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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 81545 times)
jademaxsuy
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June 26, 2019, 11:09:30 PM
 #1041

Cracking bitcoin private key would be near to impossible. The wallets are unique and if it is a cold storage though one can still have the difficulties in accessing it for bitcoin has stored in that hardware or cold storage where one could not access if it is not doing online activities. So, forget that reward OP and start earning btc in a decent way like joining a bounty campaign or investing btc with your hard earn money.

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zielar
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June 26, 2019, 11:26:36 PM
 #1042

Hello to everyone,

62. Wallet Hex Code Start: Starting with 2F (B, D, E) ......

I can assure you at this stage that you missed the right range :-)
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June 27, 2019, 07:42:27 AM
 #1043

Hello to everyone,

62. Wallet Hex Code Start: Starting with 2F (B, D, E) ......

Based on what?
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June 27, 2019, 11:04:24 AM
 #1044

Computer doing random numerical assignment? It does not easily coincide with the same step. Probability is 16%.
Check between found 1-61 Purses.
Parts missing in the binary system 10-0000 / 10-0010 / 10-0011 ....
10-1011 and 10-1111, so start 2B .... 2F ...! (2D ... highly likely to be scanned, in the past time.)
Check steps 3, missing ones. B-D-E-F (2FB-2FD-2FE-2FF)
Check steps 4, missing ones. 1-3-5-6-9-B-C-E ....

The probability assessment is between %80-90. %10 -  3, second step(2-3-5-6-7), third step(B-C-D-E-F)

I know little english, translate.
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June 27, 2019, 12:10:18 PM
 #1045

I decided to drive the statistics.
Calculate the partial repetition of a part of a public address, starting from the end or from the beginning. Bit repetitions.
I.e

10101010111......010101010111010101010
01011010100......010101001011010101010
Match the end to 11 bits.


1101011011011010101010111......010101010110101010101
1101011011011001011010100......010101001010010101010
Match from the beginning to 14 bits.

Private keys get over in a row. And looking for repetition in a public address. When repetition is found, the difference in the private key is calculated.

At the moment, found such a minimum range.
Code:
24 bits: 2970 * 128   = 380160 (+-128) pkeys
23 bits: 330 * 128    = 42240  (+-128) pkeys
22 bits: 33 * 128     = 4244    (+-128) pkeys
21 bits: 11 * 128     = 1408    (+-128) pkeys
20 bits: 10 * 128     = 1280    (+-128) pkeys
Matching 19 or less bits are found with ranges of less than 128 keys.

Of course, a jump (pass) to 380k keys, at a search speed of 1M per second, is nothing. But finding large ranges, you can make big jumps.
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June 27, 2019, 01:16:44 PM
Merited by MagicByt3 (1)
 #1046

I decided to drive the statistics.
Calculate the partial repetition of a part of a public address, starting from the end or from the beginning. Bit repetitions.
I.e

10101010111......010101010111010101010
01011010100......010101001011010101010
Match the end to 11 bits.


1101011011011010101010111......010101010110101010101
1101011011011001011010100......010101001010010101010
Match from the beginning to 14 bits.

Private keys get over in a row. And looking for repetition in a public address. When repetition is found, the difference in the private key is calculated.

At the moment, found such a minimum range.
Code:
24 bits: 2970 * 128   = 380160 (+-128) pkeys
23 bits: 330 * 128    = 42240  (+-128) pkeys
22 bits: 33 * 128     = 4244    (+-128) pkeys
21 bits: 11 * 128     = 1408    (+-128) pkeys
20 bits: 10 * 128     = 1280    (+-128) pkeys
Matching 19 or less bits are found with ranges of less than 128 keys.

Of course, a jump (pass) to 380k keys, at a search speed of 1M per second, is nothing. But finding large ranges, you can make big jumps.
Sometimes, when I look at clouds, I see that they look kind of like animals.  I have seen a cloud that looks like a rabbit.  One time I saw this cloud that looked very much like an elephant.  I have seen many various animals in clouds.  I believe that if you look hard enough at any cloud you will see that it looks like some sort of animal.

