Solve a riddle, guess a 4 char password and add 10 BTC to your xmas... SOLVED!!

[CIYAM]

I really like to thank all that made it through this marathon!

I was rather surprised that my little riddle/puzzle turned out to be *so* hard for you all (guess it shows just how differently people think - especially old-timer's like myself).

[K1773R]

-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

the privkey got cracked

K1773R
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.4.10 (GNU/Linux)

iQEcBAEBAgAGBQJQ3VurAAoJEG9KjkBCBMba/uUH/28vQfVQV/Y+3Cb5q0Bfd1Od
FIUpVbGebxq2OH+lrNI1QJitKRWxCe7iUrk4ORJybqb7XDkclDYBrDyvZ3n+A74p
m2v2ExsupVNFnIsv3HPvHezab4cXhmYDV2PTyM0/yUiGsliqqFAqDqi7tdVA3eDB
dmjpsGOe8GAOH2Eoh4p2cwu7geXmgOs3gRgCN9MOp2S+waZfeghX7Y3+pN2JRcPX
/nIO3gZuLC/oeIiahmxW9Efq8La8Ffw19m5I9zDdRf7vQ8TgMgjZ6jIFJ0GLuewR
eGmZ/M5UOfrphkS67uljBAHI9bGMCfucxeeLH8ovYTF8YYeKhRzUTNgCgQ3hezg=
=qkYa
-----END PGP SIGNATURE-----

[K1773R]

Damn It. Just found the key and it had been swept already!

Must have been just seconds....

5Hq96RktdCU3BL1U4b7Wxse2wWCiSRXierJdS68RE1Mf8BXphjg

I wonder if I could issue another spend with higher fees and have it go to me.... ha ha.
It's still unconfirmed.

yea was faster, as promised il distribute the 10BTC to all who helped.

[CIYAM]

Damn It. Just found the key and it had been swept already!

Must have been just seconds....

5Hq96RktdCU3BL1U4b7Wxse2wWCiSRXierJdS68RE1Mf8BXphjg

I wonder if I could issue another spend with higher fees and have it go to me.... ha ha.
It's still unconfirmed.

If you don't manage to pip him with that then give me a BTC address and I'll send you 1 BTC for being so close!

[sveetsnelda]

what was the four char password?

[phr33]

Congrats to the winner!

And big thanks CIYAM Pty. Ltd. for arranging this! It's been fun, exciting and educational for everyone involved!

[K1773R]

what was the four char password?

dunno, the gpg passphrase is 41087438dbac5b77c3922cbe1f4d2f529ae3a4eb96a7534e4803361f1437f7e7

[phr33]

what was the four char password?

dunno, the gpg passphrase is 41087438dbac5b77c3922cbe1f4d2f529ae3a4eb96a7534e4803361f1437f7e7

Hmm. Thanks. And the "salt algorithm"?

[dooglus]

what was the four char password?

dunno, the gpg passphrase is 41087438dbac5b77c3922cbe1f4d2f529ae3a4eb96a7534e4803361f1437f7e7

So was one of the 'equations' I PM'ed you a couple of days ago right?  I'm a bit confused by your new PM.  You seem to be suggesting that one of them was right but that you didn't test it at the time?

For those who dont have enough hashing power, u can send me patterns per PM and il test em, if they match u get a portion of the 10BTC (going to distribute it fair to all who helped, including me).

Here are a few to try:

Code:

password="${password}+${password}=2${password}@L3AsT"
password="${password}+${password}=2.${password}@L3AsT"
password="${password}+${password}=2*${password}@L3AsT"
password="${password}+${password}=2x${password}@L3AsT"
password="${password}+${password}=2 ${password}@L3AsT"
password="${password}+${password}=2 . ${password}@L3AsT"
password="${password}+${password}=2 * ${password}@L3AsT"
password="${password}+${password}=2 x ${password}@L3AsT"
password="${password}+${password}=2${password}(at least)"
password="${password}+${password}=2.${password}(at least)"
password="${password}+${password}=2*${password}(at least)"
password="${password}+${password}=2x${password}(at least)"
password="${password}+${password}=2 ${password}(at least)"
password="${password}+${password}=2 . ${password}(at least)"
password="${password}+${password}=2 * ${password}(at least)"
password="${password}+${password}=2 x ${password}(at least)"
password="${password}+${password}=2${password} (at least)"
password="${password}+${password}=2.${password} (at least)"
password="${password}+${password}=2*${password} (at least)"
password="${password}+${password}=2x${password} (at least)"
password="${password}+${password}=2 ${password} (at least)"
password="${password}+${password}=2 . ${password} (at least)"
password="${password}+${password}=2 * ${password} (at least)"
password="${password}+${password}=2 x ${password} (at least)"

[CIYAM]

Oh - and the password was 3a5Y

[robocoin]

Great exercise, thx.

[phr33]

So the modified password was this?
"13a5Y+13a5Y=2x3a5Y(at least)\n"

(without the quotes)

::EDIT
No, it does not give the right sha sum..

[CIYAM]

So the modified password was this?
"13a5Y+13a5Y=2x3a5Y(at least)\n"

(without the quotes)

that's correct - thought that I'd try and make it "twice as easy (at least)"...

[BkkCoins]

what was the four char password?

dunno, the gpg passphrase is 41087438dbac5b77c3922cbe1f4d2f529ae3a4eb96a7534e4803361f1437f7e7

Congrats!!!

I didn't get the 4 char pwd either but it would be possible now to go back and run it with logging the input pwd. I was piping them and didn't have any logging going on. It was 6m 8s into this final run at only 32k p/s. I was totally pumped and had gpg ready asking for the pwd and blockchain.info open on the import page... and when it got imported and didn't add the balance to my total I knew I was just beat.

Anyway, I wouldn't have got this far without K1773R and phr33 input. Especially about fixing the JTR program to check full pwd length. I hadn't noticed that. It was my first time using that tool.

I'd welcome any contribution to my firstbits at left but please don't feel obliged. It was fun and I was truly playing catch up all the way.

[sveetsnelda]

So the modified password was this?
"13a5Y+13a5Y=2x3a5Y(at least)\n"

No.  It was "3a5Y+3a5Y=2x3a5Y(at least)\n"

[K1773R]

So the modified password was this?
"13a5Y+13a5Y=2x3a5Y(at least)\n"

(without the quotes)

that's correct!

no its not, this is correct:

Code:

$ echo "3a5Y+3a5Y=2x3a5Y(at least)" | sha256sum 
41087438dbac5b77c3922cbe1f4d2f529ae3a4eb96a7534e4803361f1437f7e7  - 

[BkkCoins]

So the modified password was this?
"13a5Y+13a5Y=2x3a5Y(at least)\n"

(without the quotes)

that's correct - thought that I'd try and make it "twice as easy (at least)"...

Not for me - Mine has no 1 added in front, just this,

 %s+%s=2x%s(at least)

edit - oh, I'm late to this party.

I'm amazed this all worked out.

[phr33]

Doh. So without the 1:s...

Alright. Thanks everyone!

[CIYAM]

Ooops - my bad - didn't see the 1's (this event has been rather tiring for me)...

"3a5Y+3a5Y=2x3a5Y(at least)\n"

[CIYAM]

BTW - I'm compiling a summary of the results in the (reserved) second post to this topic and will be adding some "what we have learned" kind of notes to it also.