1. We need to find the remainder when $5^{99}$ is divided by $11$.

2. Since $11$ is prime and $5^{10} \equiv 1 \pmod{11}$ by Fermat's little theorem, we reduce the exponent modulo $10$.

3. Compute: $99 \equiv 9 \pmod{10}$ So, $5^{99} \equiv 5^9 \pmod{11}$

4. Now compute powers of $5$ modulo $11$: $5^2 = 25 \equiv 3 \pmod{11}$ $5^4 \equiv 3^2 = 9 \pmod{11}$ $5^8 \equiv 9^2 = 81 \equiv 4 \pmod{11}$ Thus, $5^9 \equiv 5^8 \cdot 5 \equiv 4 \cdot 5 = 20 \equiv 9 \pmod{11}$

5. Therefore, the remainder is: $\boxed{9}$

6. Comparison with stored answer:

• Derived answer: $9$
• Stored correct answer: $9$

They match.