1. **Problem statement:** We are given a number $n$ such that when its square root is divided by 11, the remainder is 6, and $6 < \sqrt{n} < 28$. We need to find $n$ and then compute $n^{125} \bmod 27$ using modular exponentiation. 2. **Understanding the problem:** If $\sqrt{n}$ divided by 11 leaves a remainder 6, then $\sqrt{n} = 11k + 6$ for some integer $k$. 3. **Constraints:** Since $6 < \sqrt{n} < 28$, we find possible values of $k$: $$6 < 11k + 6 < 28$$ Subtract 6 from all parts: $$0 < 11k < 22$$ Divide by 11: $$0 < k < 2$$ So $k$ can be 1 (since $k$ is an integer). 4. **Calculate $\sqrt{n}$:** $$\sqrt{n} = 11 \times 1 + 6 = 17$$ 5. **Calculate $n$:** $$n = (\sqrt{n})^2 = 17^2 = 289$$ 6. **Calculate $n^{125} \bmod 27$ using modular exponentiation:** We want to find: $$289^{125} \bmod 27$$ First, reduce the base modulo 27: $$289 \bmod 27 = 289 - 27 \times 10 = 289 - 270 = 19$$ So: $$19^{125} \bmod 27$$ 7. **Use Euler's theorem:** Since 27 = $3^3$, Euler's totient $\phi(27) = 27 \times (1 - \frac{1}{3}) = 18$. Because $\gcd(19,27) = 1$, Euler's theorem applies: $$19^{18} \equiv 1 \bmod 27$$ 8. **Reduce the exponent modulo 18:** $$125 \bmod 18 = 125 - 18 \times 6 = 125 - 108 = 17$$ So: $$19^{125} \equiv 19^{17} \bmod 27$$ 9. **Calculate $19^{17} \bmod 27$ by repeated squaring:** - $19^1 \equiv 19 \bmod 27$ - $19^2 = 361 \equiv 361 - 27 \times 13 = 361 - 351 = 10 \bmod 27$ - $19^4 = (19^2)^2 = 10^2 = 100 \equiv 100 - 27 \times 3 = 100 - 81 = 19 \bmod 27$ - $19^8 = (19^4)^2 = 19^2 = 10 \bmod 27$ - $19^{16} = (19^8)^2 = 10^2 = 100 \equiv 19 \bmod 27$ 10. **Combine powers:** $$19^{17} = 19^{16} \times 19^1 \equiv 19 \times 19 = 361 \equiv 10 \bmod 27$$ **Final answer:** $$n = 289$$ $$n^{125} \bmod 27 = 10$$