Bitcoin puzzle transaction ~32 BTC prize to who solves it

[mabdlmonem]

Anyone know how to get the private key for 58.5?

Sure, nothing could be easier

Just generate it from decimal 58.5 or HEX 3a.8  

Jokes aside - if you want to derive a private key from a floating-point decimal for experimental or educational purposes, you could devise your own method. Bitcoin private key from a floating-point decimal isn't standard practice, as Bitcoin private keys are typically generated using cryptographic algorithms that rely on random numbers.

good luck in whatever you're trying to achieve

I know already how to get private key that has floating 0.5 , I just curious if anyone got it

[1714258921]

1714258921

[mcdouglasx]

Anyone know how to get the private key for 58.5?

Code:

import bitcoin

target= 585

Div=10

N=115792089237316195423570985008687907852837564279074904382605163141518161494337

a=bitcoin.inv(Div,N)

b= target*a % N

print("target r:",target/Div)
print("pk:",b)

[brainless]

Anyone know how to get the private key for 58.5?

Code:

import bitcoin

target= 585

Div=10

N=115792089237316195423570985008687907852837564279074904382605163141518161494337

a=bitcoin.inv(Div,N)

b= target*a % N

print("target r:",target/Div)
print("pk:",b)

58.5

PK 0.5
57896044618658097711785492504343953926418782139537452191302581570759080747169

pk 58
58

58.5
57896044618658097711785492504343953926418782139537452191302581570759080747169
+
58
=
57896044618658097711785492504343953926418782139537452191302581570759080747227

in hex
0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20db

[mabdlmonem]

Anyone know how to get the private key for 58.5?

Code:

import bitcoin

target= 585

Div=10

N=115792089237316195423570985008687907852837564279074904382605163141518161494337

a=bitcoin.inv(Div,N)

b= target*a % N

print("target r:",target/Div)
print("pk:",b)

58.5

PK 0.5
57896044618658097711785492504343953926418782139537452191302581570759080747169

pk 58
58

58.5
57896044618658097711785492504343953926418782139537452191302581570759080747169
+
58
=
57896044618658097711785492504343953926418782139537452191302581570759080747227

in hex
0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20db

What about 58.51?)

[mcdouglasx]

Anyone know how to get the private key for 58.5?

Code:

import bitcoin

target= 585

Div=10

N=115792089237316195423570985008687907852837564279074904382605163141518161494337

a=bitcoin.inv(Div,N)

b= target*a % N

print("target r:",target/Div)
print("pk:",b)

58.5

PK 0.5
57896044618658097711785492504343953926418782139537452191302581570759080747169

pk 58
58

58.5
57896044618658097711785492504343953926418782139537452191302581570759080747169
+
58
=
57896044618658097711785492504343953926418782139537452191302581570759080747227

in hex
0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20db

What about 58.51?)

5851/100

Code:

import bitcoin

target= 5851

Div=100

N=115792089237316195423570985008687907852837564279074904382605163141518161494337

a=bitcoin.inv(Div,N)

b= target*a % N

print("target r:",target/Div)
print("pk:",b)

[brainless]

Anyone know how to get the private key for 58.5?

Code:

import bitcoin

target= 585

Div=10

N=115792089237316195423570985008687907852837564279074904382605163141518161494337

a=bitcoin.inv(Div,N)

b= target*a % N

print("target r:",target/Div)
print("pk:",b)

58.5

PK 0.5
57896044618658097711785492504343953926418782139537452191302581570759080747169

pk 58
58

58.5
57896044618658097711785492504343953926418782139537452191302581570759080747169
+
58
=
57896044618658097711785492504343953926418782139537452191302581570759080747227

in hex
0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20db

What about 58.51?)

5851/100

Code:

import bitcoin

target= 5851

Div=100

N=115792089237316195423570985008687907852837564279074904382605163141518161494337

a=bitcoin.inv(Div,N)

b= target*a % N

print("target r:",target/Div)
print("pk:",b)

If your script giving you result, then ok, other word, it's simple solutions, for play with every floating points, I am not at laptop at this time, later explain if your script not working...

[GR Sasa]

I'm very exciting to see the results of @mcdouglasx, as he said that he developed a new method and will solve #130 before the end of January.

It's 10 days left, so ! let's go!

Today is the day for mcdouglasx to reveal his secret?

[WanderingPhilospher]

I'm very exciting to see the results of @mcdouglasx, as he said that he developed a new method and will solve #130 before the end of January.

It's 10 days left, so ! let's go!

Today is the day for mcdouglasx to reveal his secret?

Sometimes calculations are off, could be +/- hours or days. And that's barring no down time via power outages, equipment failure, etc.

Hopefully he is close.

[albert0bsd]

Well, we're waiting  

puzzle 66 is still there @mcdouglasx

[GR Sasa]

Well, we're waiting  

puzzle 66 is still there @mcdouglasx

@mcdouglasx was not targeting 66, but 130 with his new mysterious invention

[albert0bsd]

Oh thats right in any case puzzle 130 is still there:

1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua

[Unplugged Taste]

Based on mcdouglas' post-claim comments, I can feel his frustration when he encountered EC's punch in face! He's been tricked and overpowered by the Illusion-of-Mastery, which made him post such a big claim. I am not demotivating him, no no!! I am happy about him that he is learning good maths here. Trying to divide like 58.5 and figuring what half would be, good good, great!!!! Majority of guys here possibly have been through lots of similar curve punches here. Someone rightly figured out here that EC cannot be broken from inside, whatsoever!!!! The only thing attached to it from outside is its point at infinity, WHICH CAN'T BE TOUCHED DUE TO BEING AT INFINITY SOMEHOW I GUESS, OR MAY BE IDK!!
Whatever secret mcdouglasx has, I am sure many of guys here have already tried & tested and accordingly slapped in ass by EC! SORRY FOR THAT!!
The possibilities of EC being broken by any BTC puzzle enthusiast are very slim! Because there are giant research & security institutes (I know few of them personally visited as well) which are rigorously and systematically approaching towards the solutions to ECDLP.
I bet Satoshi must have a good laugh today, Cheers, it is 1st of Feb 24!!!

