Bitcoin puzzle transaction ~32 BTC prize to who solves it

[nomachine]

@nomachine: you quoted him, so I see his messages... damn  it took me a while to find out about the great button IGNORE and I can highly recommend using it. You will find it under his name. Believe me or not - it helps!

Thanks, Done. 

[1715366182]

1715366182

[1715366182]

1715366182

[KiiraStella]

I guess it is not big deal. it should be possible to solve in one day or so. But it is true that it is quite boring to copy paste different number 10 hours for example.

[SDDD125]

Can anyone recover the valid key by reusing r

Priv = 00000000000000000000000000000000000000000000000000180788E47E326C
Pub=040FAAF5F3AFE58300A335874C80681CF66933E2A7AEB28387C0D28BB048BC634965455EF6AFC625E4AEEF1229F052DFD9AB299B02AAA7E659C0E010A795E49E38
r = 0x1fafd110f1b06756ee15a1781daf5fa52d2d548bc772fb61cbbae7effde374db
s1 = 0x1fafd110f1b06756ee15a1781daf5fa52d2d548bc772fb61cbbae7effde374db
z1 = 0x3f5fa221e360ceaddc2b42f03b5ebf4a5a5aa9178ee5f6c39775cfdffbc6e9b6
z2 = 0x13f5fa221e360ceaddc2b42f03b5ebf49150985fe3e2e96ff57482e6ccbfd2af7
s2 = 0xe0502eef0e4f98a911ea5e87e250a0598d81885ae7d5a4d9f417769cd252cc66
V1 Is True!
V2 Is True!
The recovered value of k is correct!
The value of k retrieved 1: 0x180788e47e326e
The recovered k value 2: 0xfffffffffffffffffffffffffffffebaaedce6af48a03bbfba5703ebb80ed3
The R point corresponding to k1: 0x1fafd110f1b06756ee15a1781daf5fa52d2d548bc772fb61cbbae7effde374db
The R point corresponding to k2: 0x1fafd110f1b06756ee15a1781daf5fa52d2d548bc772fb61cbbae7effde374db
r = 0x949bdc126ed9f366d564a3c5e25e425bedf7045ed85b855fccf7904d7b844f90
s1 = 0x4a4dee09376cf9b36ab251e2f12f212df6fb822f6c2dc2afe67bc826bdc227c8
z1 = 0xdee9ca1ba646ed1a4016f5a8d38d6389e4f2868e4489480fb373587439467758
z2 = 0x1dee9ca1ba646ed1a4016f5a8d38d63889fa16374f3d1e84b7345b701097cb899
s2 = 0xb5b211f6c893064c954dae1d0ed0ded0c3b35ab7431add8bd956966612741979
V1 Is True!
V2 Is True!
The recovered value of k is correct!
The value of k retrieved 1: 0x300f11c8fc64db
The value of k retrieved 2: 0xfffffffffffffffffffffffffffffebaaedce6af48a03bbfa24f7b0739dc66
The R point corresponding to k1: 0x949bdc126ed9f366d564a3c5e25e425bedf7045ed85b855fccf7904d7b844f90
The R point corresponding to k2: 0x949bdc126ed9f366d564a3c5e25e425bedf7045ed85b855fccf7904d7b844f90
r = 0x588c57b33f7b758b0d10d1335cee3cec17f1babadfbf4ba96258ffce6c22e34c
s1 = 0x2c462bd99fbdbac586886899ae771e760bf8dd5d6fdfa5d4b12c7fe7361171a6
z1 = 0xb118af667ef6eb161a21a266b9dc79d82fe37575bf7e9752c4b1ff9cd845c698
z2 = 0x1b118af667ef6eb161a21a266b9dc79d6ea92525c6ec7378e84845e29a87c07d9
s2 = 0xd3b9d4266042453a797797665188e188aeb5ff893f68fa670ea5dea59a24cf9b
V1 Is True!
V2 Is True!

[Woz2000]

Don't forget to ignore his imaginary friends as well.

@nomachine: you quoted him, so I see his messages... damn  it took me a while to find out about the great button IGNORE and I can highly recommend using it. You will find it under his name. Believe me or not - it helps!

Thanks, Done. 

[yellowstripes]

One does not become a university professor after posting one liner or few liner generic sh*tposts.

You still haven't answered the question of where your scientific works are. Where can we read online?

