Bitcoin Cipher/puzzle - 0.56 Prize ! Bitcoins [SOLVED]

[trior]

192 (7)-1 should be h. See the pdf file again, page 192 should be the page 208 in the pdf.

nop !! the page 192 is the start of "chapter 9 SPECIATION"  witch is page 207 in pdf (see content page 6 in pdf)

As for the hints you could be right !!

If there's no online version of the book, you could use the "look inside" feature of amazon and keep selecting "surprise me" until you get to the right pages.

I think the PDF is the Softcover version, we need to check the hardcover one I guess .
Some one should visit the library  , or buy the book (on your own risk i could be wrong) .

[1714939713]

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[dalek]

I think the PDF is the Softcover version, we need to check the hardcover one I guess .
Some one should visit the library  , or buy the book (on your own risk i could be wrong) .

The hardcover version is also available on amazon, so the look inside feature is still a possibility.

[7788bitcoin]

192 (7)-1 should be h. See the pdf file again, page 192 should be the page 208 in the pdf.

nop !! the page 192 is the start of "chapter 9 SPECIATION"  witch is page 207 in pdf (see content page 6 in pdf)

As for the hints you could be right !!

If there's no online version of the book, you could use the "look inside" feature of amazon and keep selecting "surprise me" until you get to the right pages.

I think the PDF is the Softcover version, we need to check the hardcover one I guess .
Some one should visit the library  , or buy the book (on your own risk i could be wrong) .

Hmmm.... maybe I am wrong.
This is how I determined page 208 = 192:
Page 8 = Page 24 in the pdf, therefore, Page 192 should be Page (192+16) = 208.

Apologies if this confused you...

[trior]

Hmmm.... maybe I am wrong.
This is how I determined page 208 = 192:
Page 8 = Page 24 in the pdf, therefore, Page 192 should be Page (192+16) = 208.

Apologies if this confused you...

You could be right if the OP used the same way as you to count the page numbers .
How scanned the book  skipped a page somewhere .

3: 978-0-7538-1368-3 8(11)-4- 54(2)-1- 192(7)-1-
9(1)-1- 344(10)-1-   AAaAa

puzzle page : 8    54   192    9    344
pdf page     : 24  70    207   25   360       

[Yoshi24517]

I have gone back and looked at #3 again, I think the correct answer should be EAhBp

192 (7)-1 should be h. See the pdf file again, page 192 should be the page 208 in the pdf.

Hello! i will hold on for a day or two to Announche  any right answers again

The OP purposely have a capital A in "announce" and an additional h, meaning the h is behind A => EAhBp

this is a small hint. but hints can be small. but this Hint is: Softcover
Level 3

"this is a small hint" &  "but hints can be small" => a small "h"

OP is really smart!!

Anyone willing to put their computing power into brute forcing the answer?

Well then now I have:

5Juu2CkB3EAhBpEcyH9dypNhPetergh6RQwwbo5ueBa6------E?

[coblee]

Well then now I have:

5Juu2CkB3EAhBpEcyH9dypNhPetergh6RQwwbo5ueBa6------E?

Why Peter?

[NorrisK]

I have gone back and looked at #3 again, I think the correct answer should be EAhBp

192 (7)-1 should be h. See the pdf file again, page 192 should be the page 208 in the pdf.

Hello! i will hold on for a day or two to Announche  any right answers again

The OP purposely have a capital A in "announce" and an additional h, meaning the h is behind A => EAhBp

this is a small hint. but hints can be small. but this Hint is: Softcover
Level 3

"this is a small hint" &  "but hints can be small" => a small "h"

OP is really smart!!

Anyone willing to put their computing power into brute forcing the answer?

Well then now I have:

5Juu2CkB3EAhBpEcyH9dypNhPetergh6RQwwbo5ueBa6------E?

If you where to try and bruteforce the last part, assuming the rest is correct, it would only need 50 billion tries (if I calculated this correctly..)

[7788bitcoin]

I guess we need OP to confirm the answers to #3 and #9 first, since hints have been given.
Then we move on to #6 and then 10.

[Rmcdermott927]

I have gone back and looked at #3 again, I think the correct answer should be EAhBp

192 (7)-1 should be h. See the pdf file again, page 192 should be the page 208 in the pdf.

Hello! i will hold on for a day or two to Announche  any right answers again

The OP purposely have a capital A in "announce" and an additional h, meaning the h is behind A => EAhBp

this is a small hint. but hints can be small. but this Hint is: Softcover
Level 3

"this is a small hint" &  "but hints can be small" => a small "h"

OP is really smart!!

Anyone willing to put their computing power into brute forcing the answer?

Well then now I have:

5Juu2CkB3EAhBpEcyH9dypNhPetergh6RQwwbo5ueBa6------E?

If you where to try and bruteforce the last part, assuming the rest is correct, it would only need 50 billion tries (if I calculated this correctly..)

The quickest way to tackle that would be using itertools in python.  6 places should go quick.   I'll try it tomorrow if no one has yet.

[letsplayagame]

This does look interesting. I wish I had more time to study it.

