RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

(c) ${\left(17\right)}^{200}={\left(18\mathit{–}1\right)}^{200}$
We know that
${\left(\mathit{x}+\mathit{a}\right)}^{\mathit{n}}$
$={\mathit{x}}^{\mathit{n}}+\mathit{n}{\mathit{x}}^{\mathit{n}}\mathit{–}1.\mathit{a}$
$+\frac{\mathit{n}\left(\mathit{n}\mathit{–}1\right)}{1\times 2}{\mathit{x}}^{\mathit{n}\mathit{–}2}{\mathit{a}}^2+\frac{\mathit{n}\left(\mathit{n}\mathit{–}1\right)\left(\mathit{n}\mathit{–}2\right)}{1\times 2\times 3}{\mathit{x}}^{\mathit{n}}\mathit{–}3{\mathit{a}}^3+.....+{\mathit{a}}^{\mathit{n}}$
We see that all the terms on the R.H.S. except an has x as one of its factor and hence are divisible by x. So, (x + a)n is divisible by x or not will be decided by an.
Let x = 18, a = –1 and n = 200∴${\left(18\mathit{–}1\right)}^{200}$is divisible by 18 or not will depend${\left(\mathit{–}1\right)}^{200}$ as all other terms in its expansion will be divisible by 18 because each of them will have 18 as one of their factors.
${\left(\mathit{–}1\right)}^{200}=1\left(\because 200\mathit{i}\mathit{s}\mathit{e}\mathit{v}\mathit{e}\mathit{n}\right)$ 1 is not divisible by 18 and is also less than 18.
∴1 is the remainder.