1. The problem is to find the remainder when $289^{125}$ is divided by 19, i.e., compute $289^{125} \bmod 19$. 2. We use modular arithmetic properties and Fermat's Little Theorem which states that for a prime $p$ and integer $a$ not divisible by $p$, $a^{p-1} \equiv 1 \pmod{p}$. 3. First, reduce the base modulo 19: $$289 \bmod 19 = 289 - 19 \times 15 = 289 - 285 = 4$$ 4. So the problem reduces to finding: $$4^{125} \bmod 19$$ 5. Using Fermat's Little Theorem with $p=19$, we have: $$4^{18} \equiv 1 \pmod{19}$$ 6. Express the exponent 125 in terms of 18: $$125 = 18 \times 6 + 17$$ 7. Therefore: $$4^{125} = 4^{18 \times 6 + 17} = (4^{18})^6 \times 4^{17} \equiv 1^6 \times 4^{17} = 4^{17} \pmod{19}$$ 8. Now compute $4^{17} \bmod 19$ step by step: - $4^1 = 4$ - $4^2 = 16$ - $4^3 = 4^2 \times 4 = 16 \times 4 = 64 \equiv 64 - 57 = 7 \pmod{19}$ - $4^4 = 7 \times 4 = 28 \equiv 9 \pmod{19}$ - $4^5 = 9 \times 4 = 36 \equiv 17 \pmod{19}$ - $4^6 = 17 \times 4 = 68 \equiv 11 \pmod{19}$ - $4^7 = 11 \times 4 = 44 \equiv 6 \pmod{19}$ - $4^8 = 6 \times 4 = 24 \equiv 5 \pmod{19}$ - $4^9 = 5 \times 4 = 20 \equiv 1 \pmod{19}$ 9. Since $4^9 \equiv 1 \pmod{19}$, then: $$4^{17} = 4^{9} \times 4^{8} \equiv 1 \times 5 = 5 \pmod{19}$$ 10. Therefore, the remainder is: $$\boxed{5}$$