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

[NorrisK]

i think op won't give out clues no more bc of how big the prize is ....
anyways goodluck hope you guys decode the puzzle XD

guys sad to say this but after spending a lot of hours un this I just found that he is just fooling us there is nothing in this..and this is the answer even in his 2 and 5 number he clearly stated it too

Ofcourse he is not fooling us.. That would also be quite lame to coblee who donated additional coins to the price.

There must be something we are overlooking..

[1715378808]

1715378808

[1715378808]

1715378808

[1715378808]

1715378808

[1715378808]

1715378808

[1715378808]

1715378808

[1715378808]

1715378808

[Sakarias-Corporation]

i can ensure you that i'm not fooling you.

[Jin xiou]

i say we need the book.

[trior]

Full result of what I have so far:

Code:

1. 5Juu2  (confirmed) 
2. CkB3  (confirmed) 
3. EAhBp, NAhBp
4. EcyH9  (confirmed) 
5. dypNh  (confirmed) 
6. DZ8fX  (confirmed) 
7. gh6RQ  (confirmed) 
8. wwbo5  (confirmed) 
9. teLs6, jeBa6, ueBa6, pdBa6, sdBa6, tsBj6, tsBu6
10. dGYa11E, wVLyNnE, dRNoTtE, eGYoMm5, r3C1NnE

I think there will be a reordering step at the end .

[hexafraction]

Full result of what I have so far:

Code:

1. 5Juu2  (confirmed) 
2. CkB3  (confirmed) 
3. EAhBp, NAhBp
4. EcyH9  (confirmed) 
5. dypNh  (confirmed) 
6. DZ8fX  (confirmed) 
7. gh6RQ  (confirmed) 
8. wwbo5  (confirmed) 
9. teLs6, jeBa6, ueBa6, pdBa6, sdBa6, tsBj6, tsBu6
10. dGYa11E, wVLyNnE, dRNoTtE, eGYoMm5, r3C1NnE

I think there will be a reordering step at the end .

I'm attempting a brute-force of all valid Base58Check private keys with those parameters, including the shuffling (since unshuffled had no results). Should be done within 10 minutes or so.

Edit: No match :/ Code I'm using is below. PermIterator gives all permutations of an int[] from 0 to length-1. Optimized since 5Juu2 is the only possible start to the private key.

Code:

import java.util.Arrays;

import com.google.bitcoin.core.AddressFormatException;
import com.google.bitcoin.core.Base58;

public class BrutePri {
	
//	1. 5Juu2  (confirmed) 
//	2. CkB3  (confirmed) 
//	3. EAhBp, NAhBp
//	4. EcyH9  (confirmed) 
//	5. dypNh  (confirmed) 
//	6. DZ8fX  (confirmed) 
//	7. gh6RQ  (confirmed) 
//	8. wwbo5  (confirmed) 
//	9. teLs6, jeBa6, ueBa6, pdBa6, sdBa6, tsBj6, tsBu6
//	10. dGYa11E, wVLyNnE, dRNoTtE, eGYoMm5, r3C1NnE
	String[] a0 = {"EAhBp", "NAhBp"};

	static String tkey = "5JXimtP4fhCef6disCwLnT6192QLN1rqM5bPAX3rw4CmBcRBCix";
	static int z = 0;
	public static void main(String[] args) throws Exception {
		
		String ls = "5Juu2CkB3";
		String ms = "EcyH9dypNhDZ8fXgh6RQwwbo5";
		
		try {
			Base58.decodeChecked(tkey);
			System.out.println(tkey);
			
		} catch (AddressFormatException e){
			System.err.println("Err");
		}
		
		new Thread(new Runnable(){ public void run(){
			String[] parts = new String[9];
			parts[0] = "CkB3";
			parts[1] = "ZZZ";
			parts[2] = "EcyH9";
			parts[3] = "dypNh";
			parts[4] = "DZ8fX";
			parts[5] = "gh6RQ";
			parts[6] = "wwbo5";
			parts[7] = "ZZZ";
			parts[8] = "ZZZ";
			String[] a1 = { "teLs6", "jeBa6", "ueBa6", "pdBa6", "sdBa6", "tsBj6", "tsBu6" };
			String[] a2 = { "dGYa11E", "wVLyNnE", "dRNoTtE", "eGYoMm5", "r3C1NnE" };
			for (int j = 0; j < a1.length; j++) {
				for(int k = 0; k < a2.length; k++){
					parts[1] = "EAhBp";
					parts[7] = a1[j];
					parts[8] = a2[k];
					z++;
					System.out.println("t1 iter "+z+"/70");
					PermIterator pi = new PermIterator(9);
					int[] iter = null;
					while(pi.hasNext()){
						iter = pi.next();
						StringBuilder sb = new StringBuilder("5Juu2");
						for(int q = 0; q < iter.length; q++){
							sb.append(parts[iter[q]]);
						}
						
						byte[] key;
						try {
							key = Base58.decodeChecked(sb.toString());
							System.out.println(sb.toString());
						} catch (AddressFormatException e){
							//pass
						}
					}
				}
				
