Question:

Step 1: Understand the problem and goal. We need to find the remainder when $5^{99}$ is divided by 11. This can be expressed using modular arithmetic as finding $5^{99} \pmod{11}$. Step 2: Apply Fermat's Little Theorem. Fermat's Little Theorem states that if $p$ is a prime number, then for any integer $a$ not divisible by $p$, we have $a^{p-1} \equiv 1 \pmod{p}$. In this problem, $p=11$ (which is a prime number) and $a=5$ (which is not divisible by 11). Therefore, we can write: $$5^{11-1} \equiv 5^{10} \equiv 1 \pmod{11}$$ Step 3: Simplify the exponent of the given expression. We have $5^{99}$. We can rewrite the exponent 99 in terms of 10 using division: $$99 = 10 \times 9 + 9$$ Step 4: Substitute and calculate the remainder. Now, substitute this back into the original expression: $$5^{99} = 5^{10 \times 9 + 9} = (5^{10})^9 \times 5^9$$ Taking this expression modulo 11: $$(5^{10})^9 \times 5^9 \pmod{11}$$ From Step 2, we know $5^{10} \equiv 1 \pmod{11}$. Substitute this into the expression: $$(1)^9 \times 5^9 \pmod{11}$$ $$1 \times 5^9 \pmod{11}$$ $$5^9 \pmod{11}$$ Now, we need to calculate $5^9 \pmod{11}$. We can do this by evaluating powers of 5 modulo 11: $$5^1 \equiv 5 \pmod{11}$$ $$5^2 \equiv 25 \equiv 3 \pmod{11}$$ $$5^3 \equiv 5^2 \times 5^1 \equiv 3 \times 5 \equiv 15 \equiv 4 \pmod{11}$$ $$5^4 \equiv 5^2 \times 5^2 \equiv 3 \times 3 \equiv 9 \pmod{11}$$ $$5^5 \equiv 5^4 \times 5^1 \equiv 9 \times 5 \equiv 45 \equiv 1 \pmod{11}$$ Alternatively, we could also compute $5^9 \pmod{11}$ as: $$5^9 = 5^{10} \times 5^{-1} \pmod{11}$$ Since $5^{10} \equiv 1 \pmod{11}$, we have: $$5^9 \equiv 1 \times 5^{-1} \pmod{11}$$ We need to find the multiplicative inverse of 5 modulo 11. Let $x = 5^{-1}$. $$5x \equiv 1 \pmod{11}$$ By inspection, $5 \times 9 = 45 \equiv 1 \pmod{11}$. So, $5^{-1} \equiv 9 \pmod{11}$. Therefore, $5^9 \equiv 9 \pmod{11}$. Step 5: Conclude the final remainder. The remainder on dividing $5^{99}$ by 11 is 9. The final answer is $\boxed{9}$.