1. **Problem statement:** Solve the system of congruences using the Chinese Remainder Theorem (CRT): $$x \equiv 7 \pmod{15}$$ $$x \equiv 10 \pmod{17}$$ $$x \equiv 1 \pmod{14}$$ 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 15, 17, and 14. 3. **Check coprimality:** - $\gcd(15,17) = 1$ - $\gcd(15,14) = 1$ - $\gcd(17,14) = 1$ Since all pairs are coprime, CRT applies. 4. **Calculate the product of moduli:** $$N = 15 \times 17 \times 14 = 3570$$ 5. **Calculate partial products:** $$N_1 = \frac{N}{15} = 238$$ $$N_2 = \frac{N}{17} = 210$$ $$N_3 = \frac{N}{14} = 255$$ 6. **Find modular inverses:** Find $M_i$ such that $$N_i M_i \equiv 1 \pmod{n_i}$$ - For $N_1 = 238$ mod 15: $$238 \equiv 13 \pmod{15}$$ Find $M_1$ such that $$13 M_1 \equiv 1 \pmod{15}$$ Testing values, $M_1 = 7$ since $13 \times 7 = 91 \equiv 1 \pmod{15}$. - For $N_2 = 210$ mod 17: $$210 \equiv 6 \pmod{17}$$ Find $M_2$ such that $$6 M_2 \equiv 1 \pmod{17}$$ Testing values, $M_2 = 3$ since $6 \times 3 = 18 \equiv 1 \pmod{17}$. - For $N_3 = 255$ mod 14: $$255 \equiv 3 \pmod{14}$$ Find $M_3$ such that $$3 M_3 \equiv 1 \pmod{14}$$ Testing values, $M_3 = 5$ since $3 \times 5 = 15 \equiv 1 \pmod{14}$. 7. **Construct solution:** $$x \equiv a_1 N_1 M_1 + a_2 N_2 M_2 + a_3 N_3 M_3 \pmod{N}$$ Substitute values: $$x \equiv 7 \times 238 \times 7 + 10 \times 210 \times 3 + 1 \times 255 \times 5 \pmod{3570}$$ Calculate each term: $$7 \times 238 \times 7 = 11662$$ $$10 \times 210 \times 3 = 6300$$ $$1 \times 255 \times 5 = 1275$$ Sum: $$11662 + 6300 + 1275 = 19237$$ 8. **Reduce modulo $N$:** $$x \equiv 19237 \pmod{3570}$$ Divide 19237 by 3570: $$19237 = 3570 \times 5 + 1547$$ So, $$x \equiv 1547 \pmod{3570}$$ **Final answer:** $$\boxed{x \equiv 1547 \pmod{3570}}$$ This means the solution to the system is all integers congruent to 1547 modulo 3570.