[albert0bsd]

...
I am not demotivating him, no no!! I am happy about him that he is learning good maths here.
...
Whatever secret mcdouglasx has, I am sure many of guys here have already tried & tested and accordingly slapped in ass by EC! SORRY FOR THAT!!

I agree with you, i hope to know what is he doing.

[Cryptoman2009]

EC is a spiral curve like spring

try this 3d graph generator:



import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np

# parameters
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
a = 0
b = 7
Gx = 55066263022277343669578718895168534326250603453777594175500187360389116729240
Gy = 32670510020758816978083085130507043184471273380659243275938904335757337482424

# start + end decimal key
start = int(input("input start key: "))
end = int(input("input end key: "))

# generation
x = []
y = []
z = []
for i in range(start, end + 1):
    # point calculation index i
    Px = i * Gx
    Py = i * Gy
    x.append(Px % p)
    y.append(Py % p)
    z.append(i)

# 3D chart
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(x, y, z, c='r', linewidth=2)

# add points on chart
for i, (xi, yi, zi) in enumerate(zip(x, y, z)):
    ax.text(xi, yi, zi, str(i), color='blue')

ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Index')
ax.set_title('point on elliptic curve secp256k1 (3D graph)')

plt.show()

[Cryptoman2009]

similitudes with prime numbers...

https://www.youtube.com/watch?v=pnYkK6PqUro&ab_channel=voltagedrop

[mcdouglasx]

I have developed a new method!
I have studied it 100% and it works.
My calculations tell me that before the end of January 2024 I will have unlocked puzzle #130 (if someone else doesn't solve it before).
For registration, I will send it to this address BTC bc1qxs47ttydl8tmdv8vtygp7dy76lvayz3r6rdahu

This message will not have edits for its validity.

If you want to question it, do it on February 1, 2024 if I don't send it to that address.

happy new year in advance, see you in February!, if life allows me.

blessings for all.

In summary, my search is 70%, I do not intend to take up your time waiting for a precise date, I do not take external factors into the equation when calculating the time, I live in an underdeveloped country where electrical failures are frequent, I have an i5 4GB ddr3, although I want to, I can't go faster, it is what it is, I got excited with the calculations and completely omitted the external factors, I have earned the criticism and I understand it, but it does not offend me, because my path continues.

Based on mcdouglas' post-claim comments, I can feel his frustration when he encountered EC's punch in face! He's been tricked and overpowered by the Illusion-of-Mastery, which made him post such a big claim. I am not demotivating him, no no!! I am happy about him that he is learning good maths here. Trying to divide like 58.5 and figuring what half would be, good good, great!!!! Majority of guys here possibly have been through lots of similar curve punches here. Someone rightly figured out here that EC cannot be broken from inside, whatsoever!!!! The only thing attached to it from outside is its point at infinity, WHICH CAN'T BE TOUCHED DUE TO BEING AT INFINITY SOMEHOW I GUESS, OR MAY BE IDK!!
Whatever secret mcdouglasx has, I am sure many of guys here have already tried & tested and accordingly slapped in ass by EC! SORRY FOR THAT!!
The possibilities of EC being broken by any BTC puzzle enthusiast are very slim! Because there are giant research & security institutes (I know few of them personally visited as well) which are rigorously and systematically approaching towards the solutions to ECDLP.
I bet Satoshi must have a good laugh today, Cheers, it is 1st of Feb 24!!!

no, I have not broken ecc, I never said it, otherwise, it would not take more than a month to find a key, breaking ecc is creating a formula that allows you to obtain the private key in an instant, this is not it, it is just a system of search faster than the already known ones, and it is easy to verify if a search code works or not, so I am not wrong.

edit:
It has not yet been proven that the ECDLP is computationally intractable, therefore in mathematics we cannot consider something impossible that you cannot prove to be impossible.

An algorithm has not been found that can solve the ECDLP in polynomial time, but it cannot be said that it does not have such a solution.

You can solve a problem in two ways:
1= solution.
2= no solution.
If you are going to come and tell me that ECDLP is unbreakable in polynomial time, publish the formula that shows that this is the case, otherwise don't see something as impossible just because it is difficult, if all humans had that way of thinking, society would It was still in the prehistoric era.

Yes, there are giant institutions studying this, this does not mean that you have to feel less than them, but that you can learn from them and overcome them.

And I conclude that if you try and fail, you will have learned something, if you consider yourself incapable you only gain ignorance.

[nomachine]

EC is a spiral curve like spring

Nice graphic. It won't help us much with the methods we're using here.

[Cryptoman2009]

EC is a spiral curve like spring

Nice graphic. It won't help us much with the methods we're using here.

can open your mind to new ideas such as keymath and keydivision..

[unamic]

How can i Use VanitySearch for Windows with .exe file for the Puzzle nr66? BitCrack dont work for me nothing happens after i enter the commands. Kangaroo and Keyhunt is to high for me to understand.

If i enter this command with Vanitysearch
./VanitySearch.exe -stop -t 0 -bits 66 -start 0000000000000000000000000000000000000000000000020000000000000000 -gpu 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so

i get this message:

Unexpected -bits argument

[Bill buffalo]

Theoretically speaking let's say I had a quantum computer with the computational power of a 100 qubit what algorithm would I use on qisqit python to run the program that would find the private key