Question

Consider the following systems of three equations (congruences): \( x \equiv 2 \pmod{3} \), \( x \equiv 3 \pmod{5} \), and \( x \equiv 2 \pmod{7} \). Find \( x \)?

• A. \( 33 \)
• B. \( 23 \)
• C. \( 42 \)
• D. \( 51 \)

Hint:

The Chinese Remainder Theorem (CRT) helps solve simultaneous modular equations. Compute \( M \), find each \( M_i \), determine their inverses, and sum the products for the final answer.

Answer 1

The Correct Option is B

Step 1: Understanding the given congruences We have the system of congruences: \[ x \equiv 2 \pmod{3} \] \[ x \equiv 3 \pmod{5} \] \[ x \equiv 2 \pmod{7} \]
Step 2: Applying the Chinese Remainder Theorem (CRT) Define \( x = M_i \cdot y_i \) where \( M = 3 \times 5 \times 7 = 105 \). Computing individual \( M_i \): \[ M_1 = \frac{105}{3} = 35, \quad M_2 = \frac{105}{5} = 21, \quad M_3 = \frac{105}{7} = 15. \] Solving for the multiplicative inverses: \[ y_1 \equiv 35^{-1} \pmod{3} = 2, \quad y_2 \equiv 21^{-1} \pmod{5} = 1, \quad y_3 \equiv 15^{-1} \pmod{7} = 1. \] Now, computing: \[ x = (2 \times 35 \times 2) + (3 \times 21 \times 1) + (2 \times 15 \times 1) \pmod{105} \] \[ x = (140 + 63 + 30) \pmod{105} \] \[ x = 233 \pmod{105} = 23. \]