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

128

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24

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64

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72

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Solution

The correct option is C $64$
$7^{9} + 9^{7} = {(8 - 1)}^{9} + {(8 + 1)}^{7}$

$= [{^{9} C}_{0} 8^{9} - {^{9} C}_{1} 8^{8} + {^{9} C}_{2} 8^{7} - . . . . . + {^{9} C}_{8} 8^{1} - {^{9} C}_{9} (1)] + [{^{7} C}_{0} 8^{7} + {^{7} C}_{1} 8^{6} + {^{7} C}_{2} 8^{5} + . . . . . + {^{7} C}_{6} 8^{1} + {^{7} C}_{7} (1)]$

$= 8^{2} k + 72 - 1 + 56 + 1 = 8^{2} k + 128 = 8^{2} (k + 2)$

which is divisible by $64$

Suggest Corrections

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