Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 331.

adaseb

legendary

Activity: 3808

Merit: 1723

Been out of the loop on this contest for a while.

Checking LBC looks like they are still finding the 55th key which was already solved.

Looks like someone beat them to it or just started scanning backwards.

hari1987

newbie

Activity: 21

Merit: 0

I don't think there is a pattern in this puzzle. because the owner sends bitcoin randomly. and we know that the bitcoin address is between 2^n. and after the owner knew the puzzles were a bone of contention, the owner promised to increase the reward to 10 times and he did it.
good luck guys,there is 36,028,797,018,963,968 addresses to find puzzle number 56

Amnuazka

newbie

Activity: 15

Merit: 0

Amnuazka

newbie

Activity: 15

Merit: 0

Quote from: ZafotheNinja on August 02, 2018, 07:39:34 AM

Quote from: Amnuazka on July 30, 2018, 07:11:21 PM

I found this :

-snip-

holy_ship , Andzhig, arulbero, or anyone.., can you analize the next code ( x ) for 56 ?

-snip-

Thank you.

Thanks, you successfully made a wall of text. Please actually find a pattern instead of asking everyone else to.

Even if the next number could be said for certain, it would not solve for the entire string, therefore the difficulty of finding the private key for the next puzzle would be only marginally reduced.

Keep trying, but know that holy_ship, Andzhig, arulbero, or anyone else on this thread has probably already seen the binary version on the previous private keys, and have yet to find any pattern other than the first 0 at the beginning of each.

Thank you

Coinblaze

jr. member

Activity: 49

Merit: 1

I didn't see this update posted here, so maybe most people are not aware. (edit: actually it's way back on page 13)

As of 2017-07-11 the rewards for keys with bits 161+ were removed, and others made to be 10x the amount.

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234: 1FYbLcutmRbvu4yUeLmC4TES2Q3ChhXYY 0.000000
235: 1Pp61TfkztZ9ckpdeUVCzbyA7vMqXAVsdV 0.000000
236: 1PytbQzRaf9eTpGax3c6ofKwtbxaLLSNy1 0.000000
237: 194bRtNQVRq4xi26UJZYTHexvLXijpzp3e 0.000000
238: 1DxBvzRMdvzop21DhnDJv4xJDYFGVtu7KZ 0.000000
239: 1QLHLvM6XKwygyWqo2cCPbfbf6woZzGEKH 0.000000
240: 1C8Hw6T5jypz92cNFr9Lkx4Xmr4DtP7zGA 0.000000
241: 1L7ZNx5gFPvdVPfdgFBnEUgA64woQwopqr 0.000000
242: 1EMhTbC4Kp8DYBk5zoLsTqZ2YakhiTgQYh 0.000000
243: 19CqmBBJ3H8FjxSSeSjv2Kn1cVQFXMY6AM 0.000000
244: 1KMHDrCGho2QuK59UNvfjuiPdDZSgRbneC 0.000000
245: 1HzpmnHxpu4tCJ23ZS9TfWCoX8mXfpdHiq 0.000000
246: 1LrrNP28PYg1N1z5Uo1gR8cmoPc8h4orBZ 0.000000
247: 14D2d3WnHThUxyPhGoj2AabBTxpZcoHy3t 0.000000
248: 18oqrdP6uBKsn57gvmWzLuKMAY4ShapiAU 0.000000
249: 1Bun4VzuBJ7SUoQn97dinVfDyWAS336Ldg 0.000000
250: 1Ruu3JwvGeSmhQV9GzWAnyLCz6g3evmTY 0.000000
251: 1M3u4q5Q35qtQPmDHeubVbk7APYi3VVoBX 0.000000
252: 1CaTxB3YwmXZkDnTK4rRvq61SRqs48xmui 0.000000
253: 1JqRqUPHHcQu2yrr8JZzxSYDx2jbZxEqFj 0.000000
254: 1NKkjFvXmovmjwgUujw655n3BbEvnncyza 0.000000
255: 17PEUvQmgqPkkvRkMowoR1wRDXYzre4b9Z 0.000000
256: 1FMcotmnqqE5M2x9DDX3VfPAPuBWArGisa 0.000000
Total Unclaimed: 113.400000 BTC

At today's prices, that is $855,631.35

Andzhig

jr. member

Activity: 184

Merit: 3

This he saw the ladder without fractional (seemingly). Some stalactites..

