The remainder on dividing $$5^{99}$$ by 11 is ______.

We need to find $$5^{99} \mod 11$$.

By Fermat's Little Theorem, since 11 is prime and $$\gcd(5, 11) = 1$$:

$$ 5^{10} \equiv 1 \pmod{11} $$

Dividing the exponent: $$99 = 10 \times 9 + 9$$

$$ 5^{99} = (5^{10})^9 \times 5^9 \equiv 1^9 \times 5^9 = 5^9 \pmod{11} $$

Computing $$5^9 \mod 11$$ step by step:

$$5^1 \equiv 5$$

$$5^2 \equiv 25 \equiv 3 \pmod{11}$$

$$5^4 \equiv 3^2 = 9 \pmod{11}$$

$$5^8 \equiv 9^2 = 81 \equiv 4 \pmod{11}$$

$$5^9 = 5^8 \times 5 \equiv 4 \times 5 = 20 \equiv 9 \pmod{11}$$

The remainder is 9.