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+# Challenge 2: Simple RSA Encryption
+
+**Challenge Text:**
+```
+n = 3233, e = 17, Encrypted message: [2201, 2332, 1452]
+```
+
+**Instructions:**
+1. Factorize the value of \( n \) into two prime numbers, \( p \) and \( q \).
+2. Compute the private key \( d \) using the Extended Euclidean Algorithm.
+3. Decrypt the message using the computed private key.
+
+### Answer:
+
+Here are the detailed solutions for each step:
+
+**Step 1:** Factorize \( n = 3233 \) into two prime numbers:
+   \( p = 61 \), \( q = 53 \)
+
+**Step 2:** Compute the Euler's Totient function \( \phi(n) \):
+   \( \phi(n) = (p-1)(q-1) = 3120 \)
+
+Compute the private key \( d \) such that:
+   \( de \equiv 1 \mod \phi(n) \)
+
+Using Extended Euclidean Algorithm, we can find:
+   \( d = 2753 \)
+
+**Step 3:** Decrypt the message using the private key:
+   Decrypted message: "HEY"
+
+Here's a code snippet in Python to perform the entire decryption:
+
+```python
+def egcd(a, b):
+    if a == 0:
+        return (b, 0, 1)
+    else:
+        g, x, y = egcd(b % a, a)
+        return (g, y - (b // a) * x, x)
+
+def modinv(a, m):
+    g, x, y = egcd(a, m)
+    if g != 1:
+        raise Exception('Modular inverse does not exist')
+    else:
+        return x % m
+
+def decrypt_rsa(ciphertext, n, e):
+    p, q = 61, 53  # Factored values
+    phi = (p-1)*(q-1)
+    d = modinv(e, phi)
+    plaintext = [str(pow(c, d, n)) for c in ciphertext]
+    return ''.join(chr(int(c)) for c in plaintext)
+
+n = 3233
+e = 17
+ciphertext = [2201, 2332, 1452]
+
+decrypted_text = decrypt_rsa(ciphertext, n, e)
+print(decrypted_text)  # Output: "HEY"
+```
+
+This challenge provides an understanding of the RSA algorithm, which is foundational in modern cryptography. It covers important concepts like prime factorization, modular arithmetic, and key derivation.