36028797018963968 30045390491869460 9974455244496708‬ 6763683971478124‬ 4216495639600700‬
18014398509481984 15022695245934730‬ 4987227622248354‬ 3381841985739062‬ 2108247819800350‬
9007199254740992 1 7511347622967365‬ 1 2493613811124177‬ 1 1690920992869531‬ 1054123909900175‬
4503599627370496 3755673811483682‬ 1246806905562088 ‬1 845460496434765‬ 1 527061954950087‬
2251799813685248 1 1877836905741841‬ 623403452781044‬ 422730248217382‬ 1 263530977475043‬
1125899906842624 938918452870920‬ 311701726390522‬ 1 211365124108691‬ 131765488737521‬
562949953421312 469459226435460‬ 1 155850863195261‬ 1 105682562054345‬ 65882744368760‬
281474976710656 234729613217730‬ 77925431597630‬ 52841281027172‬ 32941372184380‬
140737488355328 1 117364806608865‬ 1 38962715798815‬ 26420640513586‬ 16470686092190‬
70368744177664 58682403304432‬ 1 19481357899407‬ 1 13210320256793‬ 1 8235343046095‬
35184372088832 29341201652216‬ 1 9740678949703‬ 6605160128396‬ 1 4117671523047‬
17592186044416 14670600826108‬ 1 4870339474851‬ 3302580064198‬ 1 2058835761523‬
8796093022208 7335300413054‬ 1 2435169737425‬ 1 1651290032099‬ 1 1029417880761‬
4398046511104 1 3667650206527‬ 1217584868712‬ 1 825645016049‬ 514708940380‬
2199023255552 1 1833825103263‬ 608792434356‬ 412822508024‬ 257354470190‬
1099511627776 1 916912551631‬ 304396217178‬ 206411254012‬ 1 128677235095 ‬
549755813888 1 458456275815‬ 1 152198108589‬ 103205627006‬ 1 64338617547‬
274877906944 1 229228137907‬ 76099054294‬ 1 51602813503‬ 1 32169308773‬
137438953472 1 114614068953‬ 1 38049527147‬ 1 25801406751‬ 16084654386‬
68719476736 57307034476‬ 1 19024763573‬ 1 12900703375‬ 1 8042327193‬
34359738368 28653517238‬ 9512381786‬ 1 6450351687‬ 4021163596‬
17179869184 1 14326758619‬ 1 4756190893‬ 1 3225175843‬ 2010581798‬
8589934592 1 7163379309‬ 2378095446‬ 1 1612587921‬ 1 1005290899‬
4294967296 3581689654‬ 1 1189047723‬ 806293960‬ 1 502645449‬
2147483648 1 1790844827‬ 1 594523861‬ 403146980‬ 251322724‬
1073741824 1 895422413‬ 297261930‬ 201573490‬ 125661362‬
536870912 447711206‬ 1 148630965‬ 1 100786745‬ 1 62830681‬
268435456 1 223855603‬ 74315482‬ 50393372‬ 31415340‬
134217728 1 111927801‬ 1 37157741‬ 25196686‬ 15707670‬
67108864 55963900‬ 18578870‬ 1 12598343‬ 1 7853835‬
33554432 27981950‬ 1 9289435‬ 1 6299171‬ 1 3926917‬
16777216 1 13990975‬ 1 4644717‬ 1 3149585‬ 1963458‬
8388608 1 6995487 2322358‬ 1 1574792‬ 1 981729‬
4194304 1 3497743 1 1161179‬ 787396‬ 490864‬
2097152 1 1748871 1 580589‬ 393698‬ 245432‬
1048576 1 874435 290294‬ 1 196849‬ 122716‬
524288 1 437217 1 145147‬ 98424‬ 61358‬
262144 218608 1 72573‬ 49212‬ 1 30679‬
131072 109304 36286‬ 24606‬ 1 15339‬
65536 54652 1 18143‬ 1 12303‬ 1 7669‬
32768 27326 1 9071‬ 1 6151‬ 3834‬
16384 1 13663 1 4535‬ 1 3075‬ 1 1917‬
8192 1 6831 1 2267‬ 1 1537‬ 958‬
4096 1 3415 1 1133‬ 768‬ 1 479‬
2048 1 1707 566‬ 384‬ 1 239‬
1024 1 853 1 283‬ 192‬ 1 119‬
512 426 1 141‬ 96‬ 1 59‬
256 1 213 70‬ 48‬ 1 29‬
128 106 1 35‬ 24 14‬
64 1 53 1 17‬ 12 1 7‬
32 26 8 ‬ 6 1 3‬
16 1 13 4 1 3 1‬
8 6 2 1
4 1 3 1
2 1
1

