Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 97.

Ovixx

newbie

Activity: 30

Merit: 0

Quote from: Waveilona on April 18, 2024, 05:56:30 AM

Hi puzzle solvers , I am a user Waveilon also i have telegram @waveilon ,
So i am here to share with you my opinions ,
the first transaction from puzzle #66 will be transfered to this address 1EEs9PydaesijB9tpbFXRXpbMEuTHELSTH BTC
....................................................
Here is my request for you : if i find the key and someone could take my reward, could you send me my reward to this address: 1EEs9PydaesijB9tpbFXRXpbMEuTHELSTH BTC
You can take it from like 160 puzzle !
...................................................
Thanx all

What a good joke, I almost laughed. I have never heard anything more stupid!
There is a saying in my country, Don't sell the bear's fur from the forest!

saatoshi_falling

newbie

Activity: 12

Merit: 0

Like Bitcoin's price, this thread pumps sometimes with interesting info for a few pages, and then other times it dumps a few pages into red.
The past few pages are pretty red.
k3ntINA finally making public his Zodiac circle .gif that cracks #66 using Adobe Illustrator.
kachev87 with no proper programing skills saves some time for the would be c++ programmer to convert his ChatGTP python code to run in full GPU mode.
joseperal finding puzzle #30 is already emptied.
Waveilona who finally solved the puzzle but is holding it hostage.

averagetoaster

newbie

Activity: 6

Merit: 0

I also tried to predict #66 in a similar way but the ranges I got are still too big to do anything useful. I don't have high end hardware to test the predictions either, and the ranges I got are around 40% of the total range.
Factoring in that most likely the beginning and end of the ranges have been searched already, and doing other basic assumptions, you can reduce the range down but it's still too big.
Furthermore, now I am trying my luck on #130 because why not. It is very unlikely that #66 will get in my hands (or yours) so I just gave up on that.

Waveilona

newbie

Activity: 1

Merit: 0

Hi puzzle solvers , I am a user Waveilon also i have telegram @waveilon ,
So i am here to share with you my opinions ,
the first transaction from puzzle #66 will be transfered to one of my BTC
also i will post later here or tg groups tx id which can confirm that i solved it!
I want that the puzzle was honestly paid , That's why i am here.
All we know if public key that address broadcasts anybody can take it and find the key with bsgs or kangaroo
I don't want someone interferered to my transaction.

And here is the message to puzzle creator or investor who increased the reward to this puzzle
I worked on this project 6 months and I want to get my reward honestly,
Even miners has information about this puzzle they can just take my reward ,
Here is my request for you : if i find the key and someone could take my reward, could you send me my reward to this address: one of my BTC
You can take it from like 160 puzzle !
In my opinion if i can take my reward others can search for another puzzle without afraiding that kangaroo or bsgs even miners!
I think I could correctly express my opinion
Thanx all

zahid888

member

Activity: 275

Merit: 20

the right steps towerds the goal

Quote from: CY4NiDE on April 18, 2024, 04:50:04 AM

Quote from: jacky19790729 on April 18, 2024, 03:35:13 AM

use this code to "predict"

Hexadecimal: 0x11f774e94c1ec000 # puzzle 62 0x363d541eb611abee
Hexadecimal: 0x7d556bf6f89d2c00 # puzzle 63 0x7cce5efdaccf6808
Hexadecimal: 0xe7655f0b50acf800 # puzzle 64 0xf7051f27b09112d4
Hexadecimal: 0x1a78fd44662532000 # puzzle 65 0x1a838b13505b26867
Hexadecimal: 0x290860e0f31602000 # puzzle 66 ?

Yep. Some results get kinda close, others not much.

Could accuracy increase as we append new keys to the sequence?

Can we improve this script somehow?

Code:

spline_rep = splrep(x_values_known, sequence_decimal, k=2)

I've messed with other values for k but 2 seems to yield better results.

