$$\,{2^{16}} - 1$$   is divisible by -

Solution (By Examveda Team)

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