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$$\,{2^{16}} - 1$$ is divisible by -
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& \Rightarrow {2^{16}} - 1\,\, \cr
& = \left( {{2^8} - {1^8}\,\,} \right)\left( {{2^8} + {1^8}\,} \right) \cr
& = \left( {{2^4} - 1\,\,} \right)\left( {{2^4} + 1\,\,} \right)\left( {{2^8} + 1\,\,} \right) \cr
& = \left( {16 - 1} \right)\left( {16 + 1} \right)\left( {{2^8} + 1\,\,} \right) \cr
& = 15 \times 17\left( {{2^8} + 1\,\,} \right) \cr
& \therefore {2^{16}} - 1\,\,{\text{is divisible by 17}} \cr} $$
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