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# Discussion

It is being given that $$(2^{32} + 1)$$ is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?

• A.$$2^{16} + 1$$
• B.$$2^{16} - 1$$
• C.$$7 \times 2^{23}$$
• D.$$2^{96} + 1$$

Let $$2^{32} = x$$.
Then, $$(2^{32} + 1) = (x + 1)$$.
Let $$(x + 1)$$ be completely divisible by the natural number N. Then,
$$(2^{96} + 1) \\= [(2^{32})^{3} + 1]\\= (x^{3} + 1)\\= (x + 1)(x^{2} - x + 1)$$
which is completely divisible by N, since $$(x + 1)$$ is divisible by N.

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