Let $$A$$ be a $$3 \times 3$$ matrix and $$\det(A) = 2$$. If $$n = \det(\underbrace{adj(adj(\ldots adj(A)))}_{\text{2024 times}})$$, then the remainder when $$n$$ is divided by 9 is equal to

We need to find the remainder when $$n = \det(\text{adj}(\text{adj}(\cdots\text{adj}(A)\cdots)))$$ (2024 times) is divided by 9, given that $$A$$ is a $$3 \times 3$$ matrix with $$\det(A) = 2$$.

To begin, we derive the formula for the determinant of the adjoint of an $$n \times n$$ matrix. Since $$\text{adj}(A) = \det(A)\cdot A^{-1}$$ when $$A$$ is invertible, taking determinants on both sides gives $$\det(\text{adj}(A)) = (\det(A))^n \cdot \det(A^{-1}) = (\det(A))^n \cdot \frac{1}{\det(A)} = (\det(A))^{n-1}.$$ For $$n = 3$$, this becomes $$\det(\text{adj}(A)) = (\det(A))^2.$$

Next, we build a recurrence for the determinant after successive adjoint operations. Let $$D_k$$ be the determinant after applying the adjoint operation $$k$$ times. We have $$D_0 = \det(A) = 2,$$ and each adjoint operation squares the determinant, so $$D_1 = D_0^2 = 2^2 = 4,$$ $$D_2 = D_1^2 = 4^2 = 2^4,$$ $$D_3 = D_2^2 = (2^4)^2 = 2^8.$$ In general, one finds $$D_k = 2^{2^k}.$$

We then compute $$D_{2024} \bmod 9$$ by evaluating $$2^{2^{2024}} \bmod 9.$$

To proceed, we note the pattern of powers of 2 modulo 9: $$2^1 = 2,\quad 2^2 = 4,\quad 2^3 = 8,\quad 2^4 = 16 \equiv 7,\quad 2^5 = 32 \equiv 5,\quad 2^6 = 64 \equiv 1 \pmod{9}.$$ Thus the cycle length is 6, so the value of $$2^m \bmod 9$$ depends only on $$m \bmod 6$$.

We next determine $$2^{2024} \bmod 6$$ by computing it modulo 2 and modulo 3 separately. Since any power of 2 with exponent at least 1 is even, we have $$2^{2024} \bmod 2 = 0.$$ Moreover, because $$2 \equiv -1 \pmod{3},$$ $$2^{2024} \equiv (-1)^{2024} = 1 \pmod{3}.$$ By the Chinese Remainder Theorem, the unique solution modulo 6 satisfying these conditions is $$2^{2024} \equiv 4 \pmod{6}.$$

It follows that $$2^{2^{2024}} \equiv 2^4 = 16 \equiv 7 \pmod{9}.$$

The answer is 7.