\[ 2^{2024} = (2^2)^{2022} = 4 \cdot (8)^{674} = 4 \cdot (9 - 1)^{674}. \]
Applying modulo 9, we get:
\[ 2^{2024} \equiv 4 \pmod{9}. \]
Thus,
\[ 2^{2024} = 9m + 4, \quad m \text{ is even}. \]
Now, consider \(2^{9m+4}\):
\[ 2^{9m+4} = 16 \cdot (2^3)^{3m} \equiv 16 \pmod{9}. \]
Thus,
\[ = 7. \]
Therefore, the answer is:
\[ 7. \]
Let I be the identity matrix of order 3 × 3 and for the matrix $ A = \begin{pmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{pmatrix} $, $ |A| = -1 $. Let B be the inverse of the matrix $ \text{adj}(A \cdot \text{adj}(A^2)) $. Then $ |(\lambda B + I)| $ is equal to _______
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
If $ y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\27 & 28 & 27 \\1 & 1 & 1 \end{vmatrix} $, $ x \in \mathbb{R} $, then $ \frac{d^2y}{dx^2} + y $ is equal to
Match List-I with List-II: List-I