			}

		}}).start();
		new Thread(new Runnable(){ public void run(){
			String[] parts = new String[9];
			parts[0] = "CkB3";
			parts[1] = "ZZZ";
			parts[2] = "EcyH9";
			parts[3] = "dypNh";
			parts[4] = "DZ8fX";
			parts[5] = "gh6RQ";
			parts[6] = "wwbo5";
			parts[7] = "ZZZ";
			parts[8] = "ZZZ";
			String[] a1 = { "teLs6", "jeBa6", "ueBa6", "pdBa6", "sdBa6", "tsBj6", "tsBu6" };
			String[] a2 = { "dGYa11E", "wVLyNnE", "dRNoTtE", "eGYoMm5", "r3C1NnE" };
			for (int j = 0; j < a1.length; j++) {
				for(int k = 0; k < a2.length; k++){
					parts[1] = "NAhBp";
					parts[7] = a1[j];
					parts[8] = a2[k];
					z++;
					System.out.println("t2 iter "+z+"/70");
					PermIterator pi = new PermIterator(9);
					int[] iter = null;
					while(pi.hasNext()){
						iter = pi.next();
						StringBuilder sb = new StringBuilder("5Juu2");
						for(int q = 0; q < iter.length; q++){
							sb.append(parts[iter[q]]);
						}
						byte[] key;
						try {
							key = Base58.decodeChecked(sb.toString());
							System.out.println(sb.toString());
							
						} catch (AddressFormatException e){
							//pass
						}
					}
				}
				
			}

		}}).start();
		// Thread.sleep(1);

	}
}

[Yoshi24517]

Uh Oh. We should actually try to figure out what #3 actually is.

[HugoTheSpider]

I think we should also try to solve #9. The casing of #3 which is XXxXx (AAaAa-ZZzZz) is more clear than the casing of #9.

[trior]

Uh Oh. We should actually try to figure out what #3 actually is.

As I stated before there is tow versions of the book Softcover and Hardcover the PDF we used is the softcover one , but there is an error somewhere in page numbers I think . If someone can get his hand on both versions It would be great  . Or else the OP give us some help.

[dalek]

Uh Oh. We should actually try to figure out what #3 actually is.

As I stated before there is tow versions of the book Softcover and Hardcover the PDF we used is the softcover one , but there is an error somewhere in page numbers I think . If someone can get his hand on both versions It would be great  . Or else the OP give us some help.

As I suggested earlier, you can use Amazon's "look inside" feature to grab the pages for both versions of the book.

[trior]

Uh Oh. We should actually try to figure out what #3 actually is.

As I stated before there is tow versions of the book Softcover and Hardcover the PDF we used is the softcover one , but there is an error somewhere in page numbers I think . If someone can get his hand on both versions It would be great  . Or else the OP give us some help.

As I suggested earlier, you can use Amazon's "look inside" feature to grab the pages for both versions of the book.

There is no "look inside" for Hardcover. 
And the look inside doesn't cover the hole book anyway.

If you can manage to get us the pages we need . It would be very nice of you .

[Yoshi24517]

I mean, I can try to do that when I get on the computer because I am on my iPod right now.

[MF Doom]

Uh Oh. We should actually try to figure out what #3 actually is.

As I stated before there is tow versions of the book Softcover and Hardcover the PDF we used is the softcover one , but there is an error somewhere in page numbers I think . If someone can get his hand on both versions It would be great  . Or else the OP give us some help.

As I suggested earlier, you can use Amazon's "look inside" feature to grab the pages for both versions of the book.

There is no "look inside" for Hardcover. 
And the look inside doesn't cover the hole book anyway.

If you can manage to get us the pages we need . It would be very nice of you .

time to head to Mr. Paperback, wait, is that place still around??

[RedDiamond]

Has anybody already checked could 3 be page(word)-character instead of page(line)-character?

