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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 407273 times)
The_Prof
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August 28, 2026, 04:45:12 PM
 #13821

Here is a tiny list of subranges scanned to 100% competition. In case anyone was scanning them, don't bother.

7e0f0
7f0f0
6f0f0
6e0f0
5e0f0
5f0f0
4e0f0
4f0f0

Why did you scanned those? You lost a lot of time for nothing...the first 7 of key do not have double 0 or tripple 0 Smiley
You will see when the key will be found Smiley)


A few months ago some poster swore up and down she randomly found the key but her computer crashed and she couldn't retrieve it in time. She said she caught a glimpse of the first few digits before the crash.

I offered to help her and just send over all of the btc if found. 

Turns out it wasn't true. Shocking.

Anyway, those were the keys and the variations of the beginning of the key she swore to God she found.

While it won't change much for me. I think thanks are seldom shown in this place. So thank you, I am sure you saved many people quite some time.

Look over there...
analyticnomad
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August 28, 2026, 04:59:56 PM
 #13822


While it won't change much for me. I think thanks are seldom shown in this place. So thank you, I am sure you saved many people quite some time.
[/quote]

Hey no worries. I figured however small, it could help people avoid those areas.
puzzle_72_worker
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August 28, 2026, 06:35:52 PM
 #13823

Why nobody took in count the length of the key? only 2-3 keys were with 33 length, and the rest are 34 length.
If you exclude the 33 length from puzzle 71, you have 20% less, so the difficulty decrease.

This is nothing more than a heuristic. It does not guarantee

I think you didn't do the math correct...if you remove them before calculating, you reduce the difficulty.
So 20% is quite a lot in speed.
eggsylacer
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August 28, 2026, 08:28:34 PM
 #13824

Why nobody took in count the length of the key? only 2-3 keys were with 33 length, and the rest are 34 length.
If you exclude the 33 length from puzzle 71, you have 20% less, so the difficulty decrease.

This is nothing more than a heuristic. It does not guarantee

I think you didn't do the math correct...if you remove them before calculating, you reduce the difficulty.
So 20% is quite a lot in speed.

You can subtract 2^n and check whether the key exists within a certain range, and this will result in the same reduction in complexity.
But once again, the range remains the same: from 2^k to 2^(k+1) - 1. Accordingly, the complexity will remain the same - these are all heuristics.
cicada 3301
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Today at 02:29:58 AM
Last edit: Today at 08:26:52 AM by cicada 3301
 #13825

So, what's been accomplished is over 18 program iterations, 14 physical and mathematical iterations. 17 patterns, 22 constants. I'll explain the essence of the problem: the k0 primary wallet, from which the k256 originates. Why did I say this is the same problem as the Einstein oscillator? The essence is that the k0 is not visible. The reason is parallax, and it's impossible to go to the side, beyond the angle of the gravitational lens. Therefore, there are only k256 wallets, only visible, 83 of which are open. If we transfer this to cosmology, these are Einstein arcs or the Einstein cross.
"At this point, I won't answer which option is here."
The essence of the task is to transfer and create a control analogue, but in the blockchain, since it is obvious that k0 is a specific point in space and time. And the task is simply to restore the primary wallet, and for this, we need to create a mathematical model, which is what I did. I'll tidy up the program today and start opening.
This is a game with topology, as well as functions.
This program can also play the role of predicting who will be a terrorist, since a terrorist is the same as k0. And for other topics as well.
scamrevealer
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Today at 03:21:45 AM
 #13826

...

This has been posted already. Please use the search function.
cicada 3301
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Today at 08:22:12 AM
 #13827

In the course of experiments, I even created my own kangoon gpu and kangoon cpu C11 invariants and Cobol as well as a calculator for operating numbers, up to 16k values container size. Working is 4k for a single number.
detechs
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Today at 09:19:26 AM
 #13828

In the course of experiments, I even created my own kangoon gpu and kangoon cpu C11 invariants and Cobol as well as a calculator for operating numbers, up to 16k values container size. Working is 4k for a single number.

Your feeding the threads info into an ai. And it's telling you it can solve it and then you're coming here and tell us without actually solving it. How do I know? Cause I also scraped the forum and anylise everyone's posts, and yours shows obvious patterns lol. Have you closed the circle yet? Cause if not you'll be chasing your tail like everyone else
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