If x is a natural numbeer, which is a perfect square, then the number `x + sqrtx` must end in :

Options

A. A)0 or 5
B. B)0 or 1 or 9
C. C)0 or 2 or 6
D. D)0 or 4 or 8

Answer

To solve the problem, we need to determine the last digit (unit digit) of the expression \( x + \sqrt{x} \) where \( x \) is a natural number that is a perfect square.

### Step-by-step Solution:

1. **Understanding Perfect Squares**:
A perfect square is a number that can be expressed as the square of an integer. The perfect squares of natural numbers starting from 1 are:
- \( 1^2 = 1 \)
- \( 2^2 = 4 \)
- \( 3^2 = 9 \)
- \( 4^2 = 16 \)
- \( 5^2 = 25 \)
- \( 6^2 = 36 \)
- \( 7^2 = 49 \)
- \( 8^2 = 64 \)
- \( 9^2 = 81 \)
- \( 10^2 = 100 \)
- and so on.

2. **Calculating \( x + \sqrt{x} \)**:
We will calculate \( x + \sqrt{x} \) for the first few perfect squares and observe the unit digits.

- For \( x = 1 \):
\[ x + \sqrt{x} = 1 + \sqrt{1} = 1 + 1 = 2 \quad \text{(unit digit is 2)} \]

- For \( x = 4 \):
\[ x + \sqrt{x} = 4 + \sqrt{4} = 4 + 2 = 6 \quad \text{(unit digit is 6)} \]

- For \( x = 9 \):
\[ x + \sqrt{x} = 9 + \sqrt{9} = 9 + 3 = 12 \quad \text{(unit digit is 2)} \]

- For \( x = 16 \):
\[ x + \sqrt{x} = 16 + \sqrt{16} = 16 + 4 = 20 \quad \text{(unit digit is 0)} \]

- For \( x = 25 \):
\[ x + \sqrt{x} = 25 + \sqrt{25} = 25 + 5 = 30 \quad \text{(unit digit is 0)} \]

- For \( x = 36 \):
\[ x + \sqrt{x} = 36 + \sqrt{36} = 36 + 6 = 42 \quad \text{(unit digit is 2)} \]

- For \( x = 49 \):
\[ x + \sqrt{x} = 49 + \sqrt{49} = 49 + 7 = 56 \quad \text{(unit digit is 6)} \]

- For \( x = 64 \):
\[ x + \sqrt{x} = 64 + \sqrt{64} = 64 + 8 = 72 \quad \text{(unit digit is 2)} \]

- For \( x = 81 \):
\[ x + \sqrt{x} = 81 + \sqrt{81} = 81 + 9 = 90 \quad \text{(unit digit is 0)} \]

- For \( x = 100 \):
\[ x + \sqrt{x} = 100 + \sqrt{100} = 100 + 10 = 110 \quad \text{(unit digit is 0)} \]

3. **Collecting Unit Digits**:
From the calculations, we observe the unit digits obtained:
- For \( x = 1 \): 2
- For \( x = 4 \): 6
- For \( x = 9 \): 2
- For \( x = 16 \): 0
- For \( x = 25 \): 0
- For \( x = 36 \): 2
- For \( x = 49 \): 6
- For \( x = 64 \): 2
- For \( x = 81 \): 0
- For \( x = 100 \): 0

The possible unit digits are 0, 2, and 6.

4. **Final Conclusion**:
Therefore, the number \( x + \sqrt{x} \) must end in either 0, 2, or 6.