Index 3: Predicted = 0xd, Actual = 0x8, Error = 5.0
Index 4: Predicted = 0x5, Actual = 0x15, Error = 15.83333333333334
Index 5: Predicted = 0x30, Actual = 0x31, Error = 0.2857142857142918
Index 6: Predicted = 0x5c, Actual = 0x4c, Error = 16.04901960784312
Index 7: Predicted = 0x63, Actual = 0xe0, Error = 124.75357443229606
Index 8: Predicted = 0x202, Actual = 0x1d3, Error = 47.404329004328815
Index 9: Predicted = 0x31c, Actual = 0x202, Error = 282.866702978386
Index 10: Predicted = 0x13c, Actual = 0x483, Error = 838.5322535426649
Index 11: Predicted = 0x9e5, Actual = 0xa7b, Error = 149.13061026670766
Index 12: Predicted = 0x1403, Actual = 0x1460, Error = 92.41323240818747
Index 13: Predicted = 0x2241, Actual = 0x2930, Error = 1774.144396004227
Index 14: Predicted = 0x4a1b, Actual = 0x68f3, Error = 7895.604944862127
Index 15: Predicted = 0xd8f3, Actual = 0xc936, Error = 4029.671642258196
Index 16: Predicted = 0x14745, Actual = 0x1764f, Error = 12041.382349990992
Index 17: Predicted = 0x2784f, Actual = 0x3080d, Error = 36797.025408181595
Index 18: Predicted = 0x59719, Actual = 0x5749f, Error = 8826.371450069128
Index 19: Predicted = 0x8b61a, Actual = 0xd2c55, Error = 292410.36592774664
Index 20: Predicted = 0x1af328, Actual = 0x1ba534, Error = 45579.312763758004
Index 21: Predicted = 0x30fdc8, Actual = 0x2de40f, Error = 203193.17374296952
Index 22: Predicted = 0x4360b7, Actual = 0x556e52, Error = 1183130.437051029
Index 23: Predicted = 0x955cee, Actual = 0xdc2a04, Error = 4640021.909114862
Index 24: Predicted = 0x1ce3cea, Actual = 0x1fa5ee5, Error = 2892282.0998125
Index 25: Predicted = 0x3b79f62, Actual = 0x340326e, Error = 7826676.155909941
Index 26: Predicted = 0x4992721, Actual = 0x6ac3875, Error = 34804051.331749916
Index 27: Predicted = 0xc998ee1, Actual = 0xd916ce8, Error = 16244230.842530549
Index 28: Predicted = 0x181a56c4, Actual = 0x17e2551e, Error = 3670438.391939521
Index 29: Predicted = 0x25955523, Actual = 0x3d94cd64, Error = 402618432.6683471
Index 30: Predicted = 0x82c6e33c, Actual = 0x7d4fe747, Error = 91683829.12309027
Index 31: Predicted = 0xd6239bac, Actual = 0xb862a62e, Error = 499185022.823071
Index 32: Predicted = 0xe9b22d07, Actual = 0x1a96ca8d8, Error = 3216669648.649395
Index 33: Predicted = 0x371532851, Actual = 0x34a65911d, Error = 653104948.3604355
Index 34: Predicted = 0x5949f8bc7, Actual = 0x4aed21170, Error = 3855448663.1672974
Index 35: Predicted = 0x5af449ee4, Actual = 0x9de820a7c, Error = 17972423575.53308
Index 36: Predicted = 0x1391413025, Actual = 0x1757756a93, Error = 16210213485.866455
Index 37: Predicted = 0x2dbf7270dd, Actual = 0x22382facd0, Error = 49513939981.24811
Index 38: Predicted = 0x2886559a9b, Actual = 0x4b5f8303e9, Error = 149672520013.72742
Index 39: Predicted = 0x98c8113e3b, Actual = 0xe9ae4933d6, Error = 347459810714.7615
Index 40: Predicted = 0x20b05d20b7e, Actual = 0x153869acc5b, Error = 788113342243.4756
Index 41: Predicted = 0x1696ccb8e1d, Actual = 0x2a221c58d8f, Error = 1343066079089.604
Index 42: Predicted = 0x50b26b4d29f, Actual = 0x6bd3b27c591, Error = 1864358884081.7373
Index 43: Predicted = 0xdef4cb268c7, Actual = 0xe02b35a358f, Error = 83326651591.11523
Index 44: Predicted = 0x1875de8190b7, Actual = 0x122fca143c05, Error = 6898060186802.75
Index 45: Predicted = 0x1230f00b831f, Actual = 0x2ec18388d544, Error = 31407275332132.58
Index 46: Predicted = 0x689e837c30da, Actual = 0x6cd610b53cba, Error = 4636639038432.0
Index 47: Predicted = 0xcd26aa6013f0, Actual = 0xade6d7ce3b9b, Error = 34358976174165.78
Index 48: Predicted = 0xec974b5f9e34, Actual = 0x174176b015f4d, Error = 148984356258072.8
Index 49: Predicted = 0x2d6a754d3f7ab, Actual = 0x22bd43c2e9354, Error = 187823628313687.25
Index 50: Predicted = 0x2b7ce396f3181, Actual = 0x75070a1a009d4, Error = 1293723206998098.5
Index 51: Predicted = 0x11abcd83d36e18, Actual = 0xefae164cb9e3c, Error = 757478132797404.0
Index 52: Predicted = 0x18b4f3390decf8, Actual = 0x180788e47e326c, Error = 190672196844172.0
Index 53: Predicted = 0x2258a643d0ea28, Actual = 0x236fb6d5ad1f43, Error = 306834910754076.0
Index 54: Predicted = 0x31634c7beb9390, Actual = 0x6abe1f9b67e114, Error = 1.6143936485346692e+16
Index 55: Predicted = 0xf7c9efde77f4e0, Actual = 0x9d18b63ac4ffdf, Error = 2.55276090216256e+16
Index 56: Predicted = 0xaaf00881ca1cd0, Actual = 0x1eb25c90795d61c, Error = 9.013109354212386e+16
Index 57: Predicted = 0x48bd5d0ecad0f40, Actual = 0x2c675b852189a21, Error = 1.2761382323882934e+17
Index 58: Predicted = 0x2e13f10ba558040, Actual = 0x7496cbb87cab44f, Error = 3.175539853443205e+17
Index 59: Predicted = 0x10359b6a07b1e600, Actual = 0xfc07a1825367bbe, Error = 3.2969207850953344e+16
Index 60: Predicted = 0x1c178523462fa100, Actual = 0x13c96a3742f64906, Error = 5.984454014555484e+17
Index 61: Predicted = 0x11f774e94c1ec000, Actual = 0x363d541eb611abee, Error = 2.6137405792622295e+18
Index 62: Predicted = 0x7d556bf6f89d2c00, Actual = 0x7cce5efdaccf6808, Error = 3.801338671410483e+16
Index 63: Predicted = 0xe7655f0b50acf800, Actual = 0xf7051f27b09112d4, Error = 1.1258296599663145e+18
Index 64: Predicted = 0x1a78fd44662532000, Actual = 0x1a838b13505b26867, Error = 4.753071358864998e+16

