Find the unit digit in `7^(65) xx 6^(41) xx 3^(57)`.

Options

A. 6
B. 7
C. 5
D. 8

Answer

To find the unit digit of the expression \( 7^{65} \times 6^{41} \times 3^{57} \), we can follow these steps:

### Step 1: Find the unit digit of \( 7^{65} \)

The unit digits of the powers of 7 follow a pattern:
- \( 7^1 = 7 \) (unit digit 7)
- \( 7^2 = 49 \) (unit digit 9)
- \( 7^3 = 343 \) (unit digit 3)
- \( 7^4 = 2401 \) (unit digit 1)
- \( 7^5 = 16807 \) (unit digit 7) - and the pattern repeats.

The pattern of unit digits is: 7, 9, 3, 1 (repeats every 4 terms).

To find the unit digit of \( 7^{65} \), we calculate \( 65 \mod 4 \):
\[ 65 \div 4 = 16 \quad \text{remainder} \quad 1 \]
Thus, \( 65 \mod 4 = 1 \). The unit digit corresponds to the first term in the pattern, which is 7.

### Step 2: Find the unit digit of \( 6^{41} \)

The unit digits of the powers of 6 are always 6:
- \( 6^1 = 6 \) (unit digit 6)
- \( 6^2 = 36 \) (unit digit 6)
- \( 6^3 = 216 \) (unit digit 6)
- ...

Thus, the unit digit of \( 6^{41} \) is 6.

### Step 3: Find the unit digit of \( 3^{57} \)

The unit digits of the powers of 3 follow a pattern:
- \( 3^1 = 3 \) (unit digit 3)
- \( 3^2 = 9 \) (unit digit 9)
- \( 3^3 = 27 \) (unit digit 7)
- \( 3^4 = 81 \) (unit digit 1)
- \( 3^5 = 243 \) (unit digit 3) - and the pattern repeats.

The pattern of unit digits is: 3, 9, 7, 1 (repeats every 4 terms).

To find the unit digit of \( 3^{57} \), we calculate \( 57 \mod 4 \):
\[ 57 \div 4 = 14 \quad \text{remainder} \quad 1 \]
Thus, \( 57 \mod 4 = 1 \). The unit digit corresponds to the first term in the pattern, which is 3.

### Step 4: Combine the unit digits

Now we have:
- Unit digit of \( 7^{65} = 7 \)
- Unit digit of \( 6^{41} = 6 \)
- Unit digit of \( 3^{57} = 3 \)

We multiply these unit digits together:
\[ 7 \times 6 \times 3 \]

Calculating step by step:
1. \( 7 \times 6 = 42 \) (unit digit 2)
2. \( 2 \times 3 = 6 \)

Thus, the unit digit of \( 7^{65} \times 6^{41} \times 3^{57} \) is **6**.

### Final Answer:
The unit digit is **6**.