basic-mod1 picoCTF 2022 Solution

Description

1. Step 1Normalize the input
   Wrap the space-separated numbers as a Python list with one sed invocation, then iterate.
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   One sed invocation, four -e expressions applied in order to the same stream: -e "s/^/[/" prepends [, -e 's/ *$//' trims trailing spaces, -e "s/$/]/" appends ], and -e "s/ /, /g" rewrites every space as , . The result is a valid Python list literal in one pass.

   You could just open the file in Python and call .split() - the sed trick is a habit, not a requirement. Useful when you want a single shell-pipeline solution.

2. Step 2Map each value
   For every number n, compute n % 37 to keep it inside the alphabet range, then index into the character set ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_.
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   Modular arithmetic reduces a number to the remainder after dividing by a modulus. Here the modulus is 37 because the custom alphabet has exactly 37 characters (26 letters + 10 digits + 1 underscore). Any integer, no matter how large, maps to a value in 0..36 after % 37.

   The encoding rule is a simple positional table:

   index 0..25  -> 'A'..'Z'
   index 26..35 -> '0'..'9'
   index 36     -> '_'
   
   Worked example for n = 306:
     306 / 37 = 8.27...     (integer quotient = 8)
     8 * 37   = 296
     306 - 296 = 10         (remainder)
     alphabet[10] = 'K'
   
   More:
     617 % 37 = 25 -> 'Z'    (16 * 37 = 592, 617 - 592 = 25)
      28 % 37 = 28 -> '2'    (digit slot: 26 + 2)

   37 is prime, which means every non-zero value has a multiplicative inverse modulo 37 - a property the follow-up challenge basic-mod2 leans on.

   Python's % follows the divisor's sign: with a positive modulus the result is always in [0, m), even for negative dividends. C does not guarantee this; in C the sign can carry from the dividend, so -1 % 37 may give -1 instead of 36. A single comprehension - [alphabet[n % 37] for n in numbers] - is enough here.

3. Step 3Assemble the flag
   Build a string starting with picoCTF{...} to produce the final answer.
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   Once every integer is decoded into its character, joining the list with ''.join(chars) and wrapping it in the standard picoCTF{} format gives the submission-ready flag. This final assembly step is trivial but teaches an important habit: always verify the structure of your output matches what the challenge expects before submitting.

   The picoCTF{} wrapper is a flag format convention. Most CTF platforms use a consistent prefix (like flag{}, CTF{}, or competition-specific variants) so automated checkers can validate submissions. Recognizing these patterns helps you quickly confirm you decoded something correctly even before you know what the inner text means.

Flag

Simple modular arithmetic plus a custom alphabet turns the numeric sequence back into the plaintext flag.