A) \[({{2}^{16}}+1)\]
B) \[({{2}^{16}}-1)\]
C) \[7\times {{2}^{33}}\]
D) \[{{2}^{96}}+1\]
Correct Answer: D
Solution :
[d] Let \[{{2}^{32}}=x\] and let \[({{2}^{32}}+1)=(x+1)\] be divisible by a number N. Then, \[({{2}^{96}}+1)={{({{2}^{32}})}^{3}}+1=({{x}^{3}}+1)\] \[=(x+1)\,({{x}^{2}}-x+1),\] which is clearly divisible by N, since \[(x+1)\] is divisible by N. |
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