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Yeah, about that!
It happens when you let third parties to manage your funds, when will people learn that bitcoin is money and since there is no central safe keeping party to insure the safety of your funds, you should never put your funds in the hands of crypto-bank wannabes.
Everytime you want to move some coins, you should generate change addresses offline, move what you have and send the change to the addresses which you generated offline using an address generator which has no access to the internet.
There is no such a thing as a safe and secure wallet in crypto world.
You are your bank, the safety of your coins depends on mathematics and you alone, don't let others to keep them safe because you'd be disappointed.
You are either up for the task, or you should stick with central banks.
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Here is some of my past works, they have helped thousands of projects to go to daa moon!
1- cool project.
2- keeping a close eye on this.
3- wow.
4- I have been waiting my entire life for this.
5- going to the moon.
Now keep asking for bounties, campaigns etc is a huge helper, especially if we fake asking for translations, for dead or alive languages alike, moreover, posting with a few high rank alts will do miracles.
I don't have a team but I can manage to work as a team, I will shill for you with 50 accounts, that would be $1000 per day, and I take at least a day's payment upfront, you can find my address in my profile, if you don't pay me in the next hour or so, I will hunt your project down and haunt your thread to under ground!
Actually now that I think about it, since I'm an extortionist (don't even know how to spell it) You have to pay me $1000 just for letting me know about your request.🤣
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My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm  . If your algorithm is working for one key, it should work for all other keys, unless it's not working, if you can solve a known key and not any other key, you are doing things in a wrong way. Manage to find a key which you don't know it's private key, then you can claim success. Hope you feel better. |
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Digaran @
You can't offer anything useful. An endless stream of insanity.!
Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.
I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit? You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation. SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7
mask125=0000000000000000000000000000000020000000000000000000000000000000
(hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7
key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814
key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6
(key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits. I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉
digaran
If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers
I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve.
There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve.
There are special numbers on the curve that can be used to get exactly 1 digit less.
A third of the number you are looking for.
A quarter of the sought number.
You can get the mirror part of the number you are looking for, which can also be compared with the original.
And this is not even a 10th part of what is there.
You can get a lot.
But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.
I wish you success.
Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle. I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods. Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣 |
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Uhh! hey there team mates! I was wondering what is the reason to select an even -n to add and subtract, if I set -n 5, what happens? Also selecting an even number which is a combination of 2 odd numbers will add/subtract odd numbers. Or is it OK to do that?
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Do people really think Satoshi is reading all the posts here every day?
I don't know if anyone here is a giver, unfortunately we are here to take the loots.
Welcome to the jungle, and happy hunting.😉
Wow guys! I just discovered something, everytime I double a key I can get double the distance with my original base key!
I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit? A good thing about mod function in EC, you could keep halving for eternity without reaching destination, as we are moving towards bitcoin halving, we should start key halving to solve some puzzles before the halving event! 😉
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@kalos, I don't know what you are doing looking for addresses in 125 bit! When you have public key.
But, here let me give you a hint
000000000000000000000000000000001e739ce739cf7bdef7bdef7bdef7be0c
0302157655af8bb1decd37d161aa92cc7050cae1d6b4f44e4aa04507e3c1855159
Add the keys above to each other and tell me what you see.
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I give up,
Hey, I don't know who you are but don't give up yet! Here I will give you some clues to keep going. Imagine puzzle 125 is number 25. Now instead of subtracting a smaller key, we'd subtract it from a bigger key, we'd subtract it from 30, aka 2^125 = 0000000000000000000000000000000020000000000000000000000000000000 And the result = 0286936a275e6d53bb2b2718c93d8a5aa44f371f6e0300abb73b89dd851d2fbe88 Now the result is around 1/5 of our target, give or take. Now forget about puzzle #125 and focus on this new key.😉 |
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guys, windows is closing. we need a new method to solve ECDLP with wallet address only.
Yeah about that, it's the other way around, we solve the address using EC only, the DLP part can be solved through public keys and not addresses.
I have an idea, why not picking a key from the middle of 66 bit range and start adding and subtracting to it using key subtract in address mode, you never know, you might get lucky and find puzzle 66, at least this way you could adjust start and end range of addition and subtraction, then you could for example, add and subtract 1 million keys to your middle of the range key with this stride "10000000000" and then add and subtract the same to another key, repeat that until you see #66 address, but we have no tool capable of doing that by itself, so you'd have to do addition and subtraction a few million times by hand, but if we could modify key subtractor to do that by itself, it would be great. I wonder who is a wondering developer around these woods? !!!😉 |
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Subtracting -70 from +50 is resulting in +120.
