Find the unit place digit in `(82)^(102)+(183)^(103)`

Options

A. 1
B. 6
C. 8
D. 9

Answer

To find the unit place digit of \( (82)^{102} + (183)^{103} \), we can focus on the unit digits of the bases, which are 2 and 3 respectively.

### Step 1: Determine the unit digit of \( (82)^{102} \)

1. The unit digit of \( 82 \) is \( 2 \).
2. We need to find the unit digit of \( 2^{102} \).
3. The unit digits of powers of \( 2 \) cycle every 4:
- \( 2^1 = 2 \) (unit digit is 2)
- \( 2^2 = 4 \) (unit digit is 4)
- \( 2^3 = 8 \) (unit digit is 8)
- \( 2^4 = 16 \) (unit digit is 6)
- Then it repeats: \( 2, 4, 8, 6 \).

4. To find \( 2^{102} \), we calculate \( 102 \mod 4 \):
- \( 102 \div 4 = 25 \) remainder \( 2 \).
- Thus, \( 102 \mod 4 = 2 \).

5. From the cycle, the unit digit corresponding to \( 2 \) is \( 4 \).

### Step 2: Determine the unit digit of \( (183)^{103} \)

1. The unit digit of \( 183 \) is \( 3 \).
2. We need to find the unit digit of \( 3^{103} \).
3. The unit digits of powers of \( 3 \) cycle every 4:
- \( 3^1 = 3 \) (unit digit is 3)
- \( 3^2 = 9 \) (unit digit is 9)
- \( 3^3 = 27 \) (unit digit is 7)
- \( 3^4 = 81 \) (unit digit is 1)
- Then it repeats: \( 3, 9, 7, 1 \).

4. To find \( 3^{103} \), we calculate \( 103 \mod 4 \):
- \( 103 \div 4 = 25 \) remainder \( 3 \).
- Thus, \( 103 \mod 4 = 3 \).

5. From the cycle, the unit digit corresponding to \( 3 \) is \( 7 \).

### Step 3: Add the unit digits

1. Now we add the unit digits obtained:
- Unit digit of \( (82)^{102} \) is \( 4 \).
- Unit digit of \( (183)^{103} \) is \( 7 \).
- Therefore, \( 4 + 7 = 11 \).

2. The unit digit of \( 11 \) is \( 1 \).

### Final Answer

The unit place digit in \( (82)^{102} + (183)^{103} \) is \( 1 \).