Which among `43^2, 67^2, 52^2, 59^2` would end with digit 1?

Options

A. `43^2`
B. `67^2`
C. `52^2`
D. `59^2`

Answer

To determine which among \(43^2\), \(67^2\), \(52^2\), and \(59^2\) ends with the digit 1, we can analyze the last digit of each number and how it affects the last digit of their squares.

### Step-by-step Solution:

1. **Identify the last digit of each number:**
- \(43\) ends with \(3\)
- \(67\) ends with \(7\)
- \(52\) ends with \(2\)
- \(59\) ends with \(9\)

2. **Determine the last digit of the square for each last digit:**
- **Last digit \(3\)**:
\[ 3^2 = 9 \quad \text{(last digit is 9)} \]
- **Last digit \(7\)**:
\[ 7^2 = 49 \quad \text{(last digit is 9)} \]
- **Last digit \(2\)**:
\[ 2^2 = 4 \quad \text{(last digit is 4)} \]
- **Last digit \(9\)**:
\[ 9^2 = 81 \quad \text{(last digit is 1)} \]

3. **Conclusion:**
- From our analysis, the only square that ends with the digit \(1\) is \(59^2\) because the last digit of \(59\) is \(9\) and \(9^2\) ends with \(1\).

### Final Answer:
The number whose square ends with the digit \(1\) is \(59\).