeZBW8dt2HK7RkRPX","1AcAmB6jmtU6AiEcXkmiNE9TNVPsj9DULf",
        "1EQJvpsmhazYCcKX5Au6AZmZKRnzarMVZu","18KsfuHuzQaBTNLASyj15hy4LuqPUo1FNB","15EJFC5ZTs9nhsdvSUeBXjLAuYq3SWaxTc",
        "1HB1iKUqeffnVsvQsbpC6dNi1XKbyNuqao","1GvgAXVCbA8FBjXfWiAms4ytFeJcKsoyhL","12JzYkkN76xkwvcPT6AWKZtGX6w2LAgsJg",
        "1824ZJQ7nKJ9QFTRBqn7z7dHV5EGpzUpH3","18A7NA9FTsnJxWgkoFfPAFbQzuQxpRtCos","1NeGn21dUDDeqFQ63xb2SpgUuXuBLA4WT4",
        "1NLbHuJebVwUZ1XqDjsAyfTRUPwDQbemfv","1MnJ6hdhvK37VLmqcdEwqC3iFxyWH2PHUV","1KNRfGWw7Q9Rmwsc6NT5zsdvEb9M2Wkj5Z",
        "1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6","1GuBBhf61rnvRe4K8zu8vdQB3kHzwFqSy7","17s2b9ksz5y7abUm92cHwG8jEPCzK3dLnT",
        "1GDSuiThEV64c166LUFC9uDcVdGjqkxKyh","1Me3ASYt5JCTAK2XaC32RMeH34PdprrfDx","1CdufMQL892A69KXgv6UNBD17ywWqYpKut",
        "1BkkGsX9ZM6iwL3zbqs7HWBV7SvosR6m8N","1PXAyUB8ZoH3WD8n5zoAthYjN15yN5CVq5","1AWCLZAjKbV1P7AHvaPNCKiB7ZWVDMxFiz",
        "1G6EFyBRU86sThN3SSt3GrHu1sA7w7nzi4","1MZ2L1gFrCtkkn6DnTT2e4PFUTHw9gNwaj","1Hz3uv3nNZzBVMXLGadCucgjiCs5W9vaGz",
        "1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua","16zRPnT8znwq42q7XeMkZUhb1bKqgRogyy","1KrU4dHE5WrW8rhWDsTRjR21r8t3dsrS3R",
        "17uDfp5r4n441xkgLFmhNoSW1KWp6xVLD","13A3JrvXmvg5w9XGvyyR4JEJqiLz8ZySY3","16RGFo6hjq9ym6Pj7N5H7L1NR1rVPJyw2v",
        "1UDHPdovvR985NrWSkdWQDEQ1xuRiTALq","15nf31J46iLuK1ZkTnqHo7WgN5cARFK3RA","1Ab4vzG6wEQBDNQM1B2bvUz4fqXXdFk2WT",
        "1Fz63c775VV9fNyj25d9Xfw3YHE6sKCxbt","1QKBaU6WAeycb3DbKbLBkX7vJiaS8r42Xo","1CD91Vm97mLQvXhrnoMChhJx4TP9MaQkJo",
        "15MnK2jXPqTMURX4xC3h4mAZxyCcaWWEDD","13N66gCzWWHEZBxhVxG18P8wyjEWF9Yoi1","1NevxKDYuDcCh1ZMMi6ftmWwGrZKC6j7Ux",
        "19GpszRNUej5yYqxXoLnbZWKew3KdVLkXg","1M7ipcdYHey2Y5RZM34MBbpugghmjaV89P","18aNhurEAJsw6BAgtANpexk5ob1aGTwSeL",
        "1FwZXt6EpRT7Fkndzv6K4b4DFoT4trbMrV","1CXvTzR6qv8wJ7eprzUKeWxyGcHwDYP1i2","1MUJSJYtGPVGkBCTqGspnxyHahpt5Te8jy",
        "13Q84TNNvgcL3HJiqQPvyBb9m4hxjS3jkV","1LuUHyrQr8PKSvbcY1v1PiuGuqFjWpDumN","18192XpzzdDi2K11QVHR7td2HcPS6Qs5vg",
        "1NgVmsCCJaKLzGyKLFJfVequnFW9ZvnMLN","1AoeP37TmHdFh8uN72fu9AqgtLrUwcv2wJ","1FTpAbQa4h8trvhQXjXnmNhqdiGBd1oraE",
        "14JHoRAdmJg3XR4RjMDh6Wed6ft6hzbQe9","19z6waranEf8CcP8FqNgdwUe1QRxvUNKBG","14u4nA5sugaswb6SZgn5av2vuChdMnD9E5",
        "174SNxfqpdMGYy5YQcfLbSTK3MRNZEePoy","1NBC8uXJy1GiJ6drkiZa1WuKn51ps7EPTv","18ZMbwUFLMHoZBbfpCjUJQTCMCbktshgpe"]





