53 lines
1.8 KiB
Markdown
53 lines
1.8 KiB
Markdown
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## MODERN CLUELESS CHILD
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The main idea finding the flag is decryption using XOR keys.
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#### Step-1:
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After reading the given message:
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```bash
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I was surfing the crimson wave and oh my gosh I was totally bugging. I also tried out the lilac hair trend but it didn't work out. That's not to say you are any better, you are a snob and a half. But let's get back to the main question here- Who am I? (You don't know my name)
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Ciphertext = "52f41f58f51f47f57f49f48f5df46f6ef53f43f57f6cf50f6df53f53f40f58f51f6ef42f56f43f41f5ef5cf4e" (hex) Key = "12123"
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```
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#### Step-2:
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I quickly removed the `f` from cipher text as looked like it was used for space. So I wrote a script `sub.py` to replace `f` with `''`.
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```python
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ciphertext = "52f41f58f51f47f57f49f48f5df46f6ef53f43f57f6cf50f6df53f53f40f58f51f6ef42f56f43f41f5ef5cf4e"
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sub = ciphertext.replace('f','')
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print(sub)
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```
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On running `python3 sub.py` this, it gave me `52415851475749485d466e5343576c506d53534058516e425643415e5c4e`.
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#### Step-3:
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I had to check if I am not missing any cipher text so I cross check the flag by XOR checks. So, I wrote this `xor1.py` script get the `csictf{` code:
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```python
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from pwn import xor
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flag = bytes.fromhex('52415851475749485d466e5343576c506d53534058516e425643415e5c4e')
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print(xor(flag, 'csictf{'.encode()))
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```
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Output:
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```bash
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b"1212312+./\r'%,\x0f#\x040'\x1d+57'%?="
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```
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#### Step-4:
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Since we got the key `1212312` means we are right path as key has cyclic property key (12123). Now it was just replacement on the key with ASCII.
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`exlpoit.py` to get flag:
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```python
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from pwn import xor
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flag = bytes.fromhex('52415851475749485d466e5343576c506d53534058516e425643415e5c4e')
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print(xor(flag, '12123'.encode()))
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```
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On running `python3 exploit.py`, Voila! I got the flag.
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```bash
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b'csictf{you_are_a_basic_person}'
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```
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#### Step-5:
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Finally the flag becomes:
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`csictf{you_are_a_basic_person}`
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