Tryhackme — c4ptur3-th3-fl4g Writeup (Bahasa Indonesia)

c4ptur3-th3-fl4g Tryhakcme — A beginner level CTF challenge

Pada room ini, kita akan belajar tentang teknik encoding-decoding data, encryption dan steganography.

note :

  1. Disarankan memahami dasar linux (lebih baik jika juga menggunakan linux).
  2. Pengetahuan teoretis tentang apa itu encoding-decoding, encryption, dan steganography tidak akan dibahas lengkap disini. Oleh sebab itu, silakan manfaatkan kuota internet kita untuk googling, baca buku, dsb., agar lebih memahami itu semua.

Answers are all case sensitive.

Task 1 — Translation and Shifting

note : Kita akan banyak menggunakan situs CyberChef pada Task 1 ini!

  • c4n y0u c4p7u23 7h3 f149?

Mudah, tinggal ubah kumpulan karakter tersebut ke dalam tulisan normal.

correct answer : can you capture the flag?

  • 01101100 01100101 01110100 01110011 00100000 01110100 01110010 01111001 00100000 01110011 01101111 01101101 01100101 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01101111 01110101 01110100 00100001

Susunan data diatas adalah sistem data basis 2 alias data biner (binary)yang terdiri dari 0 dan 1. Sistem bilangan biner adalah dasar dari segala bilangan digital. Pengelompokan bilangan biner dalam komputer selalu berjumlah 8 karakter 0 dan 1 dengan istilah Byte, sehingga 1 Byte = 8 bit. Agar lebih memahami dunia per-biner-an lebih jauh, silakan googling :)

Untuk menemukan jawaban dari soal ini, kita hanya perlu men-copy-paste bilangan biner tersebut ke CyberChef dan mengkonversikannya ke teks.

  • MJQXGZJTGIQGS4ZAON2XAZLSEBRW63LNN5XCA2LOEBBVIRRHOM======

Jika dilihat dari kumpulan karakter di atas, sekilas hanya terdapat abjad/huruf besar/kapital seperti G, L, N, dll., dan beberapa bilangan desimal seperti 2, 3, 4, 5, dan 6. Ciri-ciri seperti ini kemungkinannya hanya ada pada tipe encoding binary/data base32. Jadi, langsung saja kita konversikan…

Atau bisa juga mengkonversi base32 melalui command line di terminal dengan perintah : echo MJQXGZJTGIQGS4ZAON2XAZLSEBRW63LNN5XCA2LOEBBVIRRHOM====== | base32 -d

  • RWFjaCBCYXNlNjQgZGlnaXQgcmVwcmVzZW50cyBleGFjdGx5IDYgYml0cyBvZiBkYXRhLg==

Jika dianalisis, karakter atau simbol yang terdapat dalam kumpulan simbol tersebut terdiri dari huruf/abjad besar/kapital dan huruf kecil, serta beberapa angka desimal. Ciri-ciri seperti ini ada pada tipe encoding data base64. Sama seperti soal sebelumnya, proses konversi bisa dilakukan di CyberChef dan melalui command line di terminal linux.

  • 68 65 78 61 64 65 63 69 6d 61 6c 20 6f 72 20 62 61 73 65 31 36 3f

Karakter dan simbol yang terdapat dalam data tersebut terdiri dari angka decimal 0, 1, 2, 3, 4, 5, 6, 8, 9 serta beberapa huruf seperti c, d, dan f. Dilihat dari susunannya, data tersebut juga tersusun dari 2 digit karakter yang dipisahkan dengan spasi. Ciri tersebut banyak terdapat dalam tipe encoding base16 (hexadecimal).

  • Ebgngr zr 13 cynprf!

Jika melihat soal tersebut sepintas lalu, mungkin memang agak sulit untuk menebak tipe atau jenis encoding yang digunakan. Akan tetapi, dilihat dari bentuk polanya, boleh jadi soal di atas adalah string yang telah dienkripsi (encrypted text/chipertext). Tidak ada salahnya untuk mula-mula mengasumsikan chipertext tersebut adalah hasil enkripsi plain text dengan metode Caesar Chiper, mengingat metode Caesar merupakan metode enkripsi yang paling umum, mudah, dan pertama diajarkan di materi-materi enkripsi dan kriptografi.

Pada chipertext tersebut juga terdapat angka 13, jadi, gak ada salahnya pula untuk berasumsi bahwa “kunci” chipertext tersebut berada di pergeseran/rotasi ke-13. Ngomong-ngomong tentang rotasi, ada 2 tipe rotasi yang umum digunakan, yaitu rotasi 13 (ROT13) dan rotasi 47 (ROT47).

Untuk kasus ini, kita coba men-decrypt chipertext di atas dari ROT13.

note : untuk pengetahuan lebih lanjut tentang ROT13 dan ROT47, silakan googling :)

  • *@F DA:? >6 C:89E C@F?5 323J C:89E C@F?5 Wcf E:>6DX

Kita coba dekripsi dengan ROT47 alias rotate 47.

  • - . .-.. . -.-. — — — — ..- -. .. -.-. .- — .. — — -.

. -. -.-. — — -.. .. -. — .

Penyandian dengan ciri-ciri titik-titik pendek, sedang, dan panjang. Apalagi kalau bukan morse, di pramuka pun sudah sering di jelaskan dan dipraktekkan. Jadi, langsung saja kita decode!

  • 85 110 112 97 99 107 32 116 104 105 115 32 66 67 68

Perhatikan, karakter-karakter di atas hanya terdiri dari angka 0–9. Sudah dapat dipastikan kalau itu adalah kumpulan bilangan desimal. Jadi, langsung kita konversikan!

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Agar lebih mudah mengerti apa tipe encoding yang digunakan disini, kita akan menggunakan fitur ‘magic’ di CyberChef.

Hasilnya menyatakan bahwa kumpulan simbol tersebut dapat dikonversi menggunakan tipe encoding base64. Kita coba…

Ternyata hasilnya tidak langsung terkonversi ke bentuk teks normal, melainkan ke bentuk morse. Artinya, ini adalah tipe encoding bersusun/berjenjang/bertahap. Kita baru akan menemukan jawabannya setelah men-decode beberapa tahap lagi.

Dan setelah 5 kali melakukan proses decoding, kita berhasil menemukan jawabannya!!

Task 2 — Spectogram

Silakan download terlebih dahulu file yang tersedia.

note : Disini, saya akan membuka file ‘.wav’ tersebut dengan Sonic Visualiser.

Layer -> Add Spectogram

Task 3 — Steganography

Silakan download terlebih dahulu file yang tersedia.

note : Untuk lebih jauh mengetahui tentang seluk-beluk steganography, bisa belajar di room CC: Steganography dan baca-baca di Writeup Tryhackme — CC: Steganography Writeup (Bahasa Indonesia).

Task 4 — Security through obscurity

Silakan download terlebih dahulu file yang tersedia.

Untuk menyelesaikan 2 soal di bawah, kita bisa menggunakan command “srings” di command line linux masing-masing. Dengan command itu, kita akan diberikan seluruh string yang terdapat dalam file tersebut.

Untuk cara penggunaanya, cukup dengan mengetikkan perintah : strings meme.jpg

  • Download and get ‘inside’ the file. What is the first filename & extension?
  • Get inside the archive and inspect the file carefully. Find the hidden text.

Yakkk!! Akhirnya semua challange telah terselesaikan dengan sempurna!

Thankyou guys! See you next time!! :))

Happy Hacking!

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A college student. A writer. Cyber security enthusiast.

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Alex

Alex

A college student. A writer. Cyber security enthusiast.

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