Mathematics - Binomial Theorem Question with Solution | TestHub

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

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

Similarly,

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

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

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

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

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

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

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