at some steps added 1. In the python 3 it will look like

Quote

a1 = 1
a2 = a1 * 2 +1
a3 = a2 * 2
a4 = a3 * 2 +1
a5 = a4 * 2
a6 = a5 * 2 +1
a7 = a6 * 2
a8 = a7 * 2 +1
a9 = a8 * 2
a10 = a9 * 2 +1
a11 = a10 * 2 +1
a12 = a11 * 2 +1
a13 = a12 * 2 +1
a14 = a13 * 2 +1
a15 = a14 * 2
a16 = a15 * 2
a17 = a16 * 2
a18 = a17 * 2
a19 = a18 * 2 +1
a20 = a19 * 2 +1
a21 = a20 * 2 +1
a22 = a21 * 2 +1
a23 = a22 * 2 +1
a24 = a23 * 2 +1
a25 = a24 * 2
a26 = a25 * 2
a27 = a26 * 2 +1
a28 = a27 * 2 +1
a29 = a28 * 2
a30 = a29 * 2 +1
a31 = a30 * 2 +1
a32 = a31 * 2
a33 = a32 * 2 +1
a34 = a33 * 2 +1
a35 = a34 * 2
a36 = a35 * 2
a37 = a36 * 2 +1
a38 = a37 * 2 +1
a39 = a38 * 2 +1
a40 = a39 * 2 +1
a41 = a40 * 2 +1
a42 = a41 * 2 +1
a43 = a42 * 2
a44 = a43 * 2
a45 = a44 * 2
a46 = a45 * 2
a47 = a46 * 2 +1
a48 = a47 * 2
a49 = a48 * 2
a50 = a49 * 2
a51 = a50 * 2 +1
a52 = a51 * 2
a53 = a52 * 2 +1
a54 = a53 * 2
a55 = a54 * 2

print(a31,a55)

where a31 possible address (+ 200000000 or +500000000 not to smear) for scanning by Holy fork (1790844827 * 2 24 times = 30045390485061632 against 30045390491869460 the right.

Accordingly, placing correctly the pros can successfully jump.

Quote

from bitcoin import *

a1 = 1
a2 = a1 * 2 +1
a3 = a2 * 2
a4 = a3 * 2
a5 = a4 * 2 +1
a6 = a5 * 2 +1
a7 = a6 * 2
a8 = a7 * 2 +1
a9 = a8 * 2
a10 = a9 * 2
a11 = a10 * 2 +1
a12 = a11 * 2 +1
a13 = a12 * 2
a14 = a13 * 2
a15 = a14 * 2
a16 = a15 * 2 +1
a17 = a16 * 2 +1
a18 = a17 * 2 +1
a19 = a18 * 2 +1
a20 = a19 * 2
a21 = a20 * 2
a22 = a21 * 2
a23 = a22 * 2 +1
a24 = a23 * 2 +1
a25 = a24 * 2
a26 = a25 * 2 +1
a27 = a26 * 2 +1
a28 = a27 * 2 +1
a29 = a28 * 2 +1
a30 = a29 * 2
a31 = a30 * 2
a32 = a31 * 2 +1 # fork scan point *2>24
a33 = a32 * 2
a34 = a33 * 2
a35 = a34 * 2
a36 = a35 * 2
a37 = a36 * 2
a38 = a37 * 2
a39 = a38 * 2
a40 = a39 * 2
a41 = a40 * 2
a42 = a41 * 2
a43 = a42 * 2
a44 = a43 * 2
a45 = a44 * 2
a46 = a45 * 2
a47 = a46 * 2
a48 = a47 * 2
a49 = a48 * 2
a50 = a49 * 2
a51 = a50 * 2
a52 = a51 * 2
a53 = a52 * 2
a54 = a53 * 2
a55 = a54 * 2
a56 = a55 * 2

scan = a32*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2+200000000
ran = scan
myhex = "%064x" % ran
myhex = myhex[:64]
priv = myhex
pub = privtopub(priv)
pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
addr = pubtoaddr(pubkey1)

print(a32,scan,a56,addr,myhex)

Or manually do not put up pluses

Quote

import secrets
from bitcoin import *

b1 = secrets.choice([0,1])
b2 = secrets.choice([0,1])
b3 = secrets.choice([0,1])
b4 = secrets.choice([0,1])
b5 = secrets.choice([0,1])
b6 = secrets.choice([0,1])
b7 = secrets.choice([0,1])
b8 = secrets.choice([0,1])
b9 = secrets.choice([0,1])
b10 = secrets.choice([0,1])
b11 = secrets.choice([0,1])
b12 = secrets.choice([0,1])
b13 = secrets.choice([0,1])
b14 = secrets.choice([0,1])
b15 = secrets.choice([0,1])
b16 = secrets.choice([0,1])
b17 = secrets.choice([0,1])
b18 = secrets.choice([0,1])
b19 = secrets.choice([0,1])
b20 = secrets.choice([0,1])
b21 = secrets.choice([0,1])
b22 = secrets.choice([0,1])
b23 = secrets.choice([0,1])
b24 = secrets.choice([0,1])
b25 = secrets.choice([0,1])
b26 = secrets.choice([0,1])
b27 = secrets.choice([0,1])
b28 = secrets.choice([0,1])
b29 = secrets.choice([0,1])
b30 = secrets.choice([0,1])
b31 = secrets.choice([0,1])

