A number 1 < N < 100 is such that it is a perfect square and perfect cube both, then the sum of digits of N is :

Options

A. 1
B. 10
C. 20
D. Can't be determined

Answer

To solve the problem step by step, we need to find a number \( N \) such that \( 1 < N < 100 \) and \( N \) is both a perfect square and a perfect cube.

### Step-by-Step Solution:

1. **Understanding Perfect Squares and Perfect Cubes**:
- A perfect square is a number that can be expressed as the square of an integer (e.g., \( 1^2, 2^2, 3^2, \ldots \)).
- A perfect cube is a number that can be expressed as the cube of an integer (e.g., \( 1^3, 2^3, 3^3, \ldots \)).
- We need to find numbers that are both perfect squares and perfect cubes.

2. **Finding Perfect Squares Between 1 and 100**:
- The perfect squares between 1 and 100 are:
- \( 2^2 = 4 \)
- \( 3^2 = 9 \)
- \( 4^2 = 16 \)
- \( 5^2 = 25 \)
- \( 6^2 = 36 \)
- \( 7^2 = 49 \)
- \( 8^2 = 64 \)
- \( 9^2 = 81 \)
- \( 10^2 = 100 \) (not included as it is not less than 100)

3. **Finding Perfect Cubes Between 1 and 100**:
- The perfect cubes between 1 and 100 are:
- \( 2^3 = 8 \)
- \( 3^3 = 27 \)
- \( 4^3 = 64 \)
- \( 5^3 = 125 \) (not included as it is greater than 100)

4. **Finding Common Numbers**:
- Now, we look for numbers that are both perfect squares and perfect cubes from the lists above.
- The only number that appears in both lists is \( 64 \).

5. **Conclusion**:
- Therefore, \( N = 64 \).

6. **Calculating the Sum of Digits**:
- The digits of \( 64 \) are \( 6 \) and \( 4 \).
- The sum of the digits is \( 6 + 4 = 10 \).

### Final Answer:
The sum of the digits of \( N \) is **10**.

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