I am still not getting trangular thing, can you explain with a simple example pl
Lol, I corrected my mistake. As for showing examples, if I had a working example, I would have used it already to solve the puzzle! But here how it works in my mind, we have 2 and 4 as our base known points, and two unknown keys, we pick 15 and 20 for simplicity, we start by subtracting 2 from 15 and 4 from 15. Now we have two new mutual keys, 11 and 13, if we subtract 11 from 20 we will have 9 which by subtracting 2*4 = 8 from it, we will have half of our first base point "2" ( aka 1). Second step, we subtract 13 from 20 to have 7, now if we subtract 2+4 = 6 from it, we still have the half of our first base point "2" ( aka 1). Therefore we have a triangle 2, 4 and 1, our known keys, and then we have 7, 9, 11, 13, 15, 20! Wait that turned into a threesome of triangles, 🤣 anyways the quantity of triangles is insignificant compared to the quality of our equation. The important part, if we do 2+4 = 6 *4 = 24, we now have 20 + 4 = 24, because we know 4, by adding it to x= 20(unknown key) we can observe whether any of our results are mutually related or not. More complicated equation, we do the following, 2*4 = 8*2 = 16, now adding 4 to it, we will observe our x= 20 again. Trick is picking your base points carefully, otherwise you will be doing things blindly. Find mutual keys that's all.😉 |
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I guess you are way off the map! 2 trillion keys? The actual number is 5 to 6 public keys in order to form 2 triangles and solve the key, more than that will confuse you, as it has confused me with a few million keys, I see hexadecimal characters wherever I look, after working on them for hours. Lol.
LOL, these hexadecimals do often come into my dreams and dancing around as well. I sure didn't got your point of TRIANGLES. What I was referring to trillion thing is this.. Suppose we are searching the private key which is within bit range of 16:32. I made a list of initial 6 public keys like I have a file in which 1-6 public keys are stored. Now I subtract 16 from the puzzle public key. Suppose the hidden private key is 21. So when I subtract 16 from it the resulting key at the same position but in lower bit size. Now 21-16=5 and now I check whether this public key is in my file in which I saved initial public keys. So this is simple thing I am referring as 2 trillions... Can you explain your TRIANGLE THING?  ?? Well, you are guessing the number, you could as well be subtracting 2 from 21 and not knowing it, and then going to search below 10 in hopes of finding your "5". About triangulation, it is similar to how GPS works, but instead of sending and receiving signals, we add and subtract, you basically need only to know the value of 2 keys out of 6 keys, you have to find mutually related keys and find your way by adding and subtracting your 2 known keys to and from 4 unknown mutually related keys to find one of them! It is rather complex, if I had the whole idea pictured in front of me, I could have explained it better, this however is easy to do on #125 because we know the first character is 1, and if I manage to solve it, the method won't work for any other puzzle, unless another puzzle starts with 1. Lol.
I'm going to call this method, "PPP" short for PinPointing Position! I have managed to figure out that #125 starts with obviously 1 and the second character is definitely greater than 4, but I know for sure it is greater than 14000000000000000000000000000000 I'm still confused about the way numbers work in EC, but if for example I use 2^124 aka 10000000000000000000000000000000 as one of my 2 base point keys, I just need to add and subtract other secret keys by the same numbers to see if I can find a few of their mutual friends, that is how I'd know which one is smaller than my target. But now I could be subtracting a -n key from my target and actually be adding their difference to my target, e.g adding -70 to +50 will result in -20. |
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Lol. This dogecoin @digaran actually trying to solve 125 bits which is currently the most hardest puzzle ever with some weird incorrect math HAHA.
He doesn't know that he needs to bruteforce at least 63 bits for a CHANCE that he gets the correct key using algorithm. Idk how subtraction and Multiplication would help u in such large range address to solve it lol.
it make absolutely no sense. Even if the pub key is known.
Dogecoin, really? To know that my math is incorrect you have to know the correct math, how else could you determine it is incorrect? But yes, LOL indeed 2^125 times.
All puzzles starts with "1". You can go through all keys, and subtract 2^N for each N-th key.
Maybe you mean they all start with 1 in binary code? Otherwise puzzle key ranges differ from one another.
Now you need to understand that amount of numbers or p.keys from 1 to 2^124 is equal to the amount between 2^124 : 2^125.
