E-CTF-2025 - Crypto

Step 1: Analyzing the Challenge Link to heading

Screenshot

The challenge provides a series of ASCII numbers along with an intriguing phrase:

108 100 111 109 123 85 99 49 122 95 106 53 95 79 111 51 95 88 52 116 95 48 109 95 51 111 88 121 90 107 97 106 48 105 125 10 10 69 98 111 98 32 102 112 32 118 108 114 111 32 104 98 118 44 32 100 108 108 97 32 105 114 122 104 32 58 32 72 66 86 72 66 86 10 10 87 101 108 108 32 100 111 110 101 44 32 98 117 116 32 110 111 119 32 100 111 32 121 111 117 32 107 110 111 119 32 97 98 111 117 116 32 116 104 101 32 103 117 121 32 119 104 111 32 103 111 116 32 115 116 97 98 98 101 100 32 50 51 32 116 105 109 101 115 32 63

Step 2: Decoding ASCII Link to heading

I used DCode to convert the ASCII numbers into text.

Decoded ASCII output:

ldom{Uc1z_j5_Oo3_X4t_0m_3oXyZkaj0i}
Ebob fp vlro hbv, dlla irzh : HBVHBV
Well done, but now do you know about the guy who got stabbed 23 times ?

The last sentence seems to be a hint, mentioning “the guy who got stabbed 23 times,” likely a historical reference.

Analysis: Link to heading

  • ldom{Uc1z_j5_Oo3_X4t_0m_3oXyZkaj0i} looks like a flag but is probably encrypted.
  • The phrase “Well done, but now do you know about the guy who got stabbed 23 times?” gives an important clue: ROT23.

Step 3: ROT23 Decryption Link to heading

I used DCode to apply ROT23 to the flag.

ROT23 output:

ogrp{Xf1c_m5_Rr3_A4w_0p_3rAbCndm0l}
Here is your key, good luck : KEYKEY
Zhoo grqh, exw qrz gr brx nqrz derxw wkh jxb zkr jrw vwdeehg 23 wlphv ?

Analysis: Link to heading

  • The flag remains encrypted.
  • A new sentence appears: “Here is your key, good luck : KEYKEY”.
  • We now have a key (“KEYKEY”) for a key-based encryption.

Step 4: Vigenère Decryption Link to heading

I tried several classic key-based algorithms, and Vigenère encryption yielded a good result.

I used DCode, entering ogrp{Xf1c_m5_Rr3_A4w_0p_3rAbCndm0l} and the key KEYKEY.

Output:

ectf{Th1s_i5_Th3_W4y_0f_3nCrYpti0n}

Flag found