1. **Problem statement:** Solve the system of congruences using the Chinese Remainder Theorem (CRT): $$x \equiv 2 \pmod{11}$$ $$x \equiv 4 \pmod{15}$$ $$x \equiv 9 \pmod{14}$$ $$x \equiv 7 \pmod{13}$$ 2. **Formula and explanation:** The CRT states that if the moduli are pairwise coprime, there exists a unique solution modulo the product of the moduli. Here, the moduli are 11, 15, 14, and 13. Check pairwise coprimality: - gcd(11,15)=1 - gcd(11,14)=1 - gcd(11,13)=1 - gcd(15,14)=1 - gcd(15,13)=1 - gcd(14,13)=1 All are coprime. 3. **Calculate the product of moduli:** $$N = 11 \times 15 \times 14 \times 13 = 30030$$ 4. **Calculate each partial product:** $$N_1 = \frac{N}{11} = 2730$$ $$N_2 = \frac{N}{15} = 2002$$ $$N_3 = \frac{N}{14} = 2145$$ $$N_4 = \frac{N}{13} = 2310$$ 5. **Find the modular inverses $M_i$ such that:** $$N_i M_i \equiv 1 \pmod{n_i}$$ - For $N_1=2730$ mod 11: Calculate $M_1$ such that $2730 M_1 \equiv 1 \pmod{11}$. Since $2730 \equiv 1 \pmod{11}$ (because $2730 = 11 \times 248 + 2$, actually $2730 \mod 11 = 2730 - 11 \times 248 = 2730 - 2728 = 2$), correct calculation: $2730 \mod 11 = 2$ So solve $2 M_1 \equiv 1 \pmod{11}$. Try $M_1=6$ since $2 \times 6=12 \equiv 1 \pmod{11}$. So $M_1=6$. - For $N_2=2002$ mod 15: Calculate $M_2$ such that $2002 M_2 \equiv 1 \pmod{15}$. $2002 \mod 15 = 2002 - 15 \times 133 = 2002 - 1995 = 7$ So solve $7 M_2 \equiv 1 \pmod{15}$. Try $M_2=13$ since $7 \times 13=91 \equiv 1 \pmod{15}$ (because $91 - 15 \times 6=91-90=1$). So $M_2=13$. - For $N_3=2145$ mod 14: Calculate $M_3$ such that $2145 M_3 \equiv 1 \pmod{14}$. $2145 \mod 14 = 2145 - 14 \times 153 = 2145 - 2142 = 3$ So solve $3 M_3 \equiv 1 \pmod{14}$. Try $M_3=5$ since $3 \times 5=15 \equiv 1 \pmod{14}$. So $M_3=5$. - For $N_4=2310$ mod 13: Calculate $M_4$ such that $2310 M_4 \equiv 1 \pmod{13}$. $2310 \mod 13 = 2310 - 13 \times 177 = 2310 - 2301 = 9$ So solve $9 M_4 \equiv 1 \pmod{13}$. Try $M_4=3$ since $9 \times 3=27 \equiv 1 \pmod{13}$ (27 - 13 \times 2=27-26=1). So $M_4=3$. 6. **Construct the solution:** $$x \equiv \sum_{i=1}^4 a_i N_i M_i \pmod{N}$$ where $a_1=2$, $a_2=4$, $a_3=9$, $a_4=7$. Calculate each term: $$2 \times 2730 \times 6 = 32760$$ $$4 \times 2002 \times 13 = 104104$$ $$9 \times 2145 \times 5 = 96525$$ $$7 \times 2310 \times 3 = 48510$$ Sum: $$32760 + 104104 + 96525 + 48510 = 281899$$ 7. **Find $x$ modulo $N=30030$:** $$x \equiv 281899 \pmod{30030}$$ Calculate remainder: $$281899 - 30030 \times 9 = 281899 - 270270 = 11629$$ So the solution is: $$\boxed{x \equiv 11629 \pmod{30030}}$$ This means all integers congruent to 11629 modulo 30030 satisfy the system.