To solve the problem, we need to analyze the determinant properties of the adjugate matrix when it is repeatedly applied to a $3 \times 3$ matrix.

Key Observations:

1. Adjugate and Determinants:

   • If $A$ is a $3 \times 3$ matrix, the determinant of the adjugate of $A$ is given by: $\det(\text{adj}(A)) = (\det(A))^{2}.$

2. Iterative Application:

   • Repeatedly taking the adjugate multiplies the determinant's power:
     • $\det(\text{adj}^k(A)) = (\det(A))^{2^k}.$

3. Given:

   • $\det(A) = 2$.
   • We need to find: $n = \det(\text{adj}^{2024}(A)),$ which will be: $n = (\det(A))^{2^{2024}} = 2^{2^{2024}}.$

4. Remainder modulo 9:

   • To simplify $2^{2^{2024}} \mod 9$, note that powers of 2 repeat in cycles modulo 9: $2^1 \equiv 2 \mod 9,\; 2^2 \equiv 4 \mod 9,\; 2^3 \equiv 8 \equiv -1 \mod 9,\; 2^4 \equiv 7 \equiv -2 \mod 9,\; 2^5 \equiv 5 \mod 9,\; 2^6 \equiv 1 \mod 9.$ The cycle repeats every 6 terms.

5. Exponent Simplification:

   • $2^{2024} \mod 6$:
     • Since $2^{2024}$ is a power of 2, it is divisible by 4, so $2^{2024} \mod 6 = 4$.
   • Thus: $2^{2^{2024}} \equiv 2^4 \mod 9.$

6. Final Calculation:

   • $2^4 = 16 \equiv 7 \mod 9$.

Answer:

The remainder when $n = 2^{2^{2024}}$ is divided by 9 is: $\boxed{7}.$

Let me know if you'd like a detailed breakdown of any step or further clarification! Here are some related questions:

1. What is the general pattern of powers of 2 modulo $n$ for different values of $n$?
2. How does the adjugate of a matrix behave under different dimensions?
3. Can we generalize this approach for $m \times m$ matrices?
4. What would change if $\det(A)$ were negative or not an integer?
5. How can this method be applied in cryptography or computational problems?

Tip: Always look for cyclical patterns in modular arithmetic to simplify large exponents.