def func():

    gennum1 = random.randint(0,18)
    gennum = gennum1
    print(gennum)
    number = int(gennum)
    
    lst = []
    tmp = (0, number)
    lst.append(tmp)
    tmp = (number, 0)
    lst.append(tmp)

    for j in range(number):
        for i in range(number):
            
            
            if j + i  == number:
                tmp = (j, i)
                if tmp not in lst:
                    lst.append(tmp)
                    #print(*tmp)
    return lst

a1= [(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7)]
a2= func()
a3= func()
a4= func()
a5= func()
a6= func()
a7= func()
a8= func()
a9= func()
a10=func()

print(a1)
print(a2)
print(a3)
print(a4)
print(a5)
print(a6)
print(a7)
print(a8)
print(a9)
print(a10)

time.sleep(3.0)
count = 0

for b1 in a1:
    for b2 in a2:
        for b3 in a3:
            for b4 in a4:
                for b5 in a5:
                    for b6 in a6:
                        for b7 in a7:
                            for b8 in a8:
                                for b9 in a9:
                                    for b10 in a10:
                                        
                                        count += 1
                                        
                                        d = b1+b2+b3+b4+b5+b6+b7+b8+b9+b10
                                        
                                        dd = ''.join(str(v) for v in d)
                                        
                                        ran = int(dd)
                                        key1 = Key.from_int(ran)
                                        addr1 = key1.address
                                                                    
                                        if addr1 in list:

                                            print (ran,"found!!!")

                                            s5 = str(ran)
                                            f=open(u"C:/a.txt","a")
                                            f.write(s5 + '\n')
                                            f.close()

                                            break

                                        else:
                                                                            
                                            #pass
                                            print(count,ran,addr1,d,len(dd))
                                        

[/size]

***

for pz 120 kangooru can do this

664613997892457936451903530140172287-1329227995784915872903807060280344575

1329227995784915872903807060280344575

[['1', '3'], ['2', '9'], ['2', '2'], ['7', '9'], ['9', '5'], ['7', '8'], ['4', '9'], ['1', '5'], ['8', '7'], ['2', '9'], ['0', '3'], ['8', '0'], ['7', '0'], ['6', '0'], ['2', '8'], ['0', '3'], ['4', '4'], ['5', '7'], ['5']]
[4, 11, 4, 16, 14, 15, 13, 6, 15, 11, 3, 8, 7, 6, 10, 3, 8, 12, 5]
[15, 20, 29, 19, 26, 11, 13, 13, 20, 5]
35 48 26
83 26
109

[4, 11, 4, 16, 14, 15, 13, 6, 15, 11, 3, 8, 7, 6, 10, 3, 8, 12, 5]  x19 by 18

all possible 708235345355337676357632 steps.

708235345355337676357632 / 2 = 354117672677668838178816

354117672677668838178816 × 3753633603581 = 1329227995784795106519698655335940096
354117672677668838178816 × 3753633603581 = 1329227995784440988847020986497761280
354117672677668838178816 × 3753633603579 = 1329227995784086871174343317659582464
354117672677668838178816 × 3753633603578 = 1329227995783732753501665648821403648

kangaroo jumping   from 3753633603581 to ~3003633603581 if performance allows there should be a chance.

