If $9^{7} + 7^{9}$ is divisible by $2^{n}$ then the greatest value of $n,$ where $n \in N$

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Solution

The correct option is A 6
$7^{9} + 9^{7} = - (1 - 8)^{9} + (1 + 8)^{7} = - [1 -^{9} C_{1} 8 +^{9} C_{2} 8^{2} - . . . . . . . . -^{9} C_{9} 8^{9}] + [1 +^{7} C_{1} 8^{1} +^{7} C_{2} 8^{2} + . . . . . . . . +^{7} C_{7} 8^{7}] = 16 \times 8 + 64 [- (^{9} C_{2} - . . . . . . . . . -^{9} C_{9} 8^{7}) + (^{7} C_{2} + . . . . . . . . +^{7} C_{7} 8^{9})] = 64 (s o m e i n t e g e r) h e n c e i t i s d i v i s i b l e b y 64 ∴ 2^{n} = 64 o r n = 6$

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