Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 268. (Read 229433 times)

Anyone who has a max prefix of puzzle 64, please post here.. it will definitely help me with my new experiments soon I will share it..  

No discussion for a long time  Why

#66     2a183647529c49b3c     13zbaHbtmYwRPRSBMVwtQ3GSnrFXsPijso
#67     59265959791de58c7     1BY8Lw3n6TYv3ARN538RRu4VMFVYfkvTW9
#68     9e8b4c743400da30     1MVDUrdV1fbyu5N23qqmn5AHyVp9HW6tvZ

thanks

Yes that setup you're using is correct.

Hello everyone I decided to join you, but I have a weak hardware, so I won't be able to win by consistently, hope for a random. Did I set up the program correctly?   

Continuation "road to calvary".

looking for a 20 digit number, example 30568377312064202855 by 2 > 30 56 83 77 31 20 64 20 28 55 (10 pieces for mixing)

30568377312064202855 by 2 > 30 56 83 77 31 20 64 20 28 55

take the first 100 numbers power of 2 (4096-32768 or 4096-30000000)

these first 100 digits split by 2 (50 elements in list) and look at what positions our numbers (30 56 83 77 31 20 64 20 28 55) will be, then we sort them from smallest to largest,

some like this...

1   [0, 6, 7, 10, 16, 21, 24, 34, 35, 40]
2   [3, 4, 5, 11, 12, 23, 27, 40, 43, 48]
3   [8, 9, 10, 13, 17, 24, 27, 28, 35, 42]
4   [4, 7, 8, 18, 20, 22, 24, 27, 33, 36]
5   [4, 7, 8, 11, 13, 14, 21, 29, 38, 48]
6   [2, 5, 7, 12, 19, 23, 31, 33, 40, 44]
7   [0, 3, 7, 11, 13, 18, 20, 25, 35, 43]
8   [4, 13, 15, 17, 19, 25, 30, 33, 40, 46]
9   [1, 2, 4, 15, 18, 29, 35, 40, 43, 45]
10   [0, 3, 4, 5, 14, 17, 22, 32, 36, 38]

and see the difference between the numbers.

some like this...

95465   [0, 1, 10, 11, 14, 15, 22, 31, 40, 44, 47]     1 9 1 3 1 7 9 9 4 3
273880   [0, 2, 12, 15, 19, 21, 23, 31, 32, 37, 44]    2 10 3 4 2 2 8 1 5 7
310872   [0, 9, 10, 19, 23, 27, 28, 29, 41, 43, 47]    9 1 9 4 4 1 1 12 2 4
324831   [6, 10, 19, 20, 22, 23, 26, 33, 34, 35, 40]  4 9 1 2 1 3 7 1 1 5
327766   [1, 2, 5, 11, 14, 24, 27, 28, 45, 46, 49]     1 3 6 3 10 3 1 17 1 3
342364   [1, 5, 13, 15, 16, 20, 21, 26, 32, 38, 43]    4 8 2 1 4 1 5 6 6 5
484256   [3, 7, 10, 11, 15, 17, 18, 20, 41, 43, 49]    4 3 1 4 2 1 2 21 2 6
519343   [0, 2, 5, 6, 10, 24, 27, 28, 29, 40, 48]       2 3 1 4 14 3 1 1 11 8
540844   [2, 3, 6, 7, 8, 13, 29, 36, 37, 39, 45]         1 3 1 1 5 16 7 1 2 6
744223   [4, 8, 10, 13, 14, 16, 18, 26, 29, 44, 48]     4 2 3 1 2 2 8 3 15 4
773619   [2, 3, 11, 12, 13, 18, 20, 28, 31, 33, 49]     1 8 1 1 5 2 8 3 2 16
804181   [2, 6, 10, 12, 16, 33, 35, 37, 39, 45, 46]     4 4 2 4 17 2 2 2 6 1
816903   [3, 6, 11, 18, 20, 22, 23, 30, 31, 38, 45]     3 5 7 2 2 1 7 1 7 7

now we filter and look for the "formula")))

here we have several components in charge of search time, the size power of 2, first difference numbers to which to stick, the difference, and possible differences + ~10000000 for mixing 10 parts by 2. i.e. some like this    

60466176 6x10  60466176×2^262144 = 15850845241344
9765625   5x10  9765625×2^262144  = 2560000000000
blablabla

besides need to do an offset (take the first 100 numbers power of 2) 0-100 num, 100-200 num, 200-300 num...