[dextronomous]

https://bitcointalk.org/index.php?topic=5218972.msg53649852#msg53649852
no sir not quite yet,

[Dobrodav]

  so this puzzle still not solve yet??? I wish I learn good math at school so I have more chance to solve this. I will try but it's no promise.

[dextronomous]

hi there andzhig,

could you make it so it searches the address starting with 16 only or show and search only the 64th range with the same way
you already beautifully described and explained thanks a lot sir.

[bigvito19]

hi there andzhig,

could you make it so it searches the address starting with 16 only or show and search only the 64th range with the same way
you already beautifully described and explained thanks a lot sir.

Maybe with vanitygen.

[Andzhig]

hi there andzhig,

could you make it so it searches the address starting with 16 only or show and search only the 64th range with the same way
you already beautifully described and explained thanks a lot sir.

16 in the sense of a number or name of an address...

to make the number start

a1= [(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7)]

a1= [(1, 6)]

all the first similarly (9, 2), (9, 3), (9, 4), (9, 5)...

especially since we do not need more than (9, 9) in sets (11, 1), (14, 0) etc...

a1= [(0, 3), (3, 0), (1, 2), (2, 1)]
a2= [(0, 11), (11, 0), (1, 10), (2, 9), (3, Cool, (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1)]
a3= [(0, 11), (11, 0), (1, 10), (2, 9), (3, Cool, (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1)]
a4= [(0, 14), (14, 0), (1, 13), (2, 12), (3, 11), (4, 10), (5, 9), (6, Cool, (7, 7), (8, 6), (9, 5), (10, 4), (11, 3), (12, 2), (13, 1)]
a5= [(0, 4), (4, 0), (1, 3), (2, 2), (3, 1)]
a6= [(0, 2), (2, 0), (1, 1)]
a7= [(0, 10), (10, 0), (1, 9), (2, Cool, (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]
a8= [(0, 2), (2, 0), (1, 1)]
a9= [(0, 10), (10, 0), (1, 9), (2, Cool, (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]
a10=[(0, 10), (10, 0), (1, 9), (2, Cool, (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]

-

a1= [(0, 3), (3, 0), (1, 2), (2, 1)]
a2= [(2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)]
a3= [(2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)]
a4= [(5, 9), (6, , (7, 7), (8, 6), (9, 5)]
a5= [(0, 4), (4, 0), (1, 3), (2, 2), (3, 1)]
a6= [(0, 2), (2, 0), (1, 1)]
a7= [(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]
a8= [(0, 2), (2, 0), (1, 1)]
a9= [(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]
a10=[(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]

instead of all brute force after discarding, 41990400 whole run, finding at 15513701 steps.

import time

a1= [(0, 3), (3, 0), (1, 2), (2, 1)]
a2= [(2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)]
a3= [(2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)]
a4= [(5, 9), (6, , (7, 7), (8, 6), (9, 5)]
a5= [(0, 4), (4, 0), (1, 3), (2, 2), (3, 1)]
a6= [(0, 2), (2, 0), (1, 1)]
a7= [(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]
a8= [(0, 2), (2, 0), (1, 1)]
a9= [(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]
a10=[(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)]

count = 0

for b1 in a1:
    for b2 in a2:
        for b3 in a3:
            for b4 in a4:
                for b5 in a5:
                    for b6 in a6:
                        for b7 in a7:
                            for b8 in a8:
                                for b9 in a9:
                                    for b10 in a10:
                                        count += 1
                                        d = b1+b2+b3+b4+b5+b6+b7+b8+b9+b10
                                        
                                        
                                        
                                        if d == (3, 0, 5, 6, 8, 3, 7, 7, 3, 1, 2, 0, 6, 4, 2, 0, 2, 8, 5, 5):
                                            print(d)
                                            print(count)

print(count)
time.sleep(260.0)

in any case, even discarding unnecessary (not need more than (9, 9) in sets (11, 1), (14, 0) etc...)