For those solving puzzle 130... I have been able to reduce 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 to a 110 bit search range and have the 1st 4 digits. I have worked on this daily for the past 2 months. and unfortunately do not have the hardware that would allow me to perform an optimal search in the range. I am willing to sell this info for the fair price of 1BTC. PM if interested. If you want a clue if what am offering is real...feel free to ask.

[yellowstripes]

For those solving puzzle 130... I have been able to reduce 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 to a 110 bit search range and have the 1st 4 digits. I have worked on this daily for the past 2 months. and unfortunately do not have the hardware that would allow me to perform an optimal search in the range. I am willing to sell this info for the fair price of 1BTC. PM if interested. If you want a clue if what am offering is real...feel free to ask.

That public key is useless, why would we need it? Besides, if you have managed to reduce the size, you should continue reducing it further, unless your imaginary method doesn't actually work, but nice try asking 1BTC for nothing.
I will give you a public key and a range, if you can tell me it's first 3 characters, then I will ask you to reveal the 4 characters you claim to know to puzzle creator and receive your coin, deal?

It actually is not useless. 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 is the difference between the public key of puzzle 130 and the end range. Once you know it, you just need to subtract the private key of 026a...from the end range of puzzle 130 and then get the key of puzzle 130.

[sssergy2705]

For those solving puzzle 130... I have been able to reduce 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 to a 110 bit search range and have the 1st 4 digits. I have worked on this daily for the past 2 months. and unfortunately do not have the hardware that would allow me to perform an optimal search in the range. I am willing to sell this info for the fair price of 1BTC. PM if interested. If you want a clue if what am offering is real...feel free to ask.

That public key is useless, why would we need it? Besides, if you have managed to reduce the size, you should continue reducing it further, unless your imaginary method doesn't actually work, but nice try asking 1BTC for nothing.
I will give you a public key and a range, if you can tell me it's first 3 characters, then I will ask you to reveal the 4 characters you claim to know to puzzle creator and receive your coin, deal?

It actually is not useless. 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 is the difference between the public key of puzzle 130 and the end range. Once you know it, you just need to subtract the private key of 026a...from the end range of puzzle 130 and then get the key of puzzle 130.

Well, decide for yourself, why do you need to share money with someone?
As for the equipment, there are no problems now. You can rent any server, even with huge RAM, or with powerful video cards.

[SDDD125]

I guess it is not big deal. it should be possible to solve in one day or so. But it is true that it is quite boring to copy paste different number 10 hours for example.

You are absolutely right, it's not that much of a BIG deal, only worth $1 trillion in total. easy peasy.
@SDDD125, please explain your steps, what script you use etc, btw, lattice attack is useless here.

I would like to know, because it is not possible to recover the private key d in this type of signature, they are valid, they reuse the same point r but using the formula print (hex(((z1*s2 - z2*s1) * modinv((r*( s1-s2)),p)) % p)) does not give the correct value

[yellowstripes]

It actually is not useless. 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 is the difference between the public key of puzzle 130 and the end range. Once you know it, you just need to subtract the private key of 026a...from the end range of puzzle 130 and then get the key of puzzle 130.

Congratulations, you just managed to subtract #130 from it's end range, I wonder is that all you learned in the past 2 month? So what the hell am I even doing here all this time if not teaching at least 1 person something useful? sigh!

Here, think of this as #130 :

Code:

000000000000000000000000000000034551231950b75fc4402da1732fc9bebf  - 00000000000000000000000000000003ffffffffffffffffffffffffffffffff  =  00000000000000000000000000000000baaedce6af48a03bbfd25e8cd0364140

Now you want to sell the third key above for 1BTC? Do you have it? then solve the puzzle already, note the difference between the first key and the third key, if you drop the first char aka "3", you'd realize the rest of the characters of the first key have turned into a mirror version in the third key. E.g, "455123" is mirrored into "baaedc" now what is 0xb? in decimal is 11, so 4 turned into 11, 5 turned into 0xa(10), 1 turned into 0xe(14), 2 turned into 0xd(13) etc.  Work on a solution based on that knowledge, figure out their difference in percent, it is a fixed value, no matter what, any number has a fixed difference value with it's mirror self.
0,1, 2, 3, 4, 5, 6, 7, 8, 9, a, b, c, d, e, f.
f, e, d, c, b, a, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0.

Work on that instead of trying to sell your calculations, then you can solve it yourself.

