The last digit of the expression
`4 xx 9^2 xx 4^3 xx 9^4 xx 4^5 xx 9^6 xx …..xx 4^(99) xx 9^(100)` is:

Options

A. 4
B. 6
C. 9
D. 1

Answer

To find the last digit of the expression \( 4 \times 9^2 \times 4^3 \times 9^4 \times 4^5 \times 9^6 \times \ldots \times 4^{99} \times 9^{100} \), we can break it down into steps focusing on the last digits of the powers of 4 and 9.

### Step-by-Step Solution:

1. **Identify the Pattern of Powers**:
The expression alternates between powers of 4 and powers of 9. The powers of 4 are odd (1, 3, 5, ..., 99) and the powers of 9 are even (2, 4, 6, ..., 100).

2. **Last Digit of Powers of 9**:
- The last digit of \( 9^1 \) is 9.
- The last digit of \( 9^2 \) is 1.
- The last digit of \( 9^3 \) is 9.
- The last digit of \( 9^4 \) is 1.
- Thus, we see that:
- If the power of 9 is odd, the last digit is 9.
- If the power of 9 is even, the last digit is 1.

Since all powers of 9 in our expression are even, the last digit for each even power of 9 will be 1.

3. **Last Digit of Powers of 4**:
- The last digit of \( 4^1 \) is 4.
- The last digit of \( 4^2 \) is 6.
- The last digit of \( 4^3 \) is 4.
- The last digit of \( 4^4 \) is 6.
- Thus, we see that:
- If the power of 4 is odd, the last digit is 4.
- If the power of 4 is even, the last digit is 6.

In our expression, the powers of 4 are 1, 3, 5, ..., 99, which are all odd. Therefore, the last digit for each odd power of 4 will be 4.

4. **Count the Number of Terms**:
- The powers of 4 are: 1, 3, 5, ..., 99. This is an arithmetic series where:
- First term \( a = 1 \)
- Last term \( l = 99 \)
- Common difference \( d = 2 \)

The number of terms \( n \) can be calculated using the formula for the \( n \)-th term of an arithmetic series:
\[ l = a + (n-1)d \]
Plugging in the values:
\[ 99 = 1 + (n-1) \cdot 2 \]
Solving for \( n \):
\[ 99 - 1 = (n-1) \cdot 2 \implies 98 = (n-1) \cdot 2 \implies n - 1 = 49 \implies n = 50 \]

5. **Combine the Last Digits**:
- We have 50 terms of 4 (from the powers of 4) and each contributes a last digit of 4.
- The last digit of the product of 50 fours is \( 4^{50} \).

6. **Determine the Last Digit of \( 4^{50} \)**:
Since 50 is even, the last digit of \( 4^{50} \) is 6.

7. **Final Calculation**:
The last digit of the entire expression is therefore \( 6 \).

### Conclusion:
The last digit of the expression \( 4 \times 9^2 \times 4^3 \times 9^4 \times 4^5 \times 9^6 \times \ldots \times 4^{99} \times 9^{100} \) is **6**.