Flash CTF – Floating

Challenge Overview

Did you know that almost anything can be represented in IEEE-754 floating point values?

We’re given a list of numbers that don’t immediately look like text, bytes, or anything obviously encoded:

240600592
212.2753143310547
2.7884192016691608e+23
5.623021054185822e+31
17611451687157891000
8.927742989328635e-10
16391240070931153000
5.639361688736244e-8
2.115975377137147e-7

The goal is to figure out what these values really represent.

What Is a Float, Anyway?

A floating-point number (or “float”) is one way that computers represent decimal numbers using binary. The most common standard is IEEE-754, which defines how numbers are stored at the bit level.

For a 32-bit float (float32), the bits are split like this:

• 1 bit: sign (positive or negative)
• 8 bits: exponent
• 23 bits: mantissa (the significant digits)

The important thing for this challenge: A float is just 32 bits of data. We choose to interpret those bits as a number, but they could just as easily be ASCII characters, integers, or anything else.

Key Insight

These numbers aren’t meant to be calculated with, instead, each one is hiding 4 bytes of data inside its IEEE-754 representation.

If we:

1. Interpret each value as a 32-bit float
2. Use big-endian byte order
3. Concatenate the raw bytes
4. Decode them as ASCII

The flag appears cleanly!

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Solving with CyberChef (No Code)

CyberChef makes this challenge very approachable:

1. Paste the numbers into CyberChef
2. Use From Float
   • Set Endianess to Big Endian
   • Set Size to Float (4 bytes)
   • Set delimiter to Space (unless you’ve put the numbers on their own lines as this writeup does)

When the settings are correct, the output immediately becomes readable text.

Solving with Python

If you prefer scripting, here’s a minimal Python solve.

import struct

values = [
    "240600592",
    "212.2753143310547",
    "2.7884192016691608e+23",
    "5.623021054185822e+31",
    "17611451687157891000",
    "8.927742989328635e-10",
    "16391240070931153000",
    "5.639361688736244e-8",
    "2.115975377137147e-7"
]

blob = b""

for v in values:
    blob += struct.pack(">f", float(v))  # 32-bit big-endian float

print(blob.decode("ascii"))

Flag

MetaCTF{fl04t1ng_thr0ugh_cyb3r5p4c3}