Our family was terrorized by Homeland Security.  Read all about it here:  http://www.jmwagner.com/ and http://www.burtw.com/  Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!
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June 27, 2019, 03:56:41 PM
 #1047

Sometimes, when I look at clouds, I see that they look kind of like animals.  I have seen a cloud that looks like a rabbit.  One time I saw this cloud that looked very much like an elephant.  I have seen many various animals in clouds.  I believe that if you look hard enough at any cloud you will see that it looks like some sort of animal.
You too?! Oh mate, I thought that I'm the only one who see it Grin Grin Grin.
Just yesterday I saw a horse-cloud, that was beautiful!
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June 27, 2019, 04:40:25 PM
 #1048

Hello to everyone,

62. Wallet Hex Code Start: Starting with 2F (B, D, E) ......

I can assure you at this stage that you missed the right range :-)

Why did you say that? Did you scan this range?
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June 27, 2019, 06:16:40 PM
 #1049

Why did 70, 75 and 80 move?
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June 27, 2019, 07:11:12 PM
 #1050

Why did 70, 75 and 80 move?

And 85 too ....

Go back a few posts ...  Grin
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June 28, 2019, 03:22:31 PM
 #1051

...the puzzle above 70 in the next fifty years will not be solved...
We have to try continue here https://cloud.google.com/tpu/
Not with these guys is not a ride for free something to polychelate,GPU they just do not give. Grin Grin Grin
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June 28, 2019, 05:23:45 PM
 #1052

Ten years ago, Satoshi and Craig created a website where just like that from donations distributed 5 bitcoins to all those in need,and now some 0.5 bitcoins would solve my almost all problems.
I join my friend and say if God in the world let him drip on my purse incentive prize for the promotion of cryptocurrency to the masses!!! 1HqqeH1VWDY9vJq8nF4C7BHQ3Yvzxak48V
Just so you know:  begging is not allowed on this forum and you can actually be banned for begging.  Not sure if this is begging but be careful.

Our family was terrorized by Homeland Security.  Read all about it here:  http://www.jmwagner.com/ and http://www.burtw.com/  Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!
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June 29, 2019, 07:05:49 AM
 #1053

If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey / s

Confirm or disprove my calculations.
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June 29, 2019, 11:11:20 AM
 #1054

If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey / s

Confirm or disprove my calculations.
Looks like something is wrong in your calculation.
I think should be like this:

134000000 keys/s * 60 sec * 60 min * 24 hours = 11577600000000 keys/day

(4000000000000001 - 2000000000000001) / 11577600000000 = 172.75 days

Same, correct me pls if I'm wrong.
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June 29, 2019, 12:02:42 PM
Last edit: June 29, 2019, 12:18:05 PM by rud3boy
 #1055

If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey / s

Confirm or disprove my calculations.
Looks like something is wrong in your calculation.
I think should be like this:

134000000 keys/s * 60 sec * 60 min * 24 hours = 11577600000000 keys/day

(4000000000000001 - 2000000000000001) / 11577600000000 = 172.75 days

Same, correct me pls if I'm wrong.

The space is hexadecimal!
2000000000000000 hex = 2305843009213693952 decimal

134000000 Keys/ sec

199164.1626255609065782 days

545.6552400700298810362 years
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June 29, 2019, 12:26:39 PM
 #1056

If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey / s

Confirm or disprove my calculations.
Looks like something is wrong in your calculation.
I think should be like this:

134000000 keys/s * 60 sec * 60 min * 24 hours = 11577600000000 keys/day

(4000000000000001 - 2000000000000001) / 11577600000000 = 172.75 days

Same, correct me pls if I'm wrong.

The space is hexadecimal!
2000000000000000 hex = 2305843009213693952 decimal

134000000 Keys/ sec

199164.1626255609065782 days

545.6552400700298810362 years

Oh f***, that was hex value.  Roll Eyes
Then yes, you are right, it will take just only ~545 years.

But there is also a good news: imagine what will be the price of BTC after 545 years!!!  Grin Grin Grin
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June 29, 2019, 02:17:59 PM
 #1057

On a background 545 years an error per 0,5 years to make all any +/- 0,1 percents
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June 30, 2019, 09:20:41 AM
 #1058

If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey / s

Confirm or disprove my calculations.
Looks like something is wrong in your calculation.
I think should be like this:

134000000 keys/s * 60 sec * 60 min * 24 hours = 11577600000000 keys/day

(4000000000000001 - 2000000000000001) / 11577600000000 = 172.75 days

Same, correct me pls if I'm wrong.

The space is hexadecimal!
2000000000000000 hex = 2305843009213693952 decimal

134000000 Keys/ sec

199164.1626255609065782 days

545.6552400700298810362 years

Oh f***, that was hex value.  Roll Eyes
Then yes, you are right, it will take just only ~545 years.

