1. **State the problem:** Solve the system of congruences: $$x \equiv 1 \pmod{3}, \quad x \equiv 2 \pmod{5}, \quad x \equiv 3 \pmod{7}$$ 2. **Formula and method:** Use the Chinese Remainder Theorem (CRT) which states that if the moduli are pairwise coprime, there exists a unique solution modulo the product of the moduli. 3. **Calculate the product of moduli:** $$N = 3 \times 5 \times 7 = 105$$ 4. **Calculate partial products:** $$N_1 = \frac{N}{3} = 35, \quad N_2 = \frac{N}{5} = 21, \quad N_3 = \frac{N}{7} = 15$$ 5. **Find modular inverses:** Find $M_i$ such that: $$N_i M_i \equiv 1 \pmod{n_i}$$ - For $N_1=35$ mod 3: $$35 \equiv 2 \pmod{3}$$ Find $M_1$ with $2 M_1 \equiv 1 \pmod{3}$, so $M_1=2$. - For $N_2=21$ mod 5: $$21 \equiv 1 \pmod{5}$$ So $M_2=1$. - For $N_3=15$ mod 7: $$15 \equiv 1 \pmod{7}$$ So $M_3=1$. 6. **Construct solution:** $$x \equiv a_1 N_1 M_1 + a_2 N_2 M_2 + a_3 N_3 M_3 \pmod{N}$$ where $a_1=1, a_2=2, a_3=3$. Calculate: $$x \equiv 1 \times 35 \times 2 + 2 \times 21 \times 1 + 3 \times 15 \times 1 = 70 + 42 + 45 = 157 \pmod{105}$$ 7. **Simplify modulo 105:** $$157 \equiv 157 - 105 = 52 \pmod{105}$$ **Final answer:** $$x \equiv 52 \pmod{105}$$ This means all integers congruent to 52 modulo 105 satisfy the system.