It is given that `(2^32 + 1)` is completely divisible by a number. That number will certainly divide which of the following numbers?

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 divisible by the same number \( n \) that divides \( 2^{32} + 1 \).

### Step-by-Step Solution:

1. **Understand the Given Expression**:
We start with the expression \( 2^{32} + 1 \). We denote \( x = 2^{32} \), so we can rewrite the expression as \( x + 1 \).

2. **Identify the Options**:
We need to analyze the options provided to see which one can be expressed in a form that includes \( x + 1 \).

3. **Examine Option A**:
The first option is \( 2^{96} + 1 \). We can express \( 2^{96} \) as \( (2^{32})^3 = x^3 \). Thus, we rewrite the option as:
\[ 2^{96} + 1 = x^3 + 1 \]

4. **Factor the Expression**:
We can use the sum of cubes factorization:
\[ x^3 + 1 = (x + 1)(x^2 - x + 1) \]
Here, \( x + 1 \) is a factor.

5. **Divisibility**:
Since \( n \) divides \( x + 1 \) (because \( n \) divides \( 2^{32} + 1 \)), it follows that \( n \) also divides \( (x + 1)(x^2 - x + 1) \). Therefore, \( n \) divides \( 2^{96} + 1 \).

6. **Conclusion**:
Thus, we conclude that \( 2^{96} + 1 \) is divisible by \( n \).

### Final Answer:
The number that \( n \) will certainly divide is \( 2^{96} + 1 \).

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