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) \[7\times {{2}^{33}}\]
D) \[({{2}^{96}}+1)\]
Correct Answer: D
Solution :
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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