London – Der britische Geheimdienst GCHQ sucht per Code-Spiel neue Bewerber (tobiasgillen Blog berichtete). Das ‘I’ in der Überschrift stimmt zwar nicht ganz, aber mit ein bisschen Recherche findet man die Lösung von ‘Can you crack it?’ schnell raus.
Das genaue Passwort habe ich bisher noch nicht gefunden, dafür aber einen Link, mit dem man das Verfahren umgehen kann: canyoucrackit.co.uk/soyoudidit.asp. Klickt man dart drauf, kommt man auf die Seite, die im obigen Screenshot zu sehen ist. Dort kann man dann zur “Jobbörse” vom GCHQ und sich, wenn man denn will, endgültig bewerben:
Übrigens: Das Thema lässt mich nicht los! Ich will unbedingt die richtige Lösung haben (mit Erklärung und Schlussfolgerung!). Wer die Lösung (gefunden) hat, der darf sich frei fühlen, sie in den Kommentaren zu posten. Danke!
Update: In den Kommentaren ist der richtige Code aufgetaucht. Es ist “Pr0t3ct!on#cyber_security@12*12.2011+”. Danke!
Update 2: Die “Neue Züricher Zeitung” (NZZ) hat diesen Artikel verlinkt: “So knackt man den Code, ohne ihn zu knacken”
Das hier scheint meines erachtens ein Teil des Chiffriercodes zu sein. Keine Ahnung ob du damit was anfangen kannst.
Kannst dich ja mal zurückmelden wenn du was rausfindest.
mfG
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Diese Buchstaben/Zahlenkombinationen erinnern mich an HEX-Farbcodes http://www.colorschemer.com/online.html
Damit das aber sofort schlüssig ist, müsste die Zahl der Code-Teile auf der Seite durch drei teilbar sein. Naja, ist sie nicht. Wär ja aber auch ein bisschen einfach.
Es gibt keine Lösung, das ist einfach nur ein Hex Code, der in zwei Kolumnen unterteilt wird und dann als String ausgegeben wird oder so, was weiß ich, ich hab mir den Scheiß garnicht genau angeschaut, das ist nur das was mir aufn ersten Blick auffällt.
Wirklich interessieren sollte dich das aber nicht, diese Seite kommt schließlich von den Briten, einem Volk, das so dumm und nutzlos ist, dass denen jegliche Existenzberechtigung fehlt – und ca genau das spiegelt dieses “Rätsel” wieder. Ende der Geschichte. Das hat weder was mit Cracken (geschweige denn hacken) oder sonst was zu tun, das ist einfach nur ein dummes Kinderrätsel.
Sicher gibt es eine Lösung:
Pr0t3ct!on#cyber_security@12*12.2011+
Greetz
_sToRm_
Begruendung?
Sicher gibt es einen SINN ….
Es ist Brainfuck…. einfach mal probieren, hänge gerade an der letzten Zeile fest und kriegs ned zum laufen
Pr0t3ct!on#cyber_security@12*12.2011+
Schwachsinn! Es handelt sich einfach um eine x86 Implementierung einer RC4 Chiffre. Dem Programm fehlen aber noch die verschlüsselten Daten. Diese findet man aber im PNG Header im Bild selber.
Der RC4 Schlüssel ist im übrigen
DEADBEEFDEADBEEFDEADBEEFDEADBEEF…
Verschlüsselt
91 D8 F1 6D 70 20 3A AB 67 9A 0B C4 91 FB C7 66 0F FC CD CC B4 02 FA D7 77 B4 54 38 AB 1F 0E E3 8E D3 0D EB 99 C3 93 FE D1 2B 1B 11 C6 11 EF C8 CA 2F
Entschlüsselt
G E T / 1 5 b 4 3 6 d e 1 f 9 1 0 7 f 3 7 7 8 a a d 5 2 5 e 5 d 0 b 2 0 . j s H T T P / 1 . 1
Japp, aber das ist der erste Teil mein Lieber, da gehts noch weiter ;)
Siehe: http://www.canyoucrackit.co.uk/15b436de1f9107f3778aad525e5d0b20.js
Das selbe Spiel, nur muss man diesmal einen virtuellen Prozessor implementieren. Zum Glück ist der sehr sehr überschaubar.
Das Geheimnis verbirgt sich wieder in einem Maschinenprogramm ;).
