Find the unit digit of the expression.
`888^(9235!) + 222^(9235!) + 666^(2359!) + 999^(9999!)`.

Answer

To find the unit digit of the expression \( 888^{9235!} + 222^{9235!} + 666^{2359!} + 999^{9999!} \), we can focus on the unit digits of each base number and their respective powers.

### Step-by-Step Solution:

1. **Identify the unit digits of the bases:**
- The unit digit of \( 888 \) is \( 8 \).
- The unit digit of \( 222 \) is \( 2 \).
- The unit digit of \( 666 \) is \( 6 \).
- The unit digit of \( 999 \) is \( 9 \).

2. **Determine the cycles of unit digits for each base:**
- For \( 8 \): The unit digits cycle is \( 8, 4, 2, 6 \) (repeats every 4 terms).
- For \( 2 \): The unit digits cycle is \( 2, 4, 8, 6 \) (repeats every 4 terms).
- For \( 6 \): The unit digit is always \( 6 \) (does not change).
- For \( 9 \): The unit digits cycle is \( 9, 1 \) (repeats every 2 terms).

3. **Determine the power for each base:**
- Since \( 9235! \) and \( 2359! \) are both factorials, they are guaranteed to be multiples of \( 4 \) (as \( 4! = 24 \) and higher factorials include multiples of \( 4 \)).
- Therefore, for \( 8^{9235!} \) and \( 2^{9235!} \), we need the unit digit corresponding to \( 0 \) in their cycles (as \( 9235! \mod 4 = 0 \)).
- For \( 6^{2359!} \), the unit digit is \( 6 \) regardless of the power.
- For \( 9^{9999!} \), since \( 9999! \) is even, we take the second term in its cycle, which is \( 1 \).

4. **Calculate the unit digits:**
- \( 8^{9235!} \) has a unit digit of \( 6 \) (from the cycle \( 8, 4, 2, 6 \)).
- \( 2^{9235!} \) has a unit digit of \( 6 \) (from the cycle \( 2, 4, 8, 6 \)).
- \( 6^{2359!} \) has a unit digit of \( 6 \).
- \( 9^{9999!} \) has a unit digit of \( 1 \).

5. **Sum the unit digits:**
- Now we add the unit digits: \( 6 + 6 + 6 + 1 = 19 \).

6. **Find the unit digit of the sum:**
- The unit digit of \( 19 \) is \( 9 \).

### Final Answer:
The unit digit of the expression \( 888^{9235!} + 222^{9235!} + 666^{2359!} + 999^{9999!} \) is \( 9 \).