Code:

import numpy as np
from scipy.interpolate import splrep, splev
from decimal import Decimal

sequence = [1, 3, 7, 8, 21, 49, 76, 224, 467, 514, 1155, 2683, 5216, 10544, 26867, 51510, 95823, 198669, 357535, 863317, 1811764, 3007503, 5598802, 14428676, 33185509, 54538862, 111949941, 227634408, 400708894, 1033162084, 2102388551, 3093472814, 7137437912, 14133072157, 20112871792, 42387769980, 100251560595, 146971536592, 323724968937, 1003651412950, 1458252205147, 2895374552463, 7409811047825, 15404761757071, 19996463086597, 51408670348612, 119666659114170, 191206974700443, 409118905032525, 611140496167764, 2058769515153876, 4216495639600700, 6763683971478124, 9974455244496707, 30045390491869460, 44218742292676575, 138245758910846492, 199976667976342049, 525070384258266191, 1135041350219496382, 1425787542618654982, 3908372542507822062, 8993229949524469768, 17799667357578236628, 30568377312064202855]
initial_points = 3
results = []
for i in range(initial_points, len(sequence)):
    x_values_known = np.arange(i)
    sequence_decimal = [Decimal(value) for value in sequence[:i]]
    spline_rep = splrep(x_values_known, sequence_decimal, k=2)
    predicted_next_number = splev(i, spline_rep)
    actual_value = sequence[i]
    predicted_hex = hex(int(predicted_next_number))
    actual_hex = hex(actual_value)
    results.append({
        'index': i,
        'predicted': predicted_hex,
        'actual': actual_hex,
        'error': abs(float(predicted_next_number) - actual_value)
    })
for result in results:
    print(f"Index {result['index']}: Predicted = {result['predicted']}, Actual = {result['actual']}, Error = {result['error']}")