18×18×18×18×18×18×18×18×18×18 = 3570467226624 / 2 = 1785233613312 fixed combinations in which only 3 to 4 may be needed. again a performance question. If could it run 10-50 million at least in a second

1×18×18×18×18×18×18×18×18×18 1 sec
2×18×18×18×18×18×18×18×18×18 2 sec
3×18×18×18×18×18×18×18×18×18 3 sec
4×18×18×18×18×18×18×18×18×18 4 sec
18×18×18×18×18×18×18×18×18×1 blablabla sec
18×18×18×18×18×18×18×18×18×2 blablabla sec
18×18×18×18×18×1×18×18×18×18 blablabla sec
18×18×18×18×18×2×18×18×18×18 blablabla sec
...

try it on gpu (who can write the code) .

import random
from bit import Key
import time
import itertools

list = ["16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN","13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so","1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9",
        "1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ","19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG","1DJh2eHFYQfACPmrvpyWc8MSTYKh7w9eRF",
        "1PWo3JeB9jrGwfHDNpdGK54CRas7fsVzXU","1JTK7s9YVYywfm5XUH7RNhHJH1LshCaRFR","12VVRNPi4SJqUTsp6FmqDqY5sGosDtysn4",
        "1FWGcVDK3JGzCC3WtkYetULPszMaK2Jksv","1DJh2eHFYQfACPmrvpyWc8MSTYKh7w9eRF","1Bxk4CQdqL9p22JEtDfdXMsng1XacifUtE",
        "15qF6X51huDjqTmF9BJgxXdt1xcj46Jmhb","1ARk8HWJMn8js8tQmGUJeQHjSE7KRkn2t8","15qsCm78whspNQFydGJQk5rexzxTQopnHZ",
        "13zYrYhhJxp6Ui1VV7pqa5WDhNWM45ARAC","14MdEb4eFcT3MVG5sPFG4jGLuHJSnt1Dk2","1CMq3SvFcVEcpLMuuH8PUcNiqsK1oicG2D",
        "1K3x5L6G57Y494fDqBfrojD28UJv4s5JcK","1PxH3K1Shdjb7gSEoTX7UPDZ6SH4qGPrvq","16AbnZjZZipwHMkYKBSfswGWKDmXHjEpSf",
        "19QciEHbGVNY4hrhfKXmcBBCrJSBZ6TaVt","1EzVHtmbN4fs4MiNk3ppEnKKhsmXYJ4s74","1AE8NzzgKE7Yhz7BWtAcAAxiFMbPo82NB5",
        "17Q7tuG2JwFFU9rXVj3uZqRtioH3mx2Jad","1K6xGMUbs6ZTXBnhw1pippqwK6wjBWtNpL","15ANYzzCp5BFHcCnVFzXqyibpzgPLWaD8b",
        "18ywPwj39nGjqBrQJSzZVq2izR12MDpDr8","1CaBVPrwUxbQYYswu32w7Mj4HR4maNoJSX","1JWnE6p6UN7ZJBN7TtcbNDoRcjFtuDWoNL",
        "1CKCVdbDJasYmhswB6HKZHEAnNaDpK7W4n","1PXv28YxmYMaB8zxrKeZBW8dt2HK7RkRPX","1AcAmB6jmtU6AiEcXkmiNE9TNVPsj9DULf",
        "1EQJvpsmhazYCcKX5Au6AZmZKRnzarMVZu","18KsfuHuzQaBTNLASyj15hy4LuqPUo1FNB","15EJFC5ZTs9nhsdvSUeBXjLAuYq3SWaxTc",
        "1HB1iKUqeffnVsvQsbpC6dNi1XKbyNuqao","1GvgAXVCbA8FBjXfWiAms4ytFeJcKsoyhL","12JzYkkN76xkwvcPT6AWKZtGX6w2LAgsJg",
