What is the remainder when 5^ 99 is divided by 13 ?

To find the remainder when $5^{99}$ is divided by $13$, we evaluate the powers of $5$ modulo $13$.

$5^1 \equiv 5 \pmod{13}$

$5^2 = 25 \equiv -1 \pmod{13}$

Squaring both sides gives:

$5^4 \equiv (-1)^2 \equiv 1 \pmod{13}$

We can express $5^{99}$ in terms of $5^4$ as follows:

$5^{99} = 5^{96} \times 5^3 = (5^4)^{24} \times 125$

Substituting $5^4 \equiv 1 \pmod{13}$:

$5^{99} \equiv (1)^{24} \times 125 \pmod{13}$

$5^{99} \equiv 125 \pmod{13}$

Dividing $125$ by $13$ yields $125 = 13 \times 9 + 8$.

Thus, $125 \equiv 8 \pmod{13}$.

The remainder is $8$.

Answer: $8$