What's the remainder?

Method 1: By Fermat's Little Theorem $5^{12} \equiv 1 \pmod{13}$ , so

$5^{99} = 5^{12*8 + 3} \equiv 5^{3} \pmod{13} \equiv (130 - 5) \pmod{13} \equiv \boxed{8} \pmod{13}$ .

Method 2: Noting that $5^{2} \equiv -1 \pmod{13}$ , we then have that

$5^{98} \equiv (-1)^{49} \pmod{13} \equiv -1 \pmod{13} \Longrightarrow 5^{99} \equiv -5 \pmod{13} \equiv \boxed{8} \pmod{13}$ .

Observing pattern that 5^1 / 13 has remainder 5

5^2 / 13 has remainder 12

5^3 / 13 has remainder 8

5/4 / 13 has remainder 1 and the pattern repeats

Leading to the conclusion that 5^100 / 13 will have remainder 1 so 5^99 / 13 will have remainder 8