Easy1

Challenge

“The one-time pad can be secure, but not when you know the key. We give you the encrypted flag, a key, and a Vigenère table. Can you solve it?”

• Ciphertext: UFJKXQZQUNB
• Key: SOLVECRYPTO
• Goal: find the plaintext and wrap it as picoCTF{...}.

Quick intuition

This one’s a gentle warm-up. The challenge throws you a Vigenère cipher problem, one of those classic “fancy Caesar cipher” types from old-school cryptography.

If Caesar cipher shifts every letter by the same amount, Vigenère mixes things up by using a word as the key, where each letter of the key decides how much to shift the corresponding letter of the plaintext. For example, if your key is “DOG”, the first letter gets shifted by D, the next by O, and the third by G. Then the key repeats if the message is longer.

It’s like taking Caesar’s simple shift and putting it on shuffle mode.

The encryption and decryption process usually depends on a Vigenère table, or tabula recta, which looks like this: (provided inside the challange as text file)

codeA B C D E F G H I J K L M N O P Q R S T U V W X Y Z   
   +----------------------------------------------------  
A | A B C D E F G H I J K L M N O P Q R S T U V W X Y Z  
B | B C D E F G H I J K L M N O P Q R S T U V W X Y Z A  
C | C D E F G H I J K L M N O P Q R S T U V W X Y Z A B  
D | D E F G H I J K L M N O P Q R S T U V W X Y Z A B C  
E | E F G H I J K L M N O P Q R S T U V W X Y Z A B C D  
F | F G H I J K L M N O P Q R S T U V W X Y Z A B C D E  
G | G H I J K L M N O P Q R S T U V W X Y Z A B C D E F  
H | H I J K L M N O P Q R S T U V W X Y Z A B C D E F G  
I | I J K L M N O P Q R S T U V W X Y Z A B C D E F G H  
J | J K L M N O P Q R S T U V W X Y Z A B C D E F G H I  
K | K L M N O P Q R S T U V W X Y Z A B C D E F G H I J  
L | L M N O P Q R S T U V W X Y Z A B C D E F G H I J K  
M | M N O P Q R S T U V W X Y Z A B C D E F G H I J K L  
N | N O P Q R S T U V W X Y Z A B C D E F G H I J K L M  
O | O P Q R S T U V W X Y Z A B C D E F G H I J K L M N  
P | P Q R S T U V W X Y Z A B C D E F G H I J K L M N O  
Q | Q R S T U V W X Y Z A B C D E F G H I J K L M N O P  
R | R S T U V W X Y Z A B C D E F G H I J K L M N O P Q  
S | S T U V W X Y Z A B C D E F G H I J K L M N O P Q R  
T | T U V W X Y Z A B C D E F G H I J K L M N O P Q R S  
U | U V W X Y Z A B C D E F G H I J K L M N O P Q R S T  
V | V W X Y Z A B C D E F G H I J K L M N O P Q R S T U  
W | W X Y Z A B C D E F G H I J K L M N O P Q R S T U V  
X | X Y Z A B C D E F G H I J K L M N O P Q R S T U V W  
Y | Y Z A B C D E F G H I J K L M N O P Q R S T U V W X  
Z | Z A B C D E F G H I J K L M N O P Q R S T U V W X Y

When encrypting, you pick the row for the key letter and the column for the plaintext letter, the intersection gives you the ciphertext.
To decrypt, you do the opposite: find the ciphertext in the row of the key letter, then trace it back to the column header.

But here’s the funny thing the challenge already gives us the key. So, instead of “breaking” anything, we’re really just reversing the process.

Background

If we think of letters as numbers (A=0, B=1, …, Z=25), the formulas are:

Encryption:

C = (P + K) mod 26

Decryption:

P = (C - K) mod 26

Since the key and ciphertext are the same length, we can just align them and go letter by letter.

Step-by-step solve

Let’s line them up:

Ciphertext: U F J K X Q Z Q U N B
Key : S O L V E C R Y P T O

Now we subtract the key from the ciphertext (mod 26) to get our plaintext:

Plaintext: CRYPTOISFUN

Easy solve with CyberChef

This challange can easily be solved using CyberChef.

Recipe: Vigenère Decode
Key: SOLVECRYPTO
Input: UFJKXQZQUNB
Output: CRYPTOISFUN

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Flag

picoCTF{CRYPTOISFUN}

The challenge’s entire trick is that Vigenère isn’t secure when you already have the key. Once you know it, decryption is just basic math. The point here is to understand how the cipher works, not to brute-force it.