        "1824ZJQ7nKJ9QFTRBqn7z7dHV5EGpzUpH3","18A7NA9FTsnJxWgkoFfPAFbQzuQxpRtCos","1NeGn21dUDDeqFQ63xb2SpgUuXuBLA4WT4",
        "1NLbHuJebVwUZ1XqDjsAyfTRUPwDQbemfv","1MnJ6hdhvK37VLmqcdEwqC3iFxyWH2PHUV","1KNRfGWw7Q9Rmwsc6NT5zsdvEb9M2Wkj5Z",
        "1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6","1GuBBhf61rnvRe4K8zu8vdQB3kHzwFqSy7","17s2b9ksz5y7abUm92cHwG8jEPCzK3dLnT",
        "1GDSuiThEV64c166LUFC9uDcVdGjqkxKyh","1Me3ASYt5JCTAK2XaC32RMeH34PdprrfDx","1CdufMQL892A69KXgv6UNBD17ywWqYpKut",
        "1BkkGsX9ZM6iwL3zbqs7HWBV7SvosR6m8N","1PXAyUB8ZoH3WD8n5zoAthYjN15yN5CVq5","1AWCLZAjKbV1P7AHvaPNCKiB7ZWVDMxFiz",
        "1G6EFyBRU86sThN3SSt3GrHu1sA7w7nzi4","1MZ2L1gFrCtkkn6DnTT2e4PFUTHw9gNwaj","1Hz3uv3nNZzBVMXLGadCucgjiCs5W9vaGz",
        "1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua","16zRPnT8znwq42q7XeMkZUhb1bKqgRogyy","1KrU4dHE5WrW8rhWDsTRjR21r8t3dsrS3R",
        "17uDfp5r4n441xkgLFmhNoSW1KWp6xVLD","13A3JrvXmvg5w9XGvyyR4JEJqiLz8ZySY3","16RGFo6hjq9ym6Pj7N5H7L1NR1rVPJyw2v",
        "1UDHPdovvR985NrWSkdWQDEQ1xuRiTALq","15nf31J46iLuK1ZkTnqHo7WgN5cARFK3RA","1Ab4vzG6wEQBDNQM1B2bvUz4fqXXdFk2WT",
        "1Fz63c775VV9fNyj25d9Xfw3YHE6sKCxbt","1QKBaU6WAeycb3DbKbLBkX7vJiaS8r42Xo","1CD91Vm97mLQvXhrnoMChhJx4TP9MaQkJo",
        "15MnK2jXPqTMURX4xC3h4mAZxyCcaWWEDD","13N66gCzWWHEZBxhVxG18P8wyjEWF9Yoi1","1NevxKDYuDcCh1ZMMi6ftmWwGrZKC6j7Ux",
        "19GpszRNUej5yYqxXoLnbZWKew3KdVLkXg","1M7ipcdYHey2Y5RZM34MBbpugghmjaV89P","18aNhurEAJsw6BAgtANpexk5ob1aGTwSeL",
        "1FwZXt6EpRT7Fkndzv6K4b4DFoT4trbMrV","1CXvTzR6qv8wJ7eprzUKeWxyGcHwDYP1i2","1MUJSJYtGPVGkBCTqGspnxyHahpt5Te8jy",
        "13Q84TNNvgcL3HJiqQPvyBb9m4hxjS3jkV","1LuUHyrQr8PKSvbcY1v1PiuGuqFjWpDumN","18192XpzzdDi2K11QVHR7td2HcPS6Qs5vg",
        "1NgVmsCCJaKLzGyKLFJfVequnFW9ZvnMLN","1AoeP37TmHdFh8uN72fu9AqgtLrUwcv2wJ","1FTpAbQa4h8trvhQXjXnmNhqdiGBd1oraE",
        "14JHoRAdmJg3XR4RjMDh6Wed6ft6hzbQe9","19z6waranEf8CcP8FqNgdwUe1QRxvUNKBG","14u4nA5sugaswb6SZgn5av2vuChdMnD9E5",
        "174SNxfqpdMGYy5YQcfLbSTK3MRNZEePoy","1NBC8uXJy1GiJ6drkiZa1WuKn51ps7EPTv","18ZMbwUFLMHoZBbfpCjUJQTCMCbktshgpe"]



def funcc(u):
    number = int(u)
    
    lst = []
    tmp = sorted((0, number))
    lst.append(tmp)

    for j in range(number):
        for i in range(number):
            
