Find the smallest integer N such that N − 3 is divisible by each of 21, 28, 36, and 45.

Introduction / Context:
The statement “N − 3 is divisible by 21, 28, 36, and 45” means N − 3 must be a common multiple of all four numbers. The smallest such N is found by adding 3 to the least common multiple (LCM) of the set.

Given Data / Assumptions:

• Divisors: 21, 28, 36, 45.
• We want the least N such that N − 3 is a multiple of each divisor.

Concept / Approach:
Compute LCM(21, 28, 36, 45) using prime factorization. Then take N = LCM + 3. Prime factorization captures the highest needed power of each prime across the set.

Step-by-Step Solution:

Verification / Alternative check:
Check divisibility: 1260 ÷ 21 = 60; ÷ 28 = 45; ÷ 36 = 35; ÷ 45 = 28. All are integers, confirming 1260 is the LCM. Hence N = 1263 is minimal.

Why Other Options Are Wrong:

• 1260, 1257: These are N values themselves, not N − 3. Using them would not satisfy the “minus 3” condition.
• 420, 840: Smaller than the LCM; their +3 versions would not be multiples of all four divisors.

Common Pitfalls:

• Forgetting to add 3 after obtaining the LCM.
• Computing LCM with incorrect prime powers (e.g., missing 3^2 or 2^2).

Final Answer:
1263