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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.\(2^{96} + 1\)

Answer: D

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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