What 3rd key? Digaran ...please if you don't understand what I posted then ask...and nicely too for that matter. Just because our ideas and methods do not align does not mean yours is superior. And certainly put me in touch with the creator of the puzzle...I'll certainly tell him the first 4 digits of the private key of puzzle 130 for that 1 bitcoin and even gave a 110 bit range for it.

[mabdlmonem]

Code:

using System;
using System.Numerics;
using System.Security.Cryptography;
using System.Text;

public class Program
{
    public static void Main()
    {
        BigInteger x = BigInteger.Parse("55066263022277343669578718895168534326250603453777594175500187360389116729240");
        BigInteger y = BigInteger.Parse("32670510020758816978083085130507043184471273380659243275938904335757337482424");

        string compressed = CompressPoint(x, y);
        Console.WriteLine("Compressed Form: " + compressed);

        string hash160 = GetHash160(compressed);
        Console.WriteLine("Hash160: " + hash160);
    }

    public static string CompressPoint(BigInteger x, BigInteger y)
    {
        string prefix;
        if (y % 2 == 0)
        {
            prefix = "02";
        }
        else
        {
            prefix = "03";
        }

        string xHex = x.ToString("x");
        string compressed = prefix + xHex;

        return compressed;
    }

    public static string GetHash160(string data)
    {
        byte[] bytes = StringToByteArray(data);
        byte[] hashBytes = SHA256.Create().ComputeHash(bytes);

        using (RIPEMD160 ripemd160 = RIPEMD160.Create())
        {
            byte[] hash160Bytes = ripemd160.ComputeHash(hashBytes);
            return ByteArrayToString(hash160Bytes);
        }
    }

    public static byte[] StringToByteArray(string hex)
    {
        return Enumerable.Range(0, hex.Length)
                         .Where(x => x % 2 == 0)
                         .Select(x => Convert.ToByte(hex.Substring(x, 2), 16))
                         .ToArray();
    }

    public static string ByteArrayToString(byte[] bytes)
    {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < bytes.Length; i++)
        {
            sb.Append(bytes[i].ToString("x2"));
        }
        return sb.ToString();
    }
}

i would be apperciate if anyone has knowldge in c# help me to fix this code

Code:

public static string GetHash160(string data)
    {
        byte[] bytes = StringToByteArray(data);
        byte[] hashBytes = SHA256.Create().ComputeHash(bytes);

        using (RIPEMD160 ripemd160 = RIPEMD160.Create())
        {
            byte[] hash160Bytes = ripemd160.ComputeHash(hashBytes);
            return ByteArrayToString(hash160Bytes);
        }
    }

The type or namespace name 'RIPEMD160' could not be found (are you missing a using directive or an assembly reference?)
The name 'RIPEMD160' does not exist in the current context

this the problem
the expected result is
0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
751e76e8199196d454941c45d1b3a323f1433bd6
thanks again for who would help

[nomachine]

I guess it is not big deal. it should be possible to solve in one day or so.

To find a 66-bit private key within 10 days, we would need to check approximately 200 (billion) giga/hashes - addresses per second.  It doesn't matter if it's an even or odd number. The only way to make this that fast is aliens to land and solve this..

[alek76]

The only way to make this that fast is aliens to land and solve this..

Maybe this code. I don't know what else to think of

Code:

// Get Key from hash160
Int b160key;
b160key.SetInt32(0);
uint32_t *b = (uint32_t *)hash160;
//uint32_t *b = (uint32_t *)h0;
for (int i = 0; i < 5; i++) {
    b160key.bits[i] = b[i];
}
printf("\n b160key: %s", b160key.GetBase16().c_str());// for check 
//

To check, it prints in reverse order - this is correct.

[tmar777]

What 3rd key? Digaran ...please if you don't understand what I posted then ask...and nicely too for that matter. Just because our ideas and methods do not align does not mean yours is superior. And certainly put me in touch with the creator of the puzzle...I'll certainly tell him the first 4 digits of the private key of puzzle 130 for that 1 bitcoin and even gave a 110 bit range for it.

Puzzle 130 = first key. End range = second key. Their difference = third key. Maybe you should read your post again. Here is the deal, I will give you a public key, just tell me the first character of it's private key, you have 14 characters to choose from but only 1 correct guess. If you can't, then we'll have nothing to discuss on this subject.

