If m and n are whole numbers such that \(m^{n}\) = 121, the value of \((m-n)^{n+1}\) is:
Answer: D
We know that \(11^{2}\) =121.
Putting m = 11 and n = 2, we get:
\((m-1)^{n+1} = (11-1)^{(2+1)} = 10^{3} = 1000\)