Question

Is ${10}^{1000}-1$ divisible by both $9$ and $11$?

Answer 1

${10}^{1000}-1$

${{10}^{10}}^3-1$

$\because$ $99$ is divisible by both $9$ and $11.$

$999$ is divisible by $9$ but not $11.$

$9999$ is divisible by $9$ and $11.$

Hence $\left(\mathrm{999...9}\right)$ is divisible by both $,$ when $n$ is even integer$.$

And $\left({10}^{1000}\right)$ is also the form of $\left(\mathrm{99...9}\right)$

Hence it will be divisible by $9$ as well as $11.$