a1 = 1
a2 = a1 * 2 +b1
a3 = a2 * 2 +b2
a4 = a3 * 2 +b3
a5 = a4 * 2 +b4
a6 = a5 * 2 +b5
a7 = a6 * 2 +b6
a8 = a7 * 2 +b7
a9 = a8 * 2 +b8
a10 = a9 * 2 +b9
a11 = a10 * 2 +b10
a12 = a11 * 2 +b11
a13 = a12 * 2 +b12
a14 = a13 * 2 +b13
a15 = a14 * 2 +b14
a16 = a15 * 2 +b15
a17 = a16 * 2 +b16
a18 = a17 * 2 +b17
a19 = a18 * 2 +b18
a20 = a19 * 2 +b19
a21 = a20 * 2 +b20
a22 = a21 * 2 +b21
a23 = a22 * 2 +b22
a24 = a23 * 2 +b23
a25 = a24 * 2 +b24
a26 = a25 * 2 +b25
a27 = a26 * 2 +b26
a28 = a27 * 2 +b27
a29 = a28 * 2 +b28
a30 = a29 * 2 +b29
a31 = a30 * 2 +b30
a32 = a31 * 2 +b31 # fork scan point *2>24

scanpoint = a32*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2+200000000
ran = scanpoint
myhex = "%064x" % ran
myhex = myhex[:64]
priv = myhex
pub = privtopub(priv)
pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
addr = pubtoaddr(pubkey1)

print(a32,scanpoint,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13,b14,b15,b16,b17,b18,b19,b20,b21,b22,b23,b24,b25,b26,b27,b28,b29,b30,b31,addr,priv)

can make autostart fork adding (fork folder and cmd file itself lie on c:\) to loop or duplicate the repetition of the code, to whom it is more convenient, you can run several forks slightly correcting a couple of characters

Quote

import subprocess
import time

oy = """cd "C:\forkfolder1" """
ey = "\nstart /min oclvanitygen.exe -D 1:0 -C -f addresses.txt -o found.txt "
f=open("C:/testpy1.cmd","w")
f.write (oy)
f.write (ey)
f.write (priv) # adds the start address of the scan
f.close()

subprocess.Popen([r"C:/testpy1.cmd"])

time.sleep(60.0) # seconds before the scanner is closed

subprocess.call("taskkill /IM oclvanitygen.exe")

or for luck 56

Quote

from bitcoin import *
import secrets

while True:
   b1 = secrets.choice([0,1])
   b2 = secrets.choice([0,1])
   b3 = secrets.choice([0,1])
   b4 = secrets.choice([0,1])
   b5 = secrets.choice([0,1])
   b6 = secrets.choice([0,1])
   b7 = secrets.choice([0,1])
   b8 = secrets.choice([0,1])
   b9 = secrets.choice([0,1])
   b10 = secrets.choice([0,1])
   b11 = secrets.choice([0,1])
   b12 = secrets.choice([0,1])
   b13 = secrets.choice([0,1])
   b14 = secrets.choice([0,1])
   b15 = secrets.choice([0,1])
   b16 = secrets.choice([0,1])
   b17 = secrets.choice([0,1])
   b18 = secrets.choice([0,1])
   b19 = secrets.choice([0,1])
   b20 = secrets.choice([0,1])
   b21 = secrets.choice([0,1])
   b22 = secrets.choice([0,1])
   b23 = secrets.choice([0,1])
   b24 = secrets.choice([0,1])
   b25 = secrets.choice([0,1])
   b26 = secrets.choice([0,1])
   b27 = secrets.choice([0,1])
   b28 = secrets.choice([0,1])
   b29 = secrets.choice([0,1])
   b30 = secrets.choice([0,1])
   b31 = secrets.choice([0,1])
   b32 = secrets.choice([0,1])
   b33 = secrets.choice([0,1])
   b34 = secrets.choice([0,1])
   b35 = secrets.choice([0,1])
   b36 = secrets.choice([0,1])
   b37 = secrets.choice([0,1])
   b38 = secrets.choice([0,1])
   b39 = secrets.choice([0,1])
   b40 = secrets.choice([0,1])
   b41 = secrets.choice([0,1])
   b42 = secrets.choice([0,1])
   b43 = secrets.choice([0,1])
   b44 = secrets.choice([0,1])
   b45 = secrets.choice([0,1])
   b46 = secrets.choice([0,1])
   b47 = secrets.choice([0,1])
   b48 = secrets.choice([0,1])
   b49 = secrets.choice([0,1])
   b50 = secrets.choice([0,1])
   b51 = secrets.choice([0,1])
   b52 = secrets.choice([0,1])
   b53 = secrets.choice([0,1])
   b54 = secrets.choice([0,1])
   b55 = secrets.choice([0,1])