Code:

022d410abfc76bc647c29819c3349c8a59c84af25feedf59b932643d10299660f0

Edit, I tried looking in my pockets, but couldn't find the creator of this puzzle, let me check the closet. Lol. If you want to get in touch, solve a puzzle, they will find you, don't worry about the time.😉

Can you explain the logic of the difference of a key by the end of the range etc? I cannot see any valid argument in this.

[alek76]

What 3rd key? Digaran ...please if you don't understand what I posted then ask...and nicely too for that matter. Just because our ideas and methods do not align does not mean yours is superior. And certainly put me in touch with the creator of the puzzle...I'll certainly tell him the first 4 digits of the private key of puzzle 130 for that 1 bitcoin and even gave a 110 bit range for it.

Puzzle 130 = first key. End range = second key. Their difference = third key. Maybe you should read your post again. Here is the deal, I will give you a public key, just tell me the first character of it's private key, you have 14 characters to choose from but only 1 correct guess. If you can't, then we'll have nothing to discuss on this subject.

Code:

022d410abfc76bc647c29819c3349c8a59c84af25feedf59b932643d10299660f0

Edit, I tried looking in my pockets, but couldn't find the creator of this puzzle, let me check the closet. Lol. If you want to get in touch, solve a puzzle, they will find you, don't worry about the time.😉

Can you explain the logic of the difference of a key by the end of the range etc? I cannot see any valid argument in this.

No one can do this. And Digaran won't be able to.
I will try to explain what the asymmetric encryption algorithm secp128k1 means. And why, you will never find the relationship by the beginning or ending bytes, between the private and public key. For example:
1. I wished for 2 factors and tell you the remainder of the modulo division.
Can someone tell me what numbers I multiplied? No he can’t!!!
If he can, it means he broke EC
2. This operation of multiplication and division modulo is performed 130 times in this case.
3. There is only linearity in modulo multiplication and division operations.
Well, let's start breaking the EC? I guessed two factors, the result of division modulo _P is equal to:

Code:

0x9478230a3fb794b615ac8f97768ab8feb86784804ef03de53ec26f8cbfbc37f5

Digaran please tell me the multipliers, I  2 multipliers. Or anyone.  

[nomachine]

Maybe this code. I don't know what else to think of

I sent you a PM what to look at....

[btc11235]

Ok, so, when I first found out about this puzzle and started looking into it earlier this year, I was initially working under the assumption that there would be some clues or patterns that could be found to help unlock the remaining wallets... But, of course, I now know what you all already knew: There are no clues or patterns, at least nothing included intentionally by the person who created and funded these wallets...

However, I recently revisited my pattern-finding script, looking instead for any bias(es) in the data that I might be able to find (created by the pseudo-random number generator, or by assumptions built into the wallet generator's code, etc) to at least help shrink the given search spaces as much as possible, and I think I may have (possibly, maybe) found one... The problem is that even if I'm right, I just don't have the computing power to test my theory, nor the funds to buy/rent it, and won't for the foreseeable future... So I'm just going to post it here, in the hopes that, if it is useful at all, then whoever uses this to help them open any of the remaining wallets will be generous enough to kick-back a percentage of the winnings (my BTC address is bc1qmzud9l4aedq6h73efaj3vg3fdqhgv8w4ktq0pt)

Ok, so, my theory goes like this: I initially looked at all of the known pkeys (as decimal numbers, not hex) and calculated their location within the given puzzle's search space as a percentage, like this for Puzzle #11:

Code:

Hex: 0000000000000000000000000000000000000000000000000000000000000483
Hex as Decimal: 1155
Search Space: 1024 to 2047
Percent: (1155 - 1024) / (2047 - 1024) = 12.8%

Now, if that math is wrong, then I'm already screwed and you can give-up now

But after doing that for all the known pkeys, and looking at all the results, I didn't see any pattern, so I gave up and moved on to the next possible pattern I could think of... However, when I went back looking for a bias in the data instead of an outright pattern, I believe I have found that certain percentages (when just looking at the whole number... so 12% for the above example, instead of 12.8%) do (seemingly) occur more often than most...

