1. **State the problem:** Solve the system of congruences using the Chinese Remainder Theorem (CRT): $$x \equiv 11 \pmod{11}$$ $$x \equiv 2 \pmod{12}$$ $$x \equiv 4 \pmod{13}$$ 2. **Recall the Chinese Remainder Theorem:** If the moduli are pairwise coprime, the system has a unique solution modulo the product of the moduli. 3. **Check moduli:** $$11, 12, 13$$ are pairwise coprime since \(\gcd(11,12)=1\), \(\gcd(11,13)=1\), and \(\gcd(12,13)=1\). 4. **Calculate the product of moduli:** $$N = 11 \times 12 \times 13 = 1716$$ 5. **Calculate partial products:** $$N_1 = \frac{N}{11} = 156$$ $$N_2 = \frac{N}{12} = 143$$ $$N_3 = \frac{N}{13} = 132$$ 6. **Find modular inverses:** Find \(M_i\) such that \(N_i M_i \equiv 1 \pmod{n_i}\): - For \(N_1=156\) mod 11: $$156 \equiv 2 \pmod{11}$$ Solve \(2M_1 \equiv 1 \pmod{11}\) gives \(M_1 = 6\) because \(2 \times 6 = 12 \equiv 1 \pmod{11}\). - For \(N_2=143\) mod 12: $$143 \equiv 11 \pmod{12}$$ Solve \(11M_2 \equiv 1 \pmod{12}\) gives \(M_2 = 11\) because \(11 \times 11 = 121 \equiv 1 \pmod{12}\). - For \(N_3=132\) mod 13: $$132 \equiv 2 \pmod{13}$$ Solve \(2M_3 \equiv 1 \pmod{13}\) gives \(M_3 = 7\) because \(2 \times 7 = 14 \equiv 1 \pmod{13}\). 7. **Construct solution:** $$x \equiv \sum_{i=1}^3 a_i N_i M_i \pmod{N}$$ where \(a_1=11, a_2=2, a_3=4\). Calculate: $$x \equiv 11 \times 156 \times 6 + 2 \times 143 \times 11 + 4 \times 132 \times 7 \pmod{1716}$$ $$= 11 \times 936 + 2 \times 1573 + 4 \times 924$$ $$= 10296 + 3146 + 3696 = 17138$$ 8. **Reduce modulo 1716:** $$x \equiv 17138 \pmod{1716}$$ Divide 17138 by 1716: $$17138 = 1716 \times 9 + 1342$$ So, $$x \equiv 1342 \pmod{1716}$$ **Final answer:** $$\boxed{x \equiv 1342 \pmod{1716}}$$ This means all integers congruent to 1342 modulo 1716 satisfy the system.