I do realize that, one needs to understand certain things before starting to work on these puzzles! I guess you are way off the map! 2 trillion keys? The actual number is 5 to 6 public keys in order to form 2 triangles and solve the key, more than that will confuse you, as it has confused me with a few million keys, I see hexadecimal characters wherever I look, after working on them for hours. Lol. |
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Ok let me explain, imagine #125 as number 750, if we subtract 500 (2^124, half of 2^125 or in our example 2^125 = 1000) if we subtract 500 from 750 we'd have 250, now all we have to do is to figure out what number is closest to our 250 (which we don't know it's exact value) and try to subtract that guessed number from our 1/3 of #125 puzzle, and the rest would be easy. Guessing the right number is very hard. And you can always guess it wrong, subtract 800 from 750, and you will get -50. Then, assuming that your N is 1000000, you will try to break 999950, instead of trying to break 750. Also, as far as I know, people already start from some offset, so they are not trying to scan keys outside of that 500 to 1000 range, instead they just pick for example 550 to 560 range, and try to scan only that, and that alone takes a lot of time. Though if you could explain how to figure out which key is greater or smaller than the other, that would be awesome! You cannot figure it out, based only on a single public key. It is always relative to the base point. For example, if you have (basePoint,yourPoint) pair, you could have "yourPointPrivKey" as some even number, for example 250. But it is only even in a relation between this particular base point! So, only "basePoint*250=yourPoint" gives you "250" as an even number. However, you could use (basePoint*2,yourPoint) pair instead, and then your private key would be odd, and equal to 125. See? Your private key is not universally assigned to your public key alone, it is always relative to your base point. And for that reason, if there would be any algorithm that could give you a private key, based on some public key, it would take at least a pair of public keys as arguments. For the same reason, there is no such thing as "which key is greater or smaller than the other". ECDSA is working like a clock, you can give me some two public keys, and then I could calculate "(firstPubKey+secondPubKey)/2", and then say that the distance between my point, and any of your two points, is identical. Can you tell me why is this happening? Ok, if I subtract this 02A1940074961CDF60CB2A0E7BC7157A7970B05469C58EEB5AD1C0462CE0C17811 from #125 puzzle, the result is : 02ed01ff219ed5c1afc12d991a82e3063ddcee1fd53b46f7cad52a0d87a7112aed And if I subtract this 0290ad85b389d6b936463f9d0512678de208cc330b11307fffab7ac63e3fb04ed4 which is actually 2^124 public key, if I subtract it from #125 puzzle the result is : 03ed01ff219ed5c1afc12d991a82e3063ddcee1fd53b46f7cad52a0d87a7112aed I know why this is happening though, because the first subtraction is actually inverse addition, the distance between the first public key and the second one is exactly the private key of #125. This is why I'm stuck and can't solve this puzzle, and this is one of the easiest of puzzles since we know exactly it starts with a 1. But I am close God willing, it should be solved soon. |
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BTC puzzle 66 :13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so
so far found :13zb18UKn2awwhpshZUjhLXZMcGbzC5soJ
the prefix and suffix nearly same
And what is the relation between an address and a private key? There are trillions of trillions of addresses in 66 bit range, there could be millions of such addresses there, let me save you a few months of useless pursuit, stop looking for addresses, instead focus on subtracting from the known public keys to reduce their bit range.
Why do you guys think up to 120 have been solved every 5 puzzle apart? Because solving exposed public key puzzles is much more easier than brute forcing other ones. I mean what else do you guys need? Just subtract 2^124 from the #125 public key and you'd have reduced 2/3 of #125 from it. Ok let me explain, imagine #125 as number 750, if we subtract 500 (2^124, half of 2^125 or in our example 2^125 = 1000) if we subtract 500 from 750 we'd have 250, now all we have to do is to figure out what number is closest to our 250 (which we don't know it's exact value) and try to subtract that guessed number from our 1/3 of #125 puzzle, and the rest would be easy. Focus only on subtraction, stop wasting your time on randomly searching for unknown keys! 😉 Good luck and happy hunting! |
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Guys I have a clue
Don’t find the words, also have close attention to the sentences found everywhere in the picture.
Every single sentence can be used as a clue or word reference.
What are you waiting for then? Go go go tiger, it's all yours, pay really close attention to the sentences and solve it to grab the prize.😉 |
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Do you guys happen to know where Alice lives by any chance? She has received so many bitcoins as a test model over the years, she is super rich definitely! I hope she is still single!🤣
Throwback is the comments of Satoshi and Gavin on sourceforge. So many of their oldest posts have been deleted by certain forum admin.
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You had to spoil the fun right? This is a challenge and people should use their minds to figure things out. Division is possible with any number, it's not limited to n/2, n/4 etc. Though if you could explain how to figure out which key is greater or smaller than the other, that would be awesome! Lol. Note that figuring that out is equal to the end of crypto currencies! Go big or don't go at all, right guys? 😉 |
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Not clicking on any links, so how do you get some of this new garbage coin? We should buy it or.......?
I also have a token, can't recall the name! Oh activity token, I believe it can replace all other cryptocurrencies, one because I believe it, two because I say so, three because I made it.
My advice, look for a better teacher.
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Good project.
PS: This is going better than I expected.
I just noticed this masterpiece! Can someone please explain "what is going better than expected" only after 2 hours of posting the ann thread? Lol. Sometimes we need to pay attention to the details! Anyways, I think it's time to move towards the next level, we must stay on roadmap schedule and start the ICO already.
I will soon launch Howey NFT token, it will work as a second layer of our scam great project, the HL aka Howey Lightning is also under development as we speak, you guys just stay tuned for more updates. [insert skeleton meme watching the screen waiting for updates here]. |
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A BIG lol for banks! We are trying to move past you guys and let you die off as a soon to be forgotten history of the past, but now they want to come with us into the future?
Why don't you print more money and keep it safe, huh! Just mind your own business, we'll do the same as we were!
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