The last digit of the positive even number n equals the digit of `n^2`. Whichf of the following could be n?

Options

A. 10
B. 14
C. 15
D. 16

Answer

To solve the problem, we need to find the positive even number \( n \) such that the last digit of \( n \) is equal to the last digit of \( n^2 \).

### Step-by-Step Solution:

1. **Identify the last digits of even numbers**:
The last digit of an even number can be one of the following: 0, 2, 4, 6, or 8.

2. **Calculate the last digit of \( n^2 \) for each possible last digit of \( n \)**:
- If \( n \) ends with **0**:
\[ n^2 = 0^2 = 0 \quad \text{(last digit is 0)} \]
- If \( n \) ends with **2**:
\[ n^2 = 2^2 = 4 \quad \text{(last digit is 4)} \]
- If \( n \) ends with **4**:
\[ n^2 = 4^2 = 16 \quad \text{(last digit is 6)} \]
- If \( n \) ends with **6**:
\[ n^2 = 6^2 = 36 \quad \text{(last digit is 6)} \]
- If \( n \) ends with **8**:
\[ n^2 = 8^2 = 64 \quad \text{(last digit is 4)} \]

3. **Compare the last digits**:
- For \( n \) ending with **0**: last digit of \( n \) (0) equals last digit of \( n^2 \) (0).
- For \( n \) ending with **2**: last digit of \( n \) (2) does not equal last digit of \( n^2 \) (4).
- For \( n \) ending with **4**: last digit of \( n \) (4) does not equal last digit of \( n^2 \) (6).
- For \( n \) ending with **6**: last digit of \( n \) (6) equals last digit of \( n^2 \) (6).
- For \( n \) ending with **8**: last digit of \( n \) (8) does not equal last digit of \( n^2 \) (4).

4. **Conclusion**:
The only last digits of \( n \) that satisfy the condition \( \text{last digit of } n = \text{last digit of } n^2 \) are **0** and **6**. Therefore, any even number \( n \) that ends with either 0 or 6 will satisfy the condition.

### Final Answer:
Thus, the possible values for \( n \) are those that end with **0** or **6**.