Let $m = 2^{80} - 1$ b and $n = 2^{120} - 1$ then H. C. F. (m, n)is:

2^{10} - 1

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2^{20} - 1

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2^{40} - 1

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2^{60} - 1

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Solution

The correct option is C $2^{40} - 1$
$m = 2^{80} - 1 = (2^{40} + 1) (2^{40} - 1)$
$n = 2^{120} - 1 = (2^{40})^{3} - 1$
$n = (2^{40} - 1) [(2^{40})^{2} + 2^{40} + 1]$
$n = (2^{40} - 1) (2^{80} + 2^{40} + 1)$
So, the H.C.F. of (m, n)
$= (2^{40} - 1)$

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