   a1 = 1
   a2 = a1 * 2 +b1
   a3 = a2 * 2 +b2
   a4 = a3 * 2 +b3
   a5 = a4 * 2 +b4
   a6 = a5 * 2 +b5
   a7 = a6 * 2 +b6
   a8 = a7 * 2 +b7
   a9 = a8 * 2 +b8
   a10 = a9 * 2 +b9
   a11 = a10 * 2 +b10
   a12 = a11 * 2 +b11
   a13 = a12 * 2 +b12
   a14 = a13 * 2 +b13
   a15 = a14 * 2 +b14
   a16 = a15 * 2 +b15
   a17 = a16 * 2 +b16
   a18 = a17 * 2 +b17
   a19 = a18 * 2 +b18
   a20 = a19 * 2 +b19
   a21 = a20 * 2 +b20
   a22 = a21 * 2 +b21
   a23 = a22 * 2 +b22
   a24 = a23 * 2 +b23
   a25 = a24 * 2 +b24
   a26 = a25 * 2 +b25
   a27 = a26 * 2 +b26
   a28 = a27 * 2 +b27
   a29 = a28 * 2 +b28
   a30 = a29 * 2 +b29
   a31 = a30 * 2 +b30
   a32 = a31 * 2 +b31 # fork scan point *2>24
   a33 = a32 * 2 +b32
   a34 = a33 * 2 +b33
   a35 = a34 * 2 +b34
   a36 = a35 * 2 +b35
   a37 = a36 * 2 +b36
   a38 = a37 * 2 +b37
   a39 = a38 * 2 +b38
   a40 = a39 * 2 +b39
   a41 = a40 * 2 +b40
   a42 = a41 * 2 +b41
   a43 = a42 * 2 +b42
   a44 = a43 * 2 +b43
   a45 = a44 * 2 +b44
   a46 = a45 * 2 +b45
   a47 = a46 * 2 +b46
   a48 = a47 * 2 +b47
   a49 = a48 * 2 +b48
   a50 = a49 * 2 +b49
   a51 = a50 * 2 +b50
   a52 = a51 * 2 +b51
   a53 = a52 * 2 +b52
   a54 = a53 * 2 +b53
   a55 = a54 * 2 +b54
   a56 = a55 * 2 +b55
   ran = a56
   myhex = "%064x" % ran
   myhex = myhex[:64]
   priv = myhex
   pub = privtopub(priv)
   pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
   addr = pubtoaddr(pubkey1)
   if addr == "17aPYR1m6pVAacXg1PTDDU7XafvK1dxvhi":
   print ("found!!!",ran,addr,myhex)
   s1 = myhex
   s2 = addr
   f=open(u"C:/a.txt","a")
   f.write(s1)
   f.write(s2)
   f.close()
   break
   else:
   print ("searching...",ran,addr,myhex)

ZafotheNinja

jr. member

Activity: 30

Merit: 1

Quote from: Amnuazka on July 30, 2018, 07:11:21 PM

I found this :

-snip-

holy_ship , Andzhig, arulbero, or anyone.., can you analize the next code ( x ) for 56 ?

-snip-

Thank you.

Thanks, you successfully made a wall of text. Please actually find a pattern instead of asking everyone else to.

Even if the next number could be said for certain, it would not solve for the entire string, therefore the difficulty of finding the private key for the next puzzle would be only marginally reduced.

Keep trying, but know that holy_ship, Andzhig, arulbero, or anyone else on this thread has probably already seen the binary version on the previous private keys, and have yet to find any pattern other than the first 0 at the beginning of each.

Andzhig

jr. member

Activity: 184

Merit: 3

Amnuazka we so transmutation to reptilian humanoids from Zeta Reticules.

try correct for binary input

Quote

from bitcoin import *
import secrets

while True:
   c1 = secrets.choice("456")
   c2 = secrets.choice("0987654321")
   c3 = secrets.choice("0987654321")
   c4 = secrets.choice("0987654321")
   c5 = secrets.choice("0987654321")
   c6 = secrets.choice("0987654321")
   c7 = secrets.choice("0987654321")
   c8 = secrets.choice("0987654321")
   c9 = secrets.choice("0987654321")
   c10 = secrets.choice("0987654321")
   c11 = secrets.choice("0987654321")
   c12 = secrets.choice("0987654321")
   c13 = secrets.choice("0987654321")
   c14 = secrets.choice("0987654321")
   c15 = secrets.choice("0987654321")
   c16 = secrets.choice("0987654321")
   c17 = secrets.choice("0987654321")
   ran = int(c1+c2+c3+c4+c5+c6+c7+c8+c9+c10+c11+c12+c13+c14+c15+c16+c17)
   myhex = "%064x" % ran
   myhex = myhex[:64]
   priv = myhex
   pub = privtopub(priv)
   pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
   addr = pubtoaddr(pubkey1)
   if addr == "1LzhS3k3e9Ub8i2W1V8xQFdB8n2MYCHPCa":
   print ("found!!!",ran,addr,myhex)
   s1 = myhex
   s2 = addr
   f=open(u"C:/a.txt","a")
   f.write(s1)
   f.write(s2)
   f.close()
   break
   else:
   print ("searching...",ran,addr,myhex)

