It is given that `(2^(32) + 1)` is exactly divisible by a certain number. which one of the following is also definitely divisible by the same number?

Options

A. a) `2^(96)+1`
B. b) `7xx2^(33)`
C. c) `2^(16)-1`
D. d) `2^(16)+1`

Answer

To solve the problem, we need to determine which of the given options is definitely divisible by the same number that divides \(2^{32} + 1\).

### Step-by-Step Solution:

1. **Understanding the Given Expression**:
We start with the expression \(2^{32} + 1\). We know that this expression is divisible by some natural number \(n\).

2. **Rewriting the Expression**:
We can express \(2^{32} + 1\) in a different form. Notice that \(2^{96} + 1\) can be rewritten as:
\[ 2^{96} + 1 = (2^{32})^3 + 1^3 \]
This is a sum of cubes, which can be factored using the identity \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). Here, \(a = 2^{32}\) and \(b = 1\).

3. **Applying the Sum of Cubes Factorization**:
Using the sum of cubes factorization, we have:
\[ 2^{96} + 1 = (2^{32} + 1)((2^{32})^2 - 2^{32} \cdot 1 + 1^2) \]
This simplifies to:
\[ 2^{96} + 1 = (2^{32} + 1)(2^{64} - 2^{32} + 1) \]

4. **Divisibility Argument**:
Since \(2^{32} + 1\) is divisible by \(n\), and we have factored \(2^{96} + 1\) to include \(2^{32} + 1\) as one of its factors, it follows that \(2^{96} + 1\) must also be divisible by \(n\).

5. **Conclusion**:
Therefore, we conclude that \(2^{96} + 1\) is definitely divisible by the same number \(n\) that divides \(2^{32} + 1\).

### Final Answer:
The number that is also definitely divisible by the same number \(n\) is \(2^{96} + 1\).