> For the complete documentation index, see [llms.txt](https://alham-rizvi.gitbook.io/alhamrizvi/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://alham-rizvi.gitbook.io/alhamrizvi/picoctf-writeups/picoctf/cryptography/customencryption.md).

# customencryption

Alright — let’s clean this up properly because your confusion is **100% valid**. Nothing “mystical” happened, just **CTF trickery + mismatch of values**.

I’ll walk you **step-by-step clean, no BS**, so you can repeat this in exams/CTFs.

***

## 🧠 1. What the challenge ACTUALLY is

This is **NOT RSA** (they trolled you).

It’s:

#### 🔐 Step 1: Diffie-Hellman → shared key

#### 🔐 Step 2: XOR encryption (with `"trudeau"`)

#### 🔐 Step 3: Multiply by `(key * 311)`

## 🔍 3. How to solve it PROPERLY (real method)

Since `a` and `b` are unreliable → **ignore them**

***

### ✅ Step 1: Understand formula

From code:

```python
cipher[i] = ord(char) * key * 311
```

***

### ✅ Step 2: Reverse multiplication

```python
char = cipher[i] / (key * 311)
```

***

### ✅ Step 3: Problem

👉 You don’t know `key`

BUT:

* `p = 97`
* So key ∈ \[1, 96]

👉 Small range = brute force 💀

***

## 🚀 4. Final Working Method (ALWAYS WORKS)

### ✔️ Step-by-step logic

1. Try every possible key
2. Divide cipher
3. Convert to characters
4. Undo XOR
5. Reverse string
6. Check for `flag{}` pattern

***

### 💻 Final Script (clean + exam-ready)

```python
cipher = [33588, 276168, 261240, 302292, 343344, 328416, 242580, 85836, 82104, 156744, 0, 309756, 78372, 18660, 253776, 0, 82104, 320952, 3732, 231384, 89568, 100764, 22392, 22392, 63444, 22392, 97032, 190332, 119424, 182868, 97032, 26124, 44784, 63444]

text_key = "trudeau"

for key in range(1, 100):
    try:
        semi = ""

        # Step 1: undo multiplication
        for num in cipher:
            if num == 0:
                semi += "\x00"
            else:
                val = num // (key * 311)
                if val < 0 or val > 255:
                    raise
                semi += chr(val)

        # Step 2: undo XOR
        result = ""
        for i, c in enumerate(semi):
            result += chr(ord(c) ^ ord(text_key[i % len(text_key)]))

        # Step 3: reverse back
        final = result[::-1]

        # Step 4: check flag pattern
        if "{" in final and "}" in final:
            print(f"[+] Key: {key} -> {final}")

    except:
        continue
```

> If key space is small → **BRUTE FORCE FIRST, THINK LATER**

***

## ⚡ Quick Mental Model

| Situation                  | Action      |
| -------------------------- | ----------- |
| Given a, b look sketchy    | Ignore      |
| Output looks semi-readable | Logic issue |
| Output total garbage       | Math wrong  |
| Small modulus (like 97)    | brute force |

***

## Final Truth

Nothing broke.

👉 You just trusted **fake values** instead of the **cipher itself**

That’s a classic CTF trap.


---

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