something like

Quote

c1 = secrets.choice("01")
c2 = secrets.choice("01")
c3 = secrets.choice("01")
c4 = secrets.choice("01")
c5 = secrets.choice("01")
c6 = secrets.choice("01")
c7 = secrets.choice("01")
c8 = secrets.choice("01")
c9 = secrets.choice("01")
c10 = secrets.choice("01")
c11 = secrets.choice("01")
c12 = secrets.choice("01")
c13 = secrets.choice("01")
c14 = secrets.choice("01")
c15 = secrets.choice("01")
c16 = secrets.choice("01")
c17 = secrets.choice("01")
c18 = secrets.choice("01")
c19 = secrets.choice("01")
c20 = secrets.choice("01")
c21 = secrets.choice("01")
c22 = secrets.choice("01")
c23 = secrets.choice("01")
c24 = secrets.choice("01")
c25 = secrets.choice("01")
c26 = secrets.choice("01")
c27 = secrets.choice("01")
c28 = secrets.choice("01")
c29 = secrets.choice("01")
c30 = secrets.choice("01")
c31 = secrets.choice("01")
c32 = secrets.choice("01")
c33 = secrets.choice("01")
c34 = secrets.choice("01")
c35 = secrets.choice("01")
c36 = secrets.choice("01")
c37 = secrets.choice("01")
c38 = secrets.choice("01")
c39 = secrets.choice("01")
c40 = secrets.choice("01")
c41 = secrets.choice("01")
c42 = secrets.choice("01")
c43 = secrets.choice("01")
c44 = secrets.choice("01")
c45 = secrets.choice("01")
c46 = secrets.choice("01")
c47 = secrets.choice("01")
c48 = secrets.choice("01")
c49 = secrets.choice("01")
c50 = secrets.choice("01")
c51 = secrets.choice("01")
c52 = secrets.choice("01")
c53 = secrets.choice("01")
c54 = secrets.choice("01")
c55 = secrets.choice("01")
c56 = secrets.choice("01")

or from random.randint(0,1)...

if you have fractional use Decimal

Quote

from bitcoin import *
import secrets
from decimal import *

while True:
   a = str("127.")
   c1 = str (secrets.choice("1234"))
   c2 = str (secrets.choice("0123456789"))
   c3 = str (secrets.choice("0123456789"))
   c4 = str (secrets.choice("0123456789"))
   c5 = str (secrets.choice("0123456789"))
   c6 = str (secrets.choice("0123456789"))
   c7 = str (secrets.choice("0123456789"))
   c8 = str (secrets.choice("0123456789"))
   c9 = str (secrets.choice("0123456789"))
   c10 = str (secrets.choice("0123456789"))
   c11 = str (secrets.choice("0123456789"))
   c12 = str (secrets.choice("0123456789"))
   c13 = str (secrets.choice("0123456789"))
   c14 = str (secrets.choice("0123456789"))
   c15 = str (secrets.choice("0123456789"))
   c16 = str (secrets.choice("0123456789"))
   c17 = str (secrets.choice("0123456789"))
   b = Decimal(a+c1+c2+c3+c4+c5+c6+c7+c8+c9+c10+c11+c12+c13+c14+c15+c16+c17)
   rrrr = int(72057594037927936 - (((((((((b * 128) * 128) * 128)* 128) * 128) * 128)* 128) - 71494644084506624) * 128))
   ran = rrrr
   myhex = "%064x" % ran
   myhex = myhex[:64]
   priv = myhex
   pub = privtopub(priv)
   pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
   addr = pubtoaddr(pubkey1)
   if addr == "17aPYR1m6pVAacXg1PTDDU7XafvK1dxvhi":
   print ("found!!!",ran,addr,myhex)
   s1 = myhex
   s2 = addr
   f=open(u"C:/a.txt","a")
   f.write(s1)
   f.write(s2)
   f.close()
   break
   else:
   print ("searching...",b,ran,addr,myhex)

55 to bin 01101010101111100001111110011011011001111110000100010100‬ and in 56.56 digits you have 48...

Amnuazka

newbie

Activity: 15

Merit: 0

I found this :

0
01
011
0001
00101
001101
0110100
01010010
000100010
0100111110
01110011100
001000110110
0000001110001
00111000010001
001000011110010
0010101110010000
00001000111011001
000101010001011000
0101011111100100100
01111110111001010111
000010110111110101100
0111010000000010101010
01110111000111000110101
011101011110001100111111
0011110110001011000011000
01111010001111110100010000
010000111001010010101000010
0110111110010011110110011100
01101101111011101001100100100
001001010100011100011010111110
0000110100011110000101100111001
01100010011110011111111101011101
000111100110110101010011110011111
0010100000001010100010000100100101
00110000110101111100110100001000000
010000011110110000100101001010100000
0111011110000100010101111101110011000
01101100111111001100000110001001010000
001011011111101011001011011101101001110
0100010011101110100000000000111001010101
01001100110100011111001110010100001110100
11010010011111000110110101110110110000111
11110100100101001101001101110111101110100
10010101000110010011100010001110000000101
01011000100000000000000101110100010111110
11010110010111000011110110001010100100111
00101001100000100100111110110011011100010

