RRB Junior Engineer Stage 1 Model Papers 6

Remainder when ${\left(\mathit{x}-1\right)}^{\mathit{n}}$ is divided by x is ${\left(-1\right)}^{\mathit{n}}$
$\therefore {\left(17\right)}^{200}={\left(18-1\right)}^{200}$
∴ Remainder $={\left(-1\right)}^{200}=1$
Alternate Method:
${\mathit{a}}^{\mathit{n}}-{\mathit{b}}^{\mathit{n}}$ is completely divisible by a + b. If n is an even number in the case of ${17}^{200}÷18$
${17}^{200}-1^{200}$ is completely divisible by $17+1=18$
Here, 1 is remainder.
Or in other words if ${\mathit{a}}^{\mathit{n}}$ is divided by $\mathit{a}+1$ and n is even number then it always left 1 as remainder.