A number `1 lt N lt 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, we need to find a number \( N \) such that \( 1 < N < 100 \) and \( N \) is both a perfect square and a perfect cube. This means \( N \) must be a perfect sixth power (since \( \text{lcm}(2, 3) = 6 \)).

### Step-by-step Solution:

1. **Identify Perfect Sixth Powers**:
We need to find numbers of the form \( n^6 \) where \( n \) is a positive integer. We will calculate \( n^6 \) for \( n = 1, 2, 3, ... \) until \( n^6 \) exceeds 100.

- For \( n = 1 \):
\[ 1^6 = 1 \]
- For \( n = 2 \):
\[ 2^6 = 64 \]
- For \( n = 3 \):
\[ 3^6 = 729 \quad (\text{exceeds 100}) \]

Thus, the only perfect sixth powers between 1 and 100 are \( 1 \) and \( 64 \).

2. **Filter the Valid Range**:
Since we need \( 1 < N < 100 \), we discard \( 1 \) and keep \( 64 \).

3. **Calculate the Sum of Digits of \( N \)**:
Now, we find the sum of the digits of \( N = 64 \).
- The digits of \( 64 \) are \( 6 \) and \( 4 \).
- Therefore, the sum of the digits is:
\[ 6 + 4 = 10 \]

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