Find out the unit digit in the -
`888^(92335!)+222^(9235!)+666^(2359!)+9999^(9999!)`

Options

A. 5
B. 9
C. 3
D. None of these

Answer

To find the unit digit of the expression \( 888^{92335!} + 222^{9235!} + 666^{2359!} + 9999^{9999!} \), we can analyze each term separately to determine its unit digit.

### Step 1: Find the unit digit of \( 888^{92335!} \)

The unit digit of \( 888 \) is \( 8 \). We need to find the unit digit of \( 8^{92335!} \).

The unit digits of the powers of \( 8 \) cycle every 4:
- \( 8^1 \) has unit digit \( 8 \)
- \( 8^2 \) has unit digit \( 4 \)
- \( 8^3 \) has unit digit \( 2 \)
- \( 8^4 \) has unit digit \( 6 \)
- \( 8^5 \) has unit digit \( 8 \) (and the cycle repeats)

To determine which unit digit to use, we find \( 92335! \mod 4 \):
- Since \( 92335! \) is a factorial of a number greater than 4, it is divisible by \( 4 \). Thus, \( 92335! \mod 4 = 0 \).

From the cycle, if the exponent is \( 0 \mod 4 \), the unit digit is \( 6 \).

### Step 2: Find the unit digit of \( 222^{9235!} \)

The unit digit of \( 222 \) is \( 2 \). We need to find the unit digit of \( 2^{9235!} \).

The unit digits of the powers of \( 2 \) cycle every 4:
- \( 2^1 \) has unit digit \( 2 \)
- \( 2^2 \) has unit digit \( 4 \)
- \( 2^3 \) has unit digit \( 8 \)
- \( 2^4 \) has unit digit \( 6 \)
- \( 2^5 \) has unit digit \( 2 \) (and the cycle repeats)

To determine which unit digit to use, we find \( 9235! \mod 4 \):
- Since \( 9235! \) is also a factorial of a number greater than 4, it is divisible by \( 4 \). Thus, \( 9235! \mod 4 = 0 \).

From the cycle, if the exponent is \( 0 \mod 4 \), the unit digit is \( 6 \).

### Step 3: Find the unit digit of \( 666^{2359!} \)

The unit digit of \( 666 \) is \( 6 \). The unit digit of any power of \( 6 \) is always \( 6 \) (since \( 6^1 = 6, 6^2 = 36, 6^3 = 216, \ldots \)).

Thus, the unit digit of \( 666^{2359!} \) is \( 6 \).

### Step 4: Find the unit digit of \( 9999^{9999!} \)

The unit digit of \( 9999 \) is \( 9 \). We need to find the unit digit of \( 9^{9999!} \).

The unit digits of the powers of \( 9 \) cycle every 2:
- \( 9^1 \) has unit digit \( 9 \)
- \( 9^2 \) has unit digit \( 1 \)
- \( 9^3 \) has unit digit \( 9 \) (and the cycle repeats)

To determine which unit digit to use, we find \( 9999! \mod 2 \):
- Since \( 9999! \) is a factorial of a number greater than 1, it is divisible by \( 2 \). Thus, \( 9999! \mod 2 = 0 \).

From the cycle, if the exponent is \( 0 \mod 2 \), the unit digit is \( 1 \).

### Step 5: Combine the unit digits

Now we can combine the unit digits we found:
- From \( 888^{92335!} \): \( 6 \)
- From \( 222^{9235!} \): \( 6 \)
- From \( 666^{2359!} \): \( 6 \)
- From \( 9999^{9999!} \): \( 1 \)

Now, we add these unit digits:
\[ 6 + 6 + 6 + 1 = 19 \]

The unit digit of \( 19 \) is \( 9 \).

### Final Answer

The unit digit of \( 888^{92335!} + 222^{9235!} + 666^{2359!} + 9999^{9999!} \) is \( 9 \).

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