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Divisibility Rule for Composite Numbers

If m and n ar...

Question

If m and n are whole numbers such that n>1 and m $^{n}$ = 121, then the value of (m - 1) $^{n + 1}$ is :

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121

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1000

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14400

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Solution

The correct option is C
1000

We know that 11 $^{2}$ = 121. Putting m = 11 and n = 2, we get ${(m - 1)}^{n + 1} = {(11 - 1)}^{(2 + 1)} = 10^{3} = 1000.$

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