Everything is useless, I have tried many predictions like this, none of them worked Sad

NotATether

legendary

Activity: 1568

Merit: 6660

bitcoincleanup.com / bitmixlist.org

Quote from: saatoshi_falling on April 04, 2024, 06:24:50 AM

There's simply no feasible way to withdraw the funds on lower end puzzles like #66. It will be snatched up by bots. Not maybe, but it's 100% guaranteed. There will be hundreds of withdrawal transactions with varying fees all battling each other. You will simply be left in the dust.

I don't understand how it is possible for that to happen as long as the solver does not disclose the private key. I mean, #64 and #65 both have unknown keys, right?

CY4NiDE

member

Activity: 43

Merit: 10

Quote from: jacky19790729 on April 18, 2024, 03:35:13 AM

use this code to "predict"

Hexadecimal: 0x11f774e94c1ec000 # puzzle 62 0x363d541eb611abee
Hexadecimal: 0x7d556bf6f89d2c00 # puzzle 63 0x7cce5efdaccf6808
Hexadecimal: 0xe7655f0b50acf800 # puzzle 64 0xf7051f27b09112d4
Hexadecimal: 0x1a78fd44662532000 # puzzle 65 0x1a838b13505b26867
Hexadecimal: 0x290860e0f31602000 # puzzle 66 ?

Yep. Some results get kinda close, others not much.

Could accuracy increase as we append new keys to the sequence?

Can we improve this script somehow?

Code:

spline_rep = splrep(x_values_known, sequence_decimal, k=2)

I've messed with other values for k but 2 seems to yield better results.

jacky19790729

jr. member

Activity: 82

Merit: 8

Quote from: CY4NiDE on April 17, 2024, 05:26:06 PM

Next key: 3.0520846598475555e+19
Hex: 0x1a78fd44662532000
Is this jesus toast or could we use it to at least try and narrow down the first 2 characters of #66?
Your insights are much appreciated.

use this code to "predict"

Hexadecimal: 0x11f774e94c1ec000 # puzzle 62 0x363d541eb611abee
Hexadecimal: 0x7d556bf6f89d2c00 # puzzle 63 0x7cce5efdaccf6808
Hexadecimal: 0xe7655f0b50acf800 # puzzle 64 0xf7051f27b09112d4
Hexadecimal: 0x1a78fd44662532000 # puzzle 65 0x1a838b13505b26867
Hexadecimal: 0x290860e0f31602000 # puzzle 66 ?

CY4NiDE

member

Activity: 43

Merit: 10

Well, since we are here I'd like to ask what y'all think of this:

Code:

import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import splev, splrep
from decimal import Decimal

sequence = [
1, 3, 7, 8, 21, 49, 76, 224, 467, 514, 1155, 2683, 5216, 10544, 26867, 51510, 95823, 198669, 357535, 863317, 1811764, 3007503, 5598802, 14428676, 33185509, 54538862, 111949941, 227634408, 400708894, 1033162084, 2102388551, 3093472814, 7137437912, 14133072157, 20112871792, 42387769980, 100251560595, 146971536592, 323724968937, 1003651412950, 1458252205147, 2895374552463, 7409811047825, 15404761757071, 19996463086597, 51408670348612, 119666659114170, 191206974700443, 409118905032525, 611140496167764, 2058769515153876, 4216495639600700, 6763683971478124, 9974455244496707, 30045390491869460, 44218742292676575, 138245758910846492, 199976667976342049, 525070384258266191, 1135041350219496382, 1425787542618654982, 3908372542507822062, 8993229949524469768, 17799667357578236628
]

x_values_known = np.arange(len(sequence))
sequence_decimal = [Decimal(value) for value in sequence]
spline_rep = splrep(x_values_known, sequence_decimal, k=2)
extended_x_values = np.arange(len(sequence) + 1)
predicted_next_number = splev(extended_x_values[-1], spline_rep)
predicted_next_number_hex = hex(int(predicted_next_number))
plt.plot(x_values_known, sequence_decimal, label='sequence')
plt.plot(
    extended_x_values,
    splev(extended_x_values, spline_rep),
    label='Recreated Sequence',
    linestyle='dashed'
)
plt.scatter(
    extended_x_values[-1],
    float(predicted_next_number),
    color='red',
    marker='o',
    label='Predicted Next Number'
)
plt.legend()
plt.xlabel('Index')
plt.ylabel('Value')
plt.title('Original vs. Recreated sequence with Prediction')
plt.show()
print(f"Next key: {predicted_next_number}")
print(f"Hexadecimal: {predicted_next_number_hex}")