Hier noch das Programm des ersten Rätsels
0040498B /EB 04 jmp short 00404991
0040498D AF C2 BF A3
;SBOX anlegen
00404991 \81EC 00010000 sub esp, 100
00404997 31C9 xor ecx, ecx
00404999 880C0C mov byte ptr ss:[ecx+esp], cl
0040499C FEC1 inc cl
0040499E ^ 75 F9 jne short 00404999
;SBOX permutieren
;Schlüssel ist DEADBEEFDEADBEEFDEADBEEFDEADBEEF…
004049A0 31C0 xor eax, eax
004049A2 BA EFBEADDE mov edx, DEADBEEF
004049A7 02040C add al, byte ptr ss:[ecx+esp]
004049AA 00D0 add al, dl
004049AC C1CA 08 ror edx, 8
004049AF 8A1C0C mov bl, byte ptr ss:[ecx+esp]
004049B2 8A3C04 mov bh, byte ptr ss:[eax+esp]
004049B5 881C04 mov byte ptr ss:[eax+esp], bl
004049B8 883C0C mov byte ptr ss:[ecx+esp], bh
004049BB FEC1 inc cl
004049BD ^ 75 E8 jne short 004049A7
004049BF E9 5C000000 jmp 00404A20
;Konsistenz der Daten überprüfen
004049C4 89E3 mov ebx, esp
004049C6 81C3 04000000 add ebx, 4
004049CC 5C pop esp
004049CD 58 pop eax
004049CE 3D 41414141 cmp eax, 41414141
004049D3 75 43 jne short 00404A18
004049D5 58 pop eax
004049D6 3D 42424242 cmp eax, 42424242
004049DB 75 3B jne short 00404A18
004049DD 5A pop edx
;kopieren des verschlüsselten Textes
004049DE 89D1 mov ecx, edx
004049E0 89E6 mov esi, esp
004049E2 89DF mov edi, ebx
004049E4 29CF sub edi, ecx
004049E6 F3:A4 rep movs byte ptr es:[edi], byte ptr ds:
004049E8 89DE mov esi, ebx
004049EA 89D1 mov ecx, edx
004049EC 89DF mov edi, ebx
004049EE 29CF sub edi, ecx
; Entschlüsseln
004049F0 31C0 xor eax, eax
004049F2 31DB xor ebx, ebx
004049F4 31D2 xor edx, edx
004049F6 FEC0 inc al
004049F8 021C06 add bl, byte ptr ds:[eax+esi]
004049FB 8A1406 mov dl, byte ptr ds:[eax+esi]
004049FE 8A341E mov dh, byte ptr ds:[ebx+esi]
00404A01 883406 mov byte ptr ds:[eax+esi], dh
00404A04 88141E mov byte ptr ds:[ebx+esi], dl
00404A07 00F2 add dl, dh
00404A09 30F6 xor dh, dh
00404A0B 8A1C16 mov bl, byte ptr ds:[edx+esi]
00404A0E 8A17 mov dl, byte ptr ds:[edi]
00404A10 30DA xor dl, bl
00404A12 8817 mov byte ptr ds:[edi], dl
00404A14 47 inc edi
00404A15 49 dec ecx
00404A16 ^ 75 DE jne short 004049F6
; Programm beenden
00404A18 31DB xor ebx, ebx
00404A1A 89D8 mov eax, ebx
00404A1C FEC0 inc al
00404A1E CD 80 int 80
00404A20 90 nop
00404A21 90 nop
00404A22 E8 9DFFFFFF call 004049C4
00404A27 41 41 41 41
00404A2B 42 42 42 42
00404A2F 32 00 00 00
00404A33 91 D8 F1 6D 70 20 3A AB 67 9A 0B C4 91 FB C7 66 0F FC CD CC B4 02 FA D7 77 B4 54 38 AB 1F 0E E3 8E D3 0D EB 99 C3 93 FE D1 2B 1B 11 C6 11 EF C8 CA 2F 00
Hier mein Vorschlag
http://pastebin.com/n8MqC5hD
Achja, hier noch die beiden relevanten VM Maschinenprogramme. Das erste Programm entschlüsselt das zweite ab Adresse 10:00.
Am Ende wird die Zeichenkette
GET /da75370fe15c4148bd4ceec861fbdaa5.exe HTTP/1.0
erzeugt.
00:0000 31 04 movr r1, #$04
00:0002 33 aa movr r3, #$aa
00:0004 40 02 movm r0, [ds:r2]
00:0006 80 03 xor r0, r3
00:0008 52 00 movm [ds:r2], r0
00:000a 72 01 add r2, #$01
00:000c 73 01 add r3, #$01
00:000e b2 50 cmp r2, #$50
00:0010 30 14 movr r0, #$14
00:0012 c0 jmpe r0
00:0013 01 jmp r1
00:0014 80 00 xor r0, r0
00:0016 10 10 jmp 10:r0
10:0000 32 00 movr r2, #$00
10:0002 75 0c add r5, #$0c
10:0004 31 08 movr r1, #$08
10:0006 33 32 movr r3, #$32
10:0008 40 02 movm r0, [ds:r2]
10:000a 80 03 xor r0, r3
10:000c 52 00 movm [ds:r2], r0
10:000e 72 01 add r2, #$01
10:0010 73 03 add r3, #$03
10:0012 b2 00 cmp r2, #$00
10:0014 c3 jmpe r3
10:0015 b0 00 cmp r0, #$00
10:0017 30 1b movr r0, #$1b
10:0019 c0 jmpe r0
10:001a 01 jmp r1
10:001b ff hlt
lösungsweg ist interessant ;)
@ 1t2 M3:
…fail xD
Ich lach mich tot: Nachdem man das gelöst hat wollen die einen animieren bei Ihnen zu arbeiten und bieten einem CYBER SECURITY SPECIALIST ein “Hungersalär” von “Salary £25,446 (GC10) £31,152 (GC9) ” ! Ich glaub ich spinne. Fragt doch besser mal den/die Reinigungsfrau/mann ob die dafür arbeitet ! . Kein Wunder finden die keine guten Leute !
Code : Pr0t3ct!on#cyber_security@12*12.2011+
Dnke DavidJ für die Erklärung.
Naja es sind ca 36.000 €/p.a. /GC10 in GB!
Reicht nur für ne SLUM-Hütte in einem Vorort von London mit Scharia-Zone frei-Haus.
LOL
Ich hatte gleich auf X86-Machinencode am Anfang gedacht, der irgend wetwas tut, aber dachte da steht vielleicht noch nen ELF–Header davor, da Unix hier häufig verwendet wird.
Immer müder Joe