The remainder on dividing 5 ^ 99 by 11 is _____ .

Now to divide ${5}^{99}$ by $11$ we need to check the cyclicity of power of $5$ when divided by $11$,

So remainder when $\dfrac{{5}^{1}}{11} \to 5\text{as remainder}$

Similarly,

$\dfrac{{5}^{2}}{11} = \dfrac{25}{11} \to 3\text{as remainder}$

$\dfrac{{5}^{3}}{11} = \dfrac{125}{11} \to 4\text{as remainder}$

$\dfrac{{5}^{4}}{11} = \dfrac{625}{11} \to 9\text{as remainder}$

And $\dfrac{{5}^{5}}{11} = \dfrac{3125}{11} \to 1\text{as remainder}$

So, rewriting the expression we get, ${\left({5}^{5}\right)}^{19} \times {5}^{4}$

Now when $\dfrac{{\left({5}^{5}\right)}^{19} \times {5}^{4}}{11} \to 1 \times \dfrac{{5}^{4}}{11}$

Now again simplifying we get, $\dfrac{{5}^{4}}{11} \to 9\text{as remainder}$.