exemple 1

2^4096-2^32768

filt <=6 <=6 <=6 <=6 <=6 <=6 <=6 <=6 <=6

0 100
2^15047   [8, 11, 12, 15, 17, 19, 24, 24, 28, 34]     3 1 3 2 2 5 0 4 6
100 200
200 300
300 400
400 500
500 600
600 700
700 800
800 900
900 1000
2^21946   [12, 15, 15, 18, 22, 25, 31, 37, 39, 44]     3 0 3 4 3 6 6 2 5
1000 1100

exemple 2

2^4096-2^30000000

filt <=1 <=1 <=1 <=1 <=1 <=30 <=30 <=30 <=30

2^7788201   [1, 1, 2, 3, 4, 5, 19, 33, 35, 48]          0 1 1 1 1 14 14 2 13
2^20398384   [0, 0, 1, 2, 3, 4, 16, 17, 33, 41]         0 1 1 1 1 12 1 16 8
0-100
2^371466   [5, 6, 7, 7, 8, 9, 10, 11, 19, 20]            1 1 0 1 1 1 1 8 1
2^21846679   [2, 3, 4, 5, 5, 6, 21, 25, 27, 31]         1 1 1 0 1 15 4 2 4
100-200
200-300
2^23749755   [7, 8, 9, 10, 10, 11, 16, 18, 32, 36]     1 1 1 0 1 5 2 14 4
2^29029857   [5, 5, 6, 7, 8, 9, 14, 19, 35, 36]          0 1 1 1 1 5 5 16 1
300-400
2^1061949   [0, 0, 1, 2, 3, 4, 20, 38, 45, 47]            0 1 1 1 1 16 18 7 2
...


Designing and creating a puzzle seems more expensive than a reward.

Any other information for this? I haven't been able to find it anywhere else.

This game has been active for a long time now & still nobody can solve the puzzle. I’ve tried myself half-heartedly but I’m not tech gifted enough. I’m starting to think this prize will never be claimed, certainly not any time soon.

You could be just lucky and solve one of the wallets ) For example, for 64bit wallet you need only 64 "0"s or "1"s, and with luck you can find the correct private key 


It is not 32BTC, it is 100BTC at the moment. And it is not stored on one wallet, it is distributed between several wallets depending on the bit security (for 64bit wallet there is 0.64BTC, for 135bit wallet there is 1.35BTC, and so on)

Here in the topics you have already sorted out all the possible options, you can try https://bitcointalk.org/index.php?topic=5244940.1320 some kind of kangaroo is jumping somewhere...

***
maybe it makes sense to look elsewhere for vulnerabilities
http://www.righto.com/2014/02/bitcoin-mining-hard-way-algorithms.html push a suitable hash using some clever brute force method. Mining Bitcoin with pencil and paper https://www.youtube.com/watch?v=y3dqhixzGVo&feature=emb_title

Hi, can someone provide the code for searching #64? I am not a programmer, so I don't know any code. I just downloaded python and I would appreciate it if anyone can provide it.
I also only own an old computer. I am just someone who wants to try my luck.

Is there any point in this option (geometric).

take a circle ,top right 000,001,002,003 ... left 999,998,997 ...

                   000
             999        001
         998                002
    997                          003    
 ...                                    ....      


21 num length

970 436 974 005 023 690 481

by age

005 023 436 481 690 970 974

and connect the points in a circle.

angles are obtained, not less than 90 degrees (or a jump from 481 690 970 gives an acute angle? check who has the software at hand for this)

                          
                90      (working space degrees)
                 |                
                 |
___________|___________ 180

and that 90×90×90×90×90×90×90= 47829690000000 iterate step by step (for mixing).

In fact, 32BTC was the old total bounty reward. The current amount is approximately 100BTC 

All the info you need is on the first post, but you will need to read it.

Thank you so much for this offer, but we don't know how we will trust you!
There are many scammers on the internet, and they always have attractive offers.

32 BTC is a lot of money, maybe you can give us some info about you so that we can trust you.

BTCBTCI think the best existing tool currently is yours or otherwise private. I would think the best way would be to modify (ocl)vanitygen according to BurtW's suggestion. You would need to limit the random number generator to a certain amount of bits and keep the rest.

Oh I forgot about this one is in python, this was last updated last year.

https://github.com/Telariust/pollard-kangaroo

Thanks Bigvito. Its helpful.
Can you please share the same program in python

It's so amazing how long this puzzle has stood the test of time compared to that puzzle that was just published by MrBeast the other week that got solved in all of 9 hours for less money and took like $250k to develop, apparently.

If someone eventually managed to brute force Puzzle 64 (and beyond) can't someone who's monitoring the keys just use Pollard Kangaroo once the public key is broadcasted and also sweep it with a higher transaction fee since double spending is still an issue with Bitcoin?
If thats the case how can you prevent someone from sweeping the same key if you both became aware of the private key?