This is the result when we feed the script the sequence of keys up to #64, in order for it to "predict" #65:

Next key: 3.0520846598475555e+19
Hex: 0x1a78fd44662532000

Is this jesus toast or could we use it to at least try and narrow down the first 2 characters of #66?

Your insights are much appreciated.

kTimesG

member

Activity: 165

Merit: 26

Quote from: k3ntINA on April 16, 2024, 12:24:22 PM

Hello friends, it's time to make it public.
I hope these findings will help you find the keys.

Any series of strings in continuous increasing length starting from some center point and spiraled around in ascending length order will produce arms.
Why? Because they increase in same length, so for any element X(i) its length is length of X(i-1) * someConstant, so eventually their starting points on the circle's perimeter will get aligned, depending on how far from the center you position it.

This happens for any rational and non-rational number that exists, except for phi (1+sqrt(5))/2 and its family. That one is the single possible ratio that always wraps in full uniform distribution and is seen all over in nature.

satashi_nokamato

jr. member

Activity: 50

Merit: 3

Quote from: k3ntINA on April 16, 2024, 12:24:22 PM

Are we supposed to solve a 66 bit private key by looking at a gif and a picture? It more looks like a soccer ball than anything. What exactly is happening here, what should people do after looking at the image?

kachev87

newbie

Activity: 9

Merit: 0

Quote from: joseperal on April 16, 2024, 09:15:42 PM

the private key for ripemd160 hash d39c4704664e1deb76c9331e637564c257d68a08 is

000000000000000000000000000000000000000000000000000000003d94cd64

It belongs to puzzle - 30, already emptied.

address: 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps

That hash was an example...

joseperal

newbie

Activity: 56

Merit: 0

the private key for ripemd160 hash d39c4704664e1deb76c9331e637564c257d68a08 is

000000000000000000000000000000000000000000000000000000003d94cd64

It belongs to puzzle - 30, already emptied.

address: 1LHtnpd8nU5VHEMkG2TMYYNUjjLc992bps

kachev87

newbie

Activity: 9

Merit: 0

import bitcoin
import ecdsa
import base58
import random
import logging

# Function to convert private key to Wallet Import Format (WIF)
def private_key_to_wif(private_key):
wif = bitcoin.encode_privkey(bitcoin.decode_privkey(private_key, 'hex'), 'wif')
return wif

# Function to convert private key to Bitcoin address (P2PKH)
def private_key_to_address(private_key):
sk = ecdsa.SigningKey.from_string(bytes.fromhex(private_key), curve=ecdsa.SECP256k1)
vk = sk.get_verifying_key()
compressed_vk = vk.to_string('compressed').hex()
address = bitcoin.pubkey_to_address(compressed_vk)
return address

# Function to calculate Hash 160 of a Bitcoin address
def address_to_hash160(address):
decoded_address = base58.b58decode_check(address)
return decoded_address[1:].hex()

# Function to generate a Bitcoin private key and check if the corresponding address matches the target address
def generate_private_key(target_hash160):
while True:
try:
num_objects = random.randint(15,15)
random_values = random.sample(range(0, 29), num_objects)
random_values.append(29)
private_key_num = sum([2 ** power for power in random_values])
private_key = format(private_key_num, '064x')
bitcoin_address = private_key_to_address(private_key)
hash160 = address_to_hash160(bitcoin_address)

print("Private Key:", private_key)

if hash160 == target_hash160:
with open('private_key.txt', 'w') as file:
file.write(private_key)
logging.info("Private key saved to 'private_key.txt' file.")
break
except Exception as e:
logging.error(f"Error: {str(e)}")

def main():
target_hash160 = 'd39c4704664e1deb76c9331e637564c257d68a08'
logging.basicConfig(filename='bitcoin_keygen.log', level=logging.INFO, format='%(asctime)s - %(levelname)s - %(message)s')
logging.info("Generating private keys...")
generate_private_key(target_hash160)

if __name__ == "__main__":
main()