[trior]

Has anybody already checked could 3 be page(word)-character instead of page(line)-character?

Code:

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

Screenshots from the pdf to make it easier
page   8: https://i.imgur.com/AVxCBem.png
page   9: https://i.imgur.com/a7hO7Np.png
page  54: https://i.imgur.com/fIEVFGr.png
page 192: https://i.imgur.com/hTmAPsi.png
page 344: https://i.imgur.com/gd7Or3m.png


In page 192 and 344 the words count is not clear .

[Sakarias-Corporation]

For about 30 Minutes ago i received a Alert, someone has cracked the Cipher, i was also recently contacted by that person.

The person who cracked the cipher does not want the Bitcoins on the private key an only cracked it for fun, so he sent the BTC back to the Adress, for the next person to crack it! so the game is still on! (for the bitcoins that is)


https://blockchain.info/address/1CQQ9menL2fZGdEseDsdC278rQ12Ai9XfG

[trior]

For about 30 Minutes ago i received a Alert, someone has cracked the Cipher, i was also recently contacted by that person.

The person who cracked the cipher does not want the Bitcoins on the private key an only cracked it for fun, so he sent the BTC back to the Adress, for the next person to crack it! so the game is still on! (for the bitcoins that is)


https://blockchain.info/address/1CQQ9menL2fZGdEseDsdC278rQ12Ai9XfG

Well done cipher cracker , Good job . And thank you so much for letting us have another chance.

[HugoTheSpider]

Good job.

I could crack it if #3 or #9 were known. On #3 I don't have the right book, on #9 I'm out of ideas now.

My "clue" for #9 I'm now sharing is:
4(1)x5(2)x1(1)x1(2)x*6* __-__

4(1)x
5(2)x
1(1)x
1(2)x
*6*
__-__

First: 4(1) - x = small
Second: 5(2) - x = small
Third: 1(1) - x = small
Fourth: 1(2) - x = small
= xxxx "four small letters" hinting

xxxx*6* + __-__ order hint: xx6xx

Then I bruteforced all valid key ranges from:
5Juu2CkB3AAaAaEcyH9dypNhDZ8fXgh6RQwwbo5xxxxx<all> to
5Juu2CkB3ZZzZzEcyH9dypNhDZ8fXgh6RQwwbo5xxxxx<all>

with xxxxx (#9) =
je6ba
po6ja
pu6ba
su6ba
jo6ja
ja6ja
hh6do (dont make it harder ... word(char))
thdo6 (dont make it harder then ... cancel third word "it" because "-" is interpreted as "cancel it" - in "__-__ ")
so6an
ud6as (derived from Udas)
di6oh (Oh dear ... word(char))
ue6ba (derived from ueBa6 posted here)
jm6jm n(1)=person, start from left, n(2)=person, start from right
pd6ba (Judas->Udas name(char))
pdBa6 (Judas->Udas & 6 = lower case)
ueBa6 (posted)
us6jc (from "jesus" and "christus", char(word#))
ii6th (What it really is: 1=The Last Supper, 2=Leonardo da Vinci, 3=Milan, 4=Italy - 5=visitor information)
UeBa6 (ueBa6 with big U)
ay6vr (Vacation Reply)
eo6ff (Hex bytes: byte(nibble))
eoFff (Hex bytes: *6*=F -> __-__)
sd6ba
each #3 from AAaAa to ZZzZz, <all> means all valid private keys for #10

still no luck.

Apparently I'm too dumb for cracking #9 or the cracker has the right book or both

Another hint for combatants who are struggling like me: most of the chars in #10 are sitting in the checksum part of the B58 private key.

[dalek]

Another hint for combatants who are struggling like me: most of the chars in #10 are sitting in the checksum part of the B58 private key.

I'm looking at https://en.bitcoin.it/wiki/Wallet_import_format and don't see you can take advantage of that fact.
Don't you have to decode the base58 first before you can remove the 4 byte checksum?

[dalek]

Another hint for combatants who are struggling like me: most of the chars in #10 are sitting in the checksum part of the B58 private key.

I'm looking at https://en.bitcoin.it/wiki/Wallet_import_format and don't see you can take advantage of that fact.
Don't you have to decode the base58 first before you can remove the 4 byte checksum?

Better reference: https://bitcointalk.org/index.php?topic=304645.msg3270248#msg3270248

[tennozer]

I like puzzle but it looks like a bit harder than I can solve. There 15 pages and I can not read all this posts and just only I can say have a luck for you.