It begun from no 15 to no 55
15 16

0
0 1
1 1
1 1
1 0
0 1
1 1
0 0

55

0
1
1
1
1
1
0
0
0
0
0
0
1
1
0
0
1
0
0
1
0
0
1
1
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
1
0
1
0
1
0
1
0

holy_ship , Andzhig, arulbero, or anyone.., can you analize the next code ( x ) for 56 ?

x
0x
01x
011x
0001x
00101x
001101x
0110100x
01010010x
000100010x
0100111110x
01110011100x
001000110110x
0000001110001x
00111000010001x
001000011110010x
0010101110010000x
00001000111011001x
000101010001011000x
0101011111100100100x
01111110111001010111x
000010110111110101100x
0111010000000010101010x
01110111000111000110101x
011101011110001100111111x
0011110110001011000011000x
01111010001111110100010000x
010000111001010010101000010x
0110111110010011110110011100x
01101101111011101001100100100x
001001010100011100011010111110x
0000110100011110000101100111001x
01100010011110011111111101011101x
000111100110110101010011110011111x
0010100000001010100010000100100101x
00110000110101111100110100001000000x
010000011110110000100101001010100000x
0111011110000100010101111101110011000x
01101100111111001100000110001001010000x
001011011111101011001011011101101001110x
0100010011101110100000000000111001010101x
01001100110100011111001110010100001110100x
11010010011111000110110101110110110000111x
11110100100101001101001101110111101110100x
10010101000110010011100010001110000000101x
01011000100000000000000101110100010111110x
11010110010111000011110110001010100100111x
00101001100000100100111110110011011100010x

Thank you.

ZafotheNinja

jr. member

Activity: 30

Merit: 1

For anyone still trying to find a pattern in cryptographicly secure random data (me), here is a chart. X=% in Space, Y=Puzzle Number. Only did up to 40 for now, I'll get around to the rest eventually Cheesy

[spoiler]

[/spoiler]

lolobit2

newbie

Activity: 14

Merit: 0

Quote from: ?? on ??

ARULBERO help me please a private key of the my wallet lost

hi, I will help you to restore your wallet, write email in private messages

Yahkinator

newbie

Activity: 10

Merit: 0

Quote from: holy_ship on July 25, 2018, 09:47:32 PM

Andzhig, you should write your native language, because when you write in English I always get an impression that you have undertreated meningitis.

lol. I'm glad I'm not the only one that was left just scratching my head.

Hey Holy_ship, can you check on the thread about your ndv vanitygen build? I have a couple questions in there and I can't message you because I'm still a newbie on here.

ordeath

member

Activity: 228

Merit: 10

Quote from: Bulista on December 28, 2015, 11:12:47 AM

Hi guys,
In continuation to this thread: https://bitcointalksearch.org/topic/brute-force-on-bitcoin-addresses-video-of-the-action-1305887
While playing around with my bot, I found out this mysterious transaction:
https://blockchain.info/tx/08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15
those 32.896 BTC were sent to multiple addresses, all the private keys of those addresses seem to be generated by some kind of formula.
For example:
EDIT: If you find the solution feel free to leave a tip

1DPUhjHvd2K4ZkycVHEJiN6wba79j5V1u3

This is one of the things that I really love about this mysterious world of Bitcoin and cryptocurrency, you can always find something interesting in it, althought I am not going to resolve this puzzle because I am a dummy Tongue

Sample2

newbie

Activity: 27

Merit: 0

Quote from: Bitcoinpro on December 28, 2015, 01:34:53 PM

looks really easy to crack

why would reward be so high

It is not easy to crack the bitcoin address. Bitcoin cryptography is using Elliptic Curve Cryptography (ECC) which it looks like has a pattern but it is not. It looks easy to crack but it is not. So, let them go a head and we see.

holy_ship

jr. member

Activity: 115

Merit: 1

Quote from: Andzhig

phrase ip addresses (sha256) but there is also not a small figure 4294967296.

4294967296 keys to check is job for 80 seconds. OK, if they are not sequential, let it be - 1 hour. Grin

Andzhig, you should write your native language, because when you write in English I always get an impression that you have undertreated meningitis.

Andzhig

jr. member

Activity: 184

Merit: 3

Quote from: lolobit2 on July 23, 2018, 08:33:47 PM

Quote from: Andzhig on July 23, 2018, 01:45:16 PM

Andzhig

Пpивeтcтвyю ,интepecнa тeмa взлoмa бтк ,ecть нapaбoтки
нaпиши пoчтy в личкy -oбcyдим

Thank you, from me a cracker like Harry Dunn plumber in the movie "Dumb & Dumber" when he came to the house of a woman. Specially for you there is an idea scan through a phrase ip addresses (sha256) but there is also not a small figure 4294967296.