Can anyone make windows CMD program with the Idea of this script for multiple GPUs?
The changebles to be:

num_objects = random.randint(15,15) - THESE VALUES
random_values = random.sample(range(0, 29), num_objects) - THESE VALUES
random_values.append(29) - THIS VALUE
target_hash160 = 'd39c4704664e1deb76c9331e637564c257d68a08' - THIS VALUE

I am sorry for asking but I don`t have the proper programing skills to do this.
I`ve been playing with python to try to run this script with GPUs but... I just can`t ;(

k3ntINA

newbie

Activity: 18

Merit: 0

Quote from: AliBah on April 16, 2024, 01:45:36 PM

Quote from: k3ntINA on April 16, 2024, 12:24:22 PM

Do you really not know what to do?
Convert it to hex and generate an address and compare it with the addresses of the puzzle! Maybe luck is with you

AliBah

newbie

Activity: 39

Merit: 0

Quote from: k3ntINA on April 16, 2024, 12:24:22 PM

https://www.talkimg.com/images/2024/04/16/jvtt1.gif
Hello friends, it's time to make it public.
I hope these findings will help you find the keys.
The program used is Adobe Illustrator, where the keys are placed together around a spiral axis and separated by a distance.
The coloring is adjusted according to the work of zahid888.
What is strange is that with the font number 10 and the distance between the numbers of 600, we have an order and the colors are automatically placed in their own categories.
The stars selected in black color are the position of key number 66, and the next stars are placed in order up to key 70.
I have many more things and more detailed arrangements, and if I can and my condition allows, I will make them public.
Don't forget me if I helped you reach the key.
https://www.talkimg.com/images/2024/04/16/jvZq3.jpeg
1E4Shdpyfqz2vBqsnYw1Z8XT5cu7N5dEaK

and what should i do after find the numbers?

k3ntINA

newbie

Activity: 18

Merit: 0

https://www.talkimg.com/images/2024/04/16/jvtt1.gif
Hello friends, it's time to make it public.
I hope these findings will help you find the keys.
The program used is Adobe Illustrator, where the keys are placed together around a spiral axis and separated by a distance.
The coloring is adjusted according to the work of zahid888.
What is strange is that with the font number 10 and the distance between the numbers of 600, we have an order and the colors are automatically placed in their own categories.
The stars selected in black color are the position of key number 66, and the next stars are placed in order up to key 70.
I have many more things and more detailed arrangements, and if I can and my condition allows, I will make them public.
Note that the private keys are in decimal format

Don't forget me if I helped you reach the key.
1E4Shdpyfqz2vBqsnYw1Z8XT5cu7N5dEaK

https://www.talkimg.com/images/2024/04/16/jvZq3.jpeg

zahid888

member

Activity: 275

Merit: 20

the right steps towerds the goal

Happy 10x Anniversary. On this inspiring day, will Satoshi increase the value of my wallet tenfold and provide renewed motivation? Huh

ccinet

newbie

Activity: 41

Merit: 0

Quote from: kTimesG on April 14, 2024, 12:10:20 PM

Quote from: brainless on April 14, 2024, 09:42:11 AM

Quote from: zahid888 on April 14, 2024, 03:17:02 AM

e54acb08cf7e7d9be0102e2914d1a4eb643f5df386e67bb4be1bad5a05a53879

I guess he's implying that he found some <80 bits solution and he posted here some proof hash so the Collective assists him morally once the pubkey gets leaked and hundreds of TX will fight over the fee in the next 10 seconds.

A simple empty private address
1EVURWZJW38MzbqHQULons1jWDR34p96hE

kTimesG

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Activity: 165

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Quote from: brainless on April 14, 2024, 09:42:11 AM

Quote from: zahid888 on April 14, 2024, 03:17:02 AM

e54acb08cf7e7d9be0102e2914d1a4eb643f5df386e67bb4be1bad5a05a53879

I guess he's implying that he found some <80 bits solution and he posted here some proof hash so the Collective assists him morally once the pubkey gets leaked and hundreds of TX will fight over the fee in the next 10 seconds.

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 97. (Read 230098 times)