            
            if j + i  == number:
                tmp = (j, i)
                if tmp not in lst:
                    lst.append(tmp)
                    #print(*tmp)

    tmp = sorted((number, 0))
    lst.append(tmp)
    return lst

def func():
    
    
    gennum1 = random.randint(0,18)
    gennum = gennum1
    print(gennum,funcc(gennum))
    number = int(gennum)
    
    lst = []
    tmp = (0, number)
    if tmp[0] <=9:
        if tmp[1] <=9:
            lst.append(tmp)

    for j in range(number):
        for i in range(number):
            
            if j + i  == number:
                tmp = (j, i)
                if tmp[0] <=9:
                    if tmp[1] <=9:
                        if tmp not in lst:
                            lst.append(tmp)
                            #print(*tmp)
    tmp = (number, 0)
    if tmp[0] <=9:
        if tmp[1] <=9:
            lst.append(tmp)
    
    return lst

print("18×18×18×18×18×18×18×18×18×18 = 3570467226624")
print(" ")

a1= [(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7)]
a2= func()
a3= func()
a4= func()
a5= func()
a6= func()
a7= func()
a8= func()
a9= func()
a10=func()

print(" ")
print("***")
print("cut off set (18, 0), (0, 18), (17, 1), (1, 17)... = 1785233613312")
print(" ")

print(a1)
print(a2)
print(a3)
print(a4)
print(a5)
print(a6)
print(a7)
print(a8)
print(a9)
print(a10)

print(" ")
print("***")

time.sleep(3.0)
count = 0

for x in itertools.product(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10):
    count += 1

    d = ("".join(map(str, (item for sublist in x for item in sublist))))
    if len(d) <= 50:
        ran = int(d)
        key1 = Key.from_int(ran)
        addr1 = key1.address
                                                                            
        if addr1 in list:

            print (ran,"found!!!")

            s5 = str(ran)
            f=open(u"C:/a.txt","a")
            f.write(s5 + '\n')
            f.close()

            break

        else:
                                                                                    
            #pass
            print(count,ran,len(d),addr1)

[/size]

***

don’t understand how to count, we removed unnecessary and we get options 10×10×10×10×10×10×10×10×10×10 =  10000000000 possible, among which there are 400000000 (exl 8 8 6 8 10 9 7 7 3 10 = 406425600) and several thousand (8 1 3 1 3 2 3 2 7 10 = 60480)

transformation


18×18×18×18×18×18×18×18×18×18                                                   10×10×10×10×10×10×10×10×10×10

2×8×4×4×5×3×13×15×11×2        16473600      1135041350219496382    3×9×5×5×6×4×6×4×8×3     9331200
5×7×15×12×6×7×14×9×17×2      1133546400   1425787542618654982     6×8×4×7×7×8×5×10×2×3  22579200
12×8×10×7×6×5×15×4×6×2        145152000     3908372542507822062    7×9×9×8×7×6×4×5×7×3    80015040
17×12×4×18×13×7×8×15×13×8   16680867840  8993229949524469768    2×7×5×1×6×8×9×4×6×9    6531840

what can factorization give to produce exactly such in the set 3×9×5×5×6×4×6×4×8×3     9331200

how will it look to

9331201
9331202
9331203
...

[Dobrodav]

Does anyone have GitHub share code brute-forced or something relating to this puzzle?

[WanderingPhilospher]

Does anyone have GitHub share code brute-forced or something relating to this puzzle?

This program found the last 2 keys, #110 and #115

https://github.com/JeanLucPons/Kangaroo

This is one many people use to brute force for #64

https://github.com/brichard19/BitCrack