It looks like some of you are getting pretty close to solving it well in advance of the 30 day limit!

[novadragon]

if anyone had it done and get the prize 

[Rmcdermott927]

I have gone back and looked at #3 again, I think the correct answer should be EAhBp

192 (7)-1 should be h. See the pdf file again, page 192 should be the page 208 in the pdf.

Hello! i will hold on for a day or two to Announche  any right answers again

The OP purposely have a capital A in "announce" and an additional h, meaning the h is behind A => EAhBp

this is a small hint. but hints can be small. but this Hint is: Softcover
Level 3

"this is a small hint" &  "but hints can be small" => a small "h"

OP is really smart!!

Anyone willing to put their computing power into brute forcing the answer?

Well then now I have:

5Juu2CkB3EAhBpEcyH9dypNhPetergh6RQwwbo5ueBa6------E?

If you where to try and bruteforce the last part, assuming the rest is correct, it would only need 50 billion tries (if I calculated this correctly..)

The quickest way to tackle that would be using itertools in python.  6 places should go quick.   I'll try it tomorrow if no one has yet.

I ran this overnight.   I woke up to a 350+ gb unmanageable text doc and the script was still running.    If someone gets one more letter, I can run it and it will be easier.  Here is the code I used"

from itertools import product
import itertools
import sys
sys.stdout = open('c:/combo.txt', 'w')
for ele in  itertools.product("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",repeat=6):
    print("".join(ele))

[hasmukhrawal]

Cant crack.its more than logic than i thought.

[DannyHamilton]

I ran this overnight.   I woke up to a 350+ gb unmanageable text doc and the script was still running.    If someone gets one more letter, I can run it and it will be easier.  Here is the code I used"

from itertools import product
import itertools
import sys
sys.stdout = open('c:/combo.txt', 'w')
for ele in  itertools.product("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",repeat=6):
    print("".join(ele))

Note, you are including the letters I, O, l, and the number 0 which are all invalid base58check characters.

It also doesn't make much sense to store all the iterations in a file.  Instead it would be better to generate each potential key and then verify the checksum.  You only need to store those keys that have a valid checksum (which should be approximately 1 out every 4,294,967,296 keys generated).

You don't even really need to store the keys with a valid checksum.  You could just compute the address for each key that has a valid checksum as you find it and see if the address matches the winning address. If it does, then you are done.

[Rmcdermott927]

I ran this overnight.   I woke up to a 350+ gb unmanageable text doc and the script was still running.    If someone gets one more letter, I can run it and it will be easier.  Here is the code I used"

from itertools import product
import itertools
import sys
sys.stdout = open('c:/combo.txt', 'w')
for ele in  itertools.product("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",repeat=6):
    print("".join(ele))

Note, you are including the letters I, O, l, and the number 0 which are all invalid base58check characters.

It also doesn't make much sense to store all the iterations in a file.  Instead it would be better to generate each potential key and then verify the checksum.  You only need to store those keys that have a valid checksum (which should be approximately 1 out every 4,294,967,296 keys generated).

You don't even really need to store the keys with a valid checksum.  You could just compute the address for each key that has a valid checksum as you find it and see if the address matches the winning address. If it does, then you are done.

Honestly, coding that is a bit beyond my ability which was why I did it the way I did.   I'm pretty sure my brain would explode before I had a working version of what you suggested.

[DannyHamilton]

Honestly, coding that is a bit beyond my ability which was why I did it the way I did.   I'm pretty sure my brain would explode before I had a working version of what you suggested.

It's really not as difficult as you might think.  There are tools and libraries that already do most of the work.

Even if it was above your current abilities, there's still no good reason to include 4 characters in your iterations that aren't valid characters.  You've made your result set at least 4000 X bigger than it needs to be.

[thelongcoin]

Well then now I have:

5Juu2CkB3EAhBpEcyH9dypNhPetergh6RQwwbo5ueBa6------E



I bruteforced all keys associated with this pattern.  This is not correct.  I suspect the Peter needs to be changed to something else.

[coblee]

Well then now I have:

5Juu2CkB3EAhBpEcyH9dypNhPetergh6RQwwbo5ueBa6------E



I bruteforced all keys associated with this pattern.  This is not correct.  I suspect the Peter needs to be changed to something else.

Change it to "DZ8fX"

[NextTroll]

i'll keep this short, i've made a cipher, anyone who can crack it will receive 0.56 Bitcoins. Depending on how long it will take you i might increase the reward over time (timer 30 Days, after that i'll decide what to do)

here it is !


https://i.imgur.com/Djg68ip.png

Proof of funds; https://blockchain.info/address/1CQQ9menL2fZGdEseDsdC278rQ12Ai9XfG

Someone has already won?

[hexafraction]

i'll keep this short, i've made a cipher, anyone who can crack it will receive 0.56 Bitcoins. Depending on how long it will take you i might increase the reward over time (timer 30 Days, after that i'll decide what to do)

here it is !




Proof of funds; https://blockchain.info/address/1CQQ9menL2fZGdEseDsdC278rQ12Ai9XfG

Someone has already won?

No, the prize is still in that address.