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EYCC 2025 — Crypto Challenges

EYCC 2025 — Crypto Challenges#

EYCC Crypto Banner

Hey! This write-up covers the Egyptian Youth Cybersecurity Competition (EYCC) Crypto challenges that I managed to solve with steps taken to reveal the final flag.


First Challenge: Veiled Secret#

This challenge was about decrypting a secret message:

MKywL3gznaqhM2ghqzuxnzMvq-TulsD==TOR13

I noticed the TOR13 at the end, which points to starting with ROT13.

So, I copied the secret message to my go-to decoder, Dcode, and checked the output:

ROT13 Decoding

Now we’re left with Base64 encoded text, so I decoded it in the same website and got the flag:

Flag Revealed


Second Challenge: Golden Spiral#

We were given a file named goldenSpiral. It was not executable, so I made it executable:

Terminal window
chmod +x goldenSpiral

Then I tried executing it:

Terminal window
./goldenSpiral

But it threw an error:

bash: ./goldenSpiral: cannot execute: required file not found

I checked the file type:

Terminal window
file goldenSpiral

It revealed that it was an ELF executable requiring this interpreter:

/nix/store/lmn7lwydprqibdkghw7wgcn21yhllz13-glibc-2.40–66/lib/ld-linux-x86–64.so.2

After running the file with the required interpreter, an encrypted flag was returned:

fzef{7i0fGcobhxzwr4j}

The challenge mentioned a sequence that “grows almost exponentially,” suggesting the Fibonacci sequence. Since the encrypted flag seemed alphabetically shifted, I used this Fibonacci Cipher Decoder and started from 1 (because e comes right before f in the alphabet).

Fibonacci Cipher

And we got the flag!


Third Challenge: Silent Keys#

This challenge was an RSA decryption challenge.

We were given:

n = 148304669693572711157725718049458731328582148078019
e = 65537
c = 35618364216358867907731764651946346081071748936005

The hint said:

n is the product of two primes; one of them is the integer part of pi multiplied by 10^5

This challenge can be solved in two ways:

  1. Directly inputting the values in Dcode to get the flag:

RSA Decryption via Dcode

  1. Or factorizing n using factordb to get p and q:

Factoring n

Then calculating d from p, q, e:

Calculate d

Finally, decrypt the ciphertext c to reveal the flag:

Decrypted Flag


And that’s all! Don’t forget to check my other write-ups for the Web, Forensics, and OSINT challenges!

EYCC 2025 — Crypto Challenges
https://ssumix.vercel.app/posts/eycc/crypto-eycc-write-up/
Author
0xSky
Published at
2025-09-07
License
CC BY-NC-SA 4.0