1. The problem is to find the remainder when $289^{125}$ is divided by 27, i.e., compute $289^{125} \bmod 27$. 2. We use modular arithmetic properties and Euler's theorem or simplification to solve this efficiently. 3. First, reduce the base modulo 27: $$289 \bmod 27 = 289 - 27 \times 10 = 289 - 270 = 19$$ 4. So the problem reduces to finding: $$19^{125} \bmod 27$$ 5. Since 27 is $3^3$, Euler's totient function $\varphi(27) = 27 \times (1 - \frac{1}{3}) = 18$. 6. By Euler's theorem, if $\gcd(19,27) = 1$, then: $$19^{18} \equiv 1 \pmod{27}$$ 7. Check $\gcd(19,27) = 1$ (since 19 is prime and does not divide 27), so Euler's theorem applies. 8. Express the exponent 125 modulo 18: $$125 \bmod 18 = 125 - 18 \times 6 = 125 - 108 = 17$$ 9. Therefore: $$19^{125} \equiv 19^{17} \pmod{27}$$ 10. Compute $19^{17} \bmod 27$ stepwise: - $19^1 \equiv 19 \pmod{27}$ - $19^2 = 361 \equiv 361 - 27 \times 13 = 361 - 351 = 10 \pmod{27}$ - $19^4 = (19^2)^2 = 10^2 = 100 \equiv 100 - 27 \times 3 = 100 - 81 = 19 \pmod{27}$ - $19^8 = (19^4)^2 = 19^2 = 10 \pmod{27}$ (from above) - $19^{16} = (19^8)^2 = 10^2 = 100 \equiv 19 \pmod{27}$ 11. Now: $$19^{17} = 19^{16} \times 19^1 \equiv 19 \times 19 = 361 \equiv 10 \pmod{27}$$ 12. Final answer: $$\boxed{10}$$ So, $289^{125} \bmod 27 = 10$.