Find out the unit digit in the `(7^(95)-3^(58))`

Options

A. A. 7
B. B. 3
C. C. 4
D. D. 0

Answer

To find the unit digit of \(7^{95} - 3^{58}\), we will follow these steps:

### Step 1: Find the unit digit of \(7^{95}\)

1. **Identify the pattern of unit digits for powers of 7:**
- \(7^1 = 7\) (unit digit is 7)
- \(7^2 = 49\) (unit digit is 9)
- \(7^3 = 343\) (unit digit is 3)
- \(7^4 = 2401\) (unit digit is 1)
- The pattern of unit digits is: **7, 9, 3, 1** (repeats every 4 terms).

2. **Determine the position in the cycle for \(7^{95}\):**
- Divide 95 by 4:
\[ 95 \div 4 = 23 \quad \text{remainder } 3 \]
- Since the remainder is 3, the unit digit of \(7^{95}\) corresponds to the 3rd position in the cycle, which is **3**.

### Step 2: Find the unit digit of \(3^{58}\)

1. **Identify the pattern of unit digits for powers of 3:**
- \(3^1 = 3\) (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)
- The pattern of unit digits is: **3, 9, 7, 1** (repeats every 4 terms).

2. **Determine the position in the cycle for \(3^{58}\):**
- Divide 58 by 4:
\[ 58 \div 4 = 14 \quad \text{remainder } 2 \]
- Since the remainder is 2, the unit digit of \(3^{58}\) corresponds to the 2nd position in the cycle, which is **9**.

### Step 3: Calculate \(7^{95} - 3^{58}\)

1. **Subtract the unit digits:**
- The unit digit of \(7^{95}\) is 3 and the unit digit of \(3^{58}\) is 9.
- Now, calculate:
\[ 3 - 9 = -6 \]
- Since we cannot have a negative unit digit, we can adjust this by adding 10:
\[ -6 + 10 = 4 \]

### Conclusion

The unit digit of \(7^{95} - 3^{58}\) is **4**.