But there is also a good news: imagine what will be the price of BTC after 545 years!!!  Grin Grin Grin

These facts obviously have to be re-posted since you are surprised by the calculation how many resources is needed to crack just one extremely short key from this transaction:



Once again: the idea behind the creation of this puzzle transaction is to prove it can not be solved.

 
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June 30, 2019, 12:52:29 PM
 #1059

These horror stories about the number of atoms in the universe, are used for embellishment (although the numbers are large and relatively puzzle at this stage justifiably look for 1 needle in a haystack).

Let's look at the address numbers, on the number of repetitions of numbers.

1    2    3    4    5    6    7    8   9    0

8    6   8     6   9     9    6    7   6   12   88076670097206135214755096908866554407314546020283163510152690884233313095179
4    5   5    14   7     5   11   9   5   12   57701484440690220081460730827720988043707783650447848844644231359596471509755
9    7   13   3    10   13  5    8    0   9    56852111307365822553680728638432355135106461736008687516183006036037433215266
7    5   11   7    12   8   7    9   5     6    83144555669307175352868736014304552664501628893338700237812991649157337855485
11  11   7   5    11   6    6   7   5     9    105951512903139530827431508910247856221386514825762420570832790562065784213611
3    9    8   4     5   10   9  10   4    14   2800743280837621873504738280285669010665021097040374687382697063985620056327
10   9   6   7     8    6    7   7   9     8    15155401853110903276292971961426024213516891095729877044987768503568484329302
6    8    4   10  11   7    8   9    8     5   1934145264352640584468993215160278462974765570385286888570941520948795715972
4    8   10  10   6    7    8  10   2    12   28808842000054338759447213206875406362693076403134586314430882247307076152875

there are numbers in which numbers may be absent altogether (1,2,3,4,5,6,7,8,9 = 0)

13   6   6   10   0   7   7    11  11   6   14696118311433422644887998880270171111886071679929423836479971639009928414084
11   4   0   10   6   7   12  10  10   7    84167587877791947619455975590804189912148604940284674067142717911820851066978
0    7    9   9    9   7   8    16   6    6   85056283424685094879758489628438532598948587847835803373846697232020876056473
6    9   10   9   5   10  11  10   7    0   35778492181327261174949468738285433366625387652824764739658862179647493738129

or present within 20 (very rarely 21)

20  12   4   7   5   5   6   8   5   5   14168167265037749111145324098522827112187124998114101721323801125686219581042
20   3  10   8   5   7   3   9   7   5   74210210994663119334888205459886331564143036385918170716141191111683131845191
6    3   5    5   8   4   20  6  11  7   671715073647425809453041797737097177653288097773772751099955845997107768998
10   4  10   8   9   3   4   4  20   5   91330834709955852129495199148324191949367391795990099341407196639595153298354
6   3  11   5  12  20   6   6   3   5    63563038544616496835516983610048686335751646557506626310787196666355232735676

Now look at the distribution of 40000 addresses https://imgur.com/pWdzq9T

as can see, some repeat more 6 7 8 9...

the maximum length of the number we have 115792089237316195423570985008687907852837564279074904382605163141518161494336 = 78 signs

roughly speaking 78:10 = 7 ,ie numbers can be 1,2,3,4,5,6,7,8,9,0 on 7 for mixing 1111111222222233333334444444555555566666667777777888888899999990000000. (in theory, 20-30 percent of Satoshi’s addresses should be there)

generally within length 77-78 signs numbers constituting number can float 1,2,3,4,5,6,7,8,9,0-0,9,8,7,6,5,4,3,2,1 = 3-12  

3-12 or 4-13 a lot of us, we can retreat a little from the middle (7-7-7-7-7-7-7-7-7-7) 5-10 or 6-8 etc...

(and the same sequence if you cut off the first 78 characters of degrees 2)  Smiley  Grin
itod
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June 30, 2019, 03:18:54 PM
 #1060

These horror stories about the number of atoms in the universe, are used for embellishment (although the numbers are large and relatively puzzle at this stage justifiably look for 1 needle in a haystack).

Embellishment? Really? These numbers are the essence of cryptographic security. They are pure science, and still you are arguing with scientific calculations. If these calculations were off in any possible way, and you could find the private key by brute-forcing Bitcoin would be worth nothing.

 
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