What could be the possible one's digit of the square root of the number 9801?

Question

Accepted Answer

To determine the possible one's digit of the square root of the number 9801, we need to consider the properties of perfect squares and their square roots.

1. **Explanation for Option 1 (1,9):**
   - The number 9801 ends with the digit 1. When a number ends with 1, its square root can end with either 1 or 9. This is because:
     - $1^2 = 1$ (a number ending in 1 squared ends in 1)
     - $9^2 = 81$ (a number ending in 9 squared ends in 1)
   - Therefore, the possible one's digits for the square root of a number ending in 1 are 1 and 9. This makes Option 1 (1,9) the correct answer.

2. **Explanation for why other options are incorrect:**

- **Option 2 (3,9):**
     - A number ending in 3, when squared, results in a number ending in 9 (since $3^2 = 9$). Similarly, a number ending in 9, when squared, results in a number ending in 1 (since $9^2 = 81$). Therefore, a square root ending in 3 would not result in a number ending in 1, making this option incorrect.

- **Option 3 (3,3):**
     - As explained above, a number ending in 3, when squared, results in a number ending in 9. Therefore, a square root ending in 3 cannot result in a number ending in 1. Additionally, having both digits as 3 is not possible for a number ending in 1, making this option incorrect.

- **Option 4 (0,1):**
     - A number ending in 0, when squared, results in a number ending in 0 (since $0^2 = 0$). Therefore, a square root ending in 0 cannot result in a number ending in 1. While a square root ending in 1 is possible, the pair (0,1) does not cover both possible digits for a number ending in 1, making this option incorrect.

In conclusion, Option 1 (1,9) is the correct answer because it accurately reflects the possible one's digits of the square root of a number ending in 1.