The digit at the unit place in the number `19^(2005)+11^(2005)-9^(2005)` is :

Options

A. 2
B. 1
C. 0
D. 8

Answer

To find the digit at the unit place in the expression \( 19^{2005} + 11^{2005} - 9^{2005} \), we can follow these steps:

### Step 1: Find the unit digit of \( 19^{2005} \)
The unit digit of \( 19^{2005} \) is the same as the unit digit of \( 9^{2005} \) because the unit digit of a number only depends on the unit digit of the base.

The unit digits of powers of 9 follow a pattern:
- \( 9^1 = 9 \) (unit digit is 9)
- \( 9^2 = 81 \) (unit digit is 1)
- \( 9^3 = 729 \) (unit digit is 9)
- \( 9^4 = 6561 \) (unit digit is 1)

The pattern repeats every 2 terms: \( 9, 1, 9, 1, \ldots \)

Since \( 2005 \) is odd, the unit digit of \( 9^{2005} \) is \( 9 \).

### Step 2: Find the unit digit of \( 11^{2005} \)
The unit digit of \( 11^{2005} \) is the same as the unit digit of \( 1^{2005} \) because the unit digit of a number only depends on the unit digit of the base.

The unit digit of any power of 1 is always \( 1 \).

### Step 3: Combine the results
Now we can substitute back into our original expression:
- The unit digit of \( 19^{2005} \) is \( 9 \).
- The unit digit of \( 11^{2005} \) is \( 1 \).
- The unit digit of \( 9^{2005} \) is \( 9 \).

Now we compute:
\[ \text{Unit digit of } (19^{2005} + 11^{2005} - 9^{2005}) = (9 + 1 - 9) \]

### Step 4: Calculate the final unit digit
Calculating the above expression:
\[ 9 + 1 - 9 = 1 \]

### Conclusion
The digit at the unit place in the number \( 19^{2005} + 11^{2005} - 9^{2005} \) is \( 1 \).

### Final Answer
The answer is \( 1 \).

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