Quote from: holy_ship on July 24, 2018, 10:26:05 PM

Quote from: BurtW on July 23, 2018, 10:42:35 AM

The LBC project[/url] is currently running at 346.02 Mkeys/s

Actually LBC is still checking the 2^55 range. So even pasting numbers in http://brain.evilbs.com/ by hand is faster than LBC Grin

You are your "fork" in several 1080, 1070 as yet apparently also did not catch wildfowl. From where to have already gone or left this venture? I'm your program .. 49 .., 57 .., 58 ..., 59 ..., 60..., scanned in scatter, uselessly, can he be between 66-72 (30045390491869460+36028797018963968=66074187510833428 < 72057594037927936) who knows, as in step 611140496167764+1125899906842624=1737040403010388 < 2058769515153876 < 2251799813685248.

holy_ship

jr. member

Activity: 115

Merit: 1

Quote from: BurtW on July 23, 2018, 10:42:35 AM

The LBC project[/url] is currently running at 346.02 Mkeys/s

Actually LBC is still checking the 2^55 range. So even pasting numbers in http://brain.evilbs.com/ by hand is faster than LBC Grin

holy_ship

jr. member

Activity: 115

Merit: 1

Python is not hipsterish enough!
True mama's cryptokiddies choose PHP parser of directory.io
And adding right magic constant to skip "unnecessary" pages is vital!

So, first you should look at you mom's Crystal Ball to find the magic number!

lolobit2

newbie

Activity: 14

Merit: 0

Quote from: Andzhig on July 23, 2018, 01:45:16 PM

Andzhig

Пpивeтcтвyю ,интepecнa тeмa взлoмa бтк ,ecть нapaбoтки
нaпиши пoчтy в личкy -oбcyдим

Andzhig

jr. member

Activity: 184

Merit: 3

Yes it is clear that compared to LBC this is a louse (but if desired and a louse a lion can gnaw) this is so for luck, suddenly there is a magic step for a random house. There are many options you can think of, and step through a number, sort only even, odd, take each 128,256,512 ... multiply them ... floating scan area, some magical jumps, etc.

Quote

from bitcoin import *
import random
import math

while True:
f = random.random() + 15.1
g = f
b = 1152921504606846976 - ((((((((((g * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128) - 1143914305352105984) * 128)
c = math.trunc(b)
ran = c
myhex = "%064x" % ran
myhex = myhex[:64]
priv = myhex
pub = privtopub(priv)
pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
addr = pubtoaddr(pubkey1)
n = addr
if n.strip() == "1HAX2n9Uruu9YDt4cqRgYcvtGvZj1rbUyt":
print ("found!!!",addr,myhex)
s1 = myhex
s2 = addr

f=open(u"C:/a.txt","a")
f.write(s1)
f.write(s2)
f.close()
break
else:
print ("searching...",f,addr,myhex)

formula 1152921504606846976 - ((((((((((15.9 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128) - 1143914305352105984) * 128)

after 15.9xxxxxxxxxxxxxxxxxxxxxxxxxxx a random number of 29, 30 numbers can be sausage immediately at all after 7 steps of the puzzle.

Quote

1152921504606846976 - ((((((((((15,9075849271 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128) - 1143914305352105984) * 128)
147573952589676412928 - (((((((((((15,9041635330 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128) - 146421031085069565952) * 128)
18889465931478580854784 - ((((((((((((15,9058575788 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128) - 18741891978888904441856) * 128)
2417851639229258349412352 - (((((((((((((15,9058575788 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128) - 2398962173297779768557568) * 128)
309485009821345068724781056 - ((((((((((((((15,9081997978 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128) - 307067158182115810375368704) * 128)
39614081257132168796771975168 - (((((((((((((((15,9081997978 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128) - 39304596247310823728047194112) * 128)
5070602400912917605986812821504 - ((((((((((((((((15,9095945490 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128) - 5030988319655785437190040846336) * 128)
649037107316853453566312041152512 - (((((((((((((((((15,9095945490 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128) - 643966504915940535960325228331008) * 128)
83076749736557242056487941267521536 - ((((((((((((((((((15,90 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128) - 82427712629240388602921629226369024) * 128)
10633823966279326983230456482242756608 - (((((((((((((((((((15,90 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128) - 10550747216542769741173968540975235072) * 128)
1361129467683753853853498429727072845824 - ((((((((((((((((((((15,90 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128) - 1350495643717474526870267973244830089216) * 128)
174224571863520493293247799005065324265472 - (((((((((((((((((((((15,90 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128) - 172863442395836739439394300575338251419648) * 128)
22300745198530623141535718272648361505980416 - ((((((((((((((((((((((15,90 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128) - 22126520626667102648242470473643296181714944) * 128)
2854495385411919762116571938898990272765493248 - (((((((((((((((((((((((15,90 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128) - 2832194640213389138975036220626341911259512832) * 128)
365375409332725729550921208179070754913983135744 - ((((((((((((((((((((((((15,90 * 128) * 128) * 128) * 128)* 128) * 128) * 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128)* 128) - 362520913947313809788804636240171764641217642496) * 128)

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 331. (Read 239754 times)