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$9^{7} + 7^{9}$ is divisible by

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answer is C.

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Detailed Solution

9^{7} + 7^{9}

$= {(8 + 1)}^{7} + {(8 - 1)}^{9}$

$= (^{7} C_{0} 8^{7} +^{7} C_{1} 8^{6} +^{7} C_{2} 8^{5} + \dots +^{7} C_{7} 1) + (^{9} C_{0} 8^{9} +^{9} C_{1} 8^{8} +^{9} C_{2} 8^{7} + \dots +^{9} C_{9} 1)$

$= (8^{7} + {7.8}^{6} +^{7} C_{2} 8^{5} + \dots + 7.8 + 1) + (8^{9} + {9.8}^{8} +^{9} C_{2} 8^{7} + \dots + 9.8 - 1)$

$= 8^{2} (8^{5} + {7.8}^{4} +^{7} C_{2} {.8}^{3} + \dots) + 56 + 8^{2} (8^{7} - {9.8}^{6} + \dots) + 72$

$= 8^{2} (8^{5} + {7.8}^{4} +^{7} C_{2} {.8}^{3} + 1 + 8^{7} - {9.8}^{6})$

Divisible by 64.

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