\[ 2^{2024} = (2^2)^{2022} = 4 \cdot (8)^{674} = 4 \cdot (9 - 1)^{674}. \]
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:
\[ 7. \]
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)