I first grouped them by the tens (so 20 for 20-29, and 30 for 30-39, etc) and here's the results:

Code:

Tens  Count  Percents
60    13     62,63,64,64,64,65,66,66,66,67,68,69,69
30    9      31,31,32,33,33,35,36,36,38
40    9      40,43,43,44,45,45,46,46,49
10    7      10,12,13,17,17,19,19
90    7      91,92,92,95,95,96,97
80    6      82,82,82,82,82,87
20    6      22,23,23,27,28,28
50    5      50,51,51,54,57
70    5      70,72,72,75,75
0     4      0,6,8,9

So, looking at it that way, you can see that it seems like the 60-69% range hits way more often than the average... But even searching 10% of any given search space is still a crap-ton of brute-forcing to do, so I kept looking and noticed that certain individual numbers show-up more often than others, so next I grouped more simply by just the whole number:

Code:

Count  Percents
5      82
3      64,66
2      17,19,23,28,31,33,36,43,45,46,51,69,72,75,92,95
1      0,6,8,9,10,12,13,22,27,32,35,38,40,44,49,50,54,57,62,63,65,67,68,70,87,91,96,97

And here it appears as though the 82% range occurs way more often than any other, followed closely by 64% and 66% (which occur 3 times each)... Which could (again, theoretically) let you search only 1-3% of any given search space... But, unfortunately, that's still way too much brute-forcing for me to even attempt, given that there's no way to know which of the remaining pkeys will hit these ranges (and so you'd have to try several before finding a hit...)

...and that's assuming that this is a true bias in the data, and not just a coincidence that would even-out over time, which is the other possible way I can think of that this theory could completely fail: Maybe all the keys that could be found within the eighty-two-point-something percent range of their search space (for example) have already been found, leaving the others that have only hit once or twice so far to "catch up", statistically speaking...

Anyway, am I just an idiot, or does this make any sense to y'all...?

[Kamoheapohea]

Please stop. Search help. This thread became less usefull since you've entered it. Your base conversions and maths are stupid. Nobody wil ever benefit of your "works". Thank you.

[digaran]

Please stop. Search help. This thread became less usefull since you've entered it. Your base conversions and maths are stupid. Nobody wil ever benefit of your "works". Thank you.

Code:

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⣿⣿⡆⠀⠀⠀⠀⠀⠀⠀⠀⠀⣰⠟⠀⠀⠀⢨⡷⣄⠀⢿⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠸⣿⣿⣿⣿⡟⠛⠿⢿⡄⠀⠀⠀⠀⠀⠀⠀⠀⠈⠀⠐⠛⠀⠀⠀⠙⠆⢸⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠹⣿⣿⣿⡹⣄⡶⠀⠀⠀⠀⠀⠀⠀⠀⢀⣦⠀⠀⠀⣀⡤⠴⣶⠀⠀⢸⣦⣤⣄⣀⡀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⢀⣀⣀⣤⣽⣿⣿⣧⡻⣷⣦⣀⡀⠀⠀⠀⠀⠀⠈⠙⠂⢶⣯⣥⣴⣾⠏⠀⠀⣼⣿⣿⣿⣿⣿⣿⣷⣦⣤⣀
⠀⠀⢀⣠⣤⣶⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣦⣤⡰⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠉⠁⠀⠀⠀⢹⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣴⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⡄⠀⠘⢦⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠴⠆⠀⣸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣯⠳⣄⠀⠉⠓⠶⠄⠀⠀⠀⠀⠀⠀⠀⠀⢠⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⠙⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠀⠈⠓⠀⠀⠀⠀⠀⠀⠈⠓⠒⠒⠚⠙⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛ 

[alek76]

Please stop. Search help. This thread became less usefull since you've entered it. Your base conversions and maths are stupid. Nobody wil ever benefit of your "works". Thank you.

It looks like a flood and an advertisement for a mixer.
I don't comment on his posts. Im added user digaran to ignore list. My patience has run out

[citb0in]

Ok, so, when I first found out about this puzzle and started looking into it earlier this year, I was initially working under the assumption that there would be some clues or patterns that could be found to help unlock the remaining wallets...

But, of course, I now know what you all already knew: There are no clues or patterns, at least nothing included intentionally by the person who created and funded these wallets...

This already shows the irrelevance of the following miserable attempt to discover a pattern.
[---snip---]

imagine #130 is at 0.9%
if you approached this puzzle with your mindset, you would have failed miserably due to depleted resources, both financial and time. If you had not died before.

Please don't get me wrong, this is not a criticism of your skills but an indication of how people can be misled by a lack of imagination or understanding, especially when it comes to very large numbers.