Find the least perfect square that is divisible by 3, 4, 5, 6, and 8. Provide the exact number and show how it is obtained.

Introduction / Context: The task is to find the smallest number that is both a perfect square and a common multiple of the given set of integers. This tests understanding of LCM and perfect square conditions via prime exponents.

Given Data / Assumptions:

• Divisibility by 3, 4, 5, 6, and 8 required.
• Seek the least perfect square meeting these divisibility constraints.

Concept / Approach: Compute LCM of the set, then minimally adjust it to a perfect square by ensuring all prime exponents are even. Multiply by the smallest necessary factor to achieve even exponents.

Step-by-Step Solution: Prime forms: 3 = 3; 4 = 2^2; 5 = 5; 6 = 2 * 3; 8 = 2^3. LCM(3,4,5,6,8) = 2^3 * 3 * 5 = 120. To be a perfect square, each prime exponent must be even. In 120 = 2^3 * 3^1 * 5^1, exponents are odd for 2, 3, and 5. Multiply by 2 * 3 * 5 = 30 to make exponents even: 120 * 30 = 2^4 * 3^2 * 5^2 = 3600, a perfect square.

Verification / Alternative check: 3600 = 60^2 and is divisible by 3, 4, 5, 6, 8 (check quickly: 3600/8 = 450; 3600/5 = 720; etc.).

Why Other Options Are Wrong: 900 = 30^2 but 900/8 is not an integer. 1200 and 2500 are not perfect squares divisible by all listed numbