Find the smallest positive number N such that: • N leaves remainder 4 when divided by each of 16, 18, and 20; and • N is exactly divisible by 7.

Introduction / Context:This is a linear congruence with a divisibility constraint. A number that leaves the same remainder r with several moduli is congruent to r modulo the LCM of those moduli. Then, impose the divisibility requirement to select the least positive solution.

Given Data / Assumptions:

• N ≡ 4 (mod 16), N ≡ 4 (mod 18), N ≡ 4 (mod 20).
• N is divisible by 7.

Concept / Approach:First, compute L = LCM(16, 18, 20). All solutions to the congruences are N = L*k + 4 for integer k ≥ 0. Then choose the smallest k for which N is a multiple of 7. Use modular arithmetic to solve (L*k + 4) ≡ 0 (mod 7).

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• 2256, 865, 3332 fail either the remainder condition or the divisibility-by-7 condition (or both).

Common Pitfalls:

• Losing track of the LCM or using some but not all moduli in forming L.

Final Answer: