A certain number 'n' can exactly divide `(3^24-1)`, then this number can also divide the number:

Options

A. `(3^16+1)`
B. `(3^8-1)`
C. `(3^70-1)`
D. `(3^96-1)`

Answer

To solve the problem, we need to determine what number \( n \) can divide \( 3^{24} - 1 \) and what other numbers it can also divide.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( 3^{24} - 1 \). We can factor this expression using the difference of squares:
\[ 3^{24} - 1 = (3^{12} - 1)(3^{12} + 1) \]

2. **Further Factoring**:
We can further factor \( 3^{12} - 1 \) again using the difference of squares:
\[ 3^{12} - 1 = (3^6 - 1)(3^6 + 1) \]

3. **Continuing the Factorization**:
We can continue factoring \( 3^6 - 1 \):
\[ 3^6 - 1 = (3^3 - 1)(3^3 + 1) \]
And we know:
\[ 3^3 - 1 = 26 \quad \text{and} \quad 3^3 + 1 = 28 \]

4. **Writing the Complete Factorization**:
Now we have:
\[ 3^{24} - 1 = (3^{12} - 1)(3^{12} + 1) = (3^6 - 1)(3^6 + 1)(3^{12} + 1) \]
Continuing this process, we can express \( 3^{12} + 1 \) and \( 3^6 + 1 \) in terms of their factors.

5. **Identifying Divisibility**:
Since \( n \) divides \( 3^{24} - 1 \), it must also divide any linear combination or factor of \( 3^{24} - 1 \). Thus, \( n \) can also divide:
\[ 3^{12} + 1 \]
and any other factors derived from the complete factorization.

6. **Conclusion**:
Therefore, if \( n \) divides \( 3^{24} - 1 \), it can also divide \( 3^{12} + 1 \), \( 3^6 - 1 \), \( 3^6 + 1 \), and so on.

### Final Answer:
Thus, the number \( n \) can also divide \( 3^{12} + 1 \).