The unit digit of ` 17^(1983) + 11^(1983) ` is

Options

A. 4
B. 2
C. 3
D. 0

Answer

To find the unit digit of \( 17^{1983} + 11^{1983} \), we can follow these steps:

### Step 1: Identify the unit digits of the bases
The unit digit of \( 17 \) is \( 7 \) and the unit digit of \( 11 \) is \( 1 \). Therefore, we can rewrite the expression in terms of their unit digits:
\[ 17^{1983} + 11^{1983} \equiv 7^{1983} + 1^{1983} \mod 10 \]

### Step 2: Calculate the unit digit of \( 1^{1983} \)
The unit digit of \( 1^{1983} \) is straightforward:
\[ 1^{1983} = 1 \]

### Step 3: Determine the unit digit of \( 7^{1983} \)
To find the unit digit of \( 7^{1983} \), we need to observe the pattern in the unit digits of the 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 \( 1 \))

We can see that the unit digits repeat every 4 powers: \( 7, 9, 3, 1 \).

### Step 4: Find the position in the cycle
To find which unit digit corresponds to \( 7^{1983} \), we calculate \( 1983 \mod 4 \):
\[ 1983 \div 4 = 495 \quad \text{remainder } 3 \]
This means \( 1983 \equiv 3 \mod 4 \).

### Step 5: Identify the unit digit from the cycle
From our earlier observation:
- \( 7^1 \) has a unit digit of \( 7 \)
- \( 7^2 \) has a unit digit of \( 9 \)
- \( 7^3 \) has a unit digit of \( 3 \)

Since \( 1983 \equiv 3 \mod 4 \), the unit digit of \( 7^{1983} \) is \( 3 \).

### Step 6: Combine the results
Now, we can combine the unit digits we found:
\[ \text{Unit digit of } 7^{1983} + 1^{1983} = 3 + 1 = 4 \]

### Final Answer
Thus, the unit digit of \( 17^{1983} + 11^{1983} \) is \( \boxed{4} \).

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