\[ 2^{2024} = (2^2)^{2022} = 4 \cdot (8)^{674} = 4 \cdot (9 - 1)^{674}. \]
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:
\[ 7. \]
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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
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