Topic: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== - page 29. (Read 57113 times)

i have the same speed 7terrak per sec thats 7 quantillion per sec , thats why i told you my speed is slow im running bsgs on 8 core cpu, so i cant complete the entire range in 1 week or month, thats why  i ask for help here.



i understand, thanks for your help, i was thinking like that because in this post
https://bitcointalksearch.org/user/dextronomous-306088




they find same firsts caracteres hash160 of the puzzle 66  so logic its on that bit range 66
thats why when i found the same 4 hex of 110 bits of other 110 bit wallet i say that, im just asking the question and if in 2020 they cracked the 110 bit, if you have a good speed and you can search for the entire range in 1 hour you can try the public key, maybe im wrong but if its correct, you will earn smthing, you can substract the bit 125 but you wont find a hash same as you have in a specific bit range start with same digits, its luck i think when i found it but maybe im not wrong ,, if we found hashes starts with same 4 digits in all bit ranges then im wrong, that information i dont have, i dont know how the hash160 works so, i just want you to help guys maybe there is smthing.



i dont know 1 hour or 1 day, but if you want to have a test of that public key, thanks

No you aren't any closer

yea I could use iceland2k14 scalar program and make 16k keys and use albertobsd bsgs and load them all takes about 15 minutes to load and get 10 trillion/sec and do random on my 1070 but I am in Florida and the room is already too hot lol. my little 1030 is doing dp 10

I have. I have over 33GB of tame points (DP 25) in the 110 bit range lol. But I wouldn’t chase down a 4 character h160 that’s for sure lol.
Also, the “1 hour” entire range scan had me laughing 😂

I can do 1 quintillion/sec 1,000,000,000,000,000,000 but the range of 110 alone is 649037107316853453566312041152512 or
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or  7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?

You can shrink a search range via subtraction of a known public key.

If you shrink #125 down to 110 bits, you will create 2^15 new public keys. One of those will be guaranteed to be in the 110 bit range.

I do not understand the initial post of finding the first 4 hash160 characters...IMO, that does not help nor narrow down the search to that one pubkey.

How can you search for a 125 bit range public key in 109-111 bit range? And why would you care about the first 4 characters of a hash160 of another random address?
Please explain the magic here.😉

hi guys, im working on puzzle 125 for a long time, today, i have substracted the public key (im always test and substract by random) of 125 to a smaller range randomly i found a substracted key his hash160 start with the same 4 lettres as a hash160 of an adress in the range 110 bit, so, am i close to find it ? if i search this substracted public key in 110 bit range will i found it ? the probleme im runing 7TerraK per sec its slow, and if anyone can help we will split the reward of the puzzle 125,
02a0397a2b74177e0f101ab1cbe9d560ac384c01fd69d7e19593d1ab3155f9ad73
2000000000000000000000000000:3fffffffffffffffffffffffffff
if you have better speed search from 1f00000000000000000000000000:7fffffffffffffffffffffffffff from 109 to 111 bits,

if you have good speed and can finish this range in 1hour, you wont lose nothing, we will split the reward thank you

Guys I'm telling you, Satoshi solved the puzzle, few days after I pointed him to this thread, he did it with his super GPU computer, and I bet he is busy with the next one #125. I just didn't expect him to take the loots alone, we were partners.🤣 at least let me loot the garbage coins. Lol.

Not a 32 times bigger but √32 like ~5.65. this not full space search, this birthday collision check.
You can compare 'expected operations' for
115 2^58.1   
120 2^60.61
125 2^63.10
130 2^65.61
...

Nice to see you too :-)
Despite the passage of time and the fact that people who are more active in the subject know each other only from here - there is a smile on your face knowing that everything is fine with you :-)
It is also certain that no matter how much time passes - the solution of the next thresholds will certainly take place, whether it is due to the fact that the technology will develop further or because the value of assets, despite being only 4% more - will be much higher in terms of physical currency than today. Time will tell. I also believe that we will get to know the finder of solution 120 and how he did it, how much time he needed for it, or at least the private key to complete the list of keys of solved addresses. Greetings!

