$(2^{51} + 2^{52} + 2^{53} + 2^{54} + 2^{55})$ is divisible by

23

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58

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124

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127

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Solution

The correct option is A $124$
$2^{51} + 2^{52} + 2^{53} + 2^{54} + 2^{55}$

$= 2^{51} [1 + 2 + 2^{2} + 2^{3} + 2^{4}] = 2^{51} [1 + 2 + 4 + 8 + 16]$

$= 2^{51} \times 31 = 2^{49} \times 124$

Hence, the expression is divisible by $124$ .

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