Hi @zielar, nice to see you here


As we know the puzzle 125 is 32 times bigger than puzzle 120 but prize is only a 4% bigger, i doubt about that such "magic" tool exists. But we don't have way to know.

If that exists it should be a mathematical approach.


We already know that puzzles a random and there is not math relationship between them

I made some computations. Used Python 3. I was curious if the next puzzle can be ax+b=y.
d1, d2, d3 are corresponding private keys from puzzle. Needed to compute a and b.

First script for puzzle 66:


Second script for puzzle 120:


The output is however strange, because a=b!

This means the next puzzle is multiplied and added to it the same value, I mean the multiplicator and later adding a number are the same.

First script gives:


Second script gives:


What can mean that just raising the value(s) of a and b can give next puzzle private key!

There are various reasons why we still don't have the key:
1. The finder wishes to remain anonymous
2. Someone has a tool to find #125 soon
3. The withdrawal was made by someone outside the forum
4. Someone has some other purpose in it, so as not to reveal themselves.

It is certainly puzzling that BCH and BTG have not been paid out from this wallet so far, while the payment was made to the address without any other transaction before, nor was it sent further - which also raises the suspicion that the finder could use some tool that worked for a long time -> found the right key and made the withdrawal, but the initiator himself still does not know that he found solution #120. A lot of unknowns, but I believe that eventually someone will somehow provide a solution, because the solution #120 without indicating the key to the public will simply be a blow to the whole history of these puzzles.

It is a real shame that the solver of #120 still didn't give us the private key.
Dude, if you're out there, please find a way to communicate it

Do you expected a new Cicada 3301 ?

I really want some challenges like that, but diffucults please.

You think? Lol.


Convert each of them into hash160 and look at the last hex characters, change their values in an orderly fashion to be amazed by the beauty of order amongst the chaos.
Have you seen websites providing an online(totally unsafe) service where you could just insert your desired private key in hex format and see all the corresponding properties such as addresses, public/ private keys even for many coins etc? It would be great if we could build an algorithm specifically designed to do the same for addresses only, this way we wouldn't need to count from 1 up to 2^256, we'd only need to represent 2^160.8 (more or less) total number of addresses with hexadecimal values, so when you search for the address #0 you'd get for #1 you'd get and so on until you reach the end of the address range. This is just for entertainment and would have no use case for storing bitcoins because you don't know the private keys, you might only know the private keys which you had generated before, i.e, using a wallet etc.

If we could manage to do something like that, imagine the awe of the beholder! Being present at such a mesmerizing sight!

Just don't forget to include my address on the footer of the site for donations, make sure to name me as the cofounder . Lol.
And remember, no offensive words, no sacrilegious words, we wouldn't wanna upset the boss up stairs, so we should skip 4 to 8 percent of addresses.

---

This message is for @satoshi aka the author of this challenge, "Live long in prosperity under the bluish violet dome".

Now could you create a new challenge with riddles and equations for us to seek and solve please? We're bored honestly and it could light up the mood in crypto-community ushering new methods and never ending possibilities. Though not something that someone could solve them all with one go, maybe make them in episode where each riddle's answer is just a small piece of the greater puzzle. ❤👏💯

the same I am impressed because only those who ignore the difficulty do not express themselves while it remains a feat / an event !!
I just say congratulations to the one who succeeded but it would be nice if he at least gave the key!
you zielar  you said how you succeeded congratulations to you

+1 I am also interested but I think it's just playing with the ripemd160 hash

Congratulations to the person who found key #120.
I am impressed and I will be grateful if this person publicly discloses the private key and provides information on how he found the key, or provided this information in a private message if he wants to remain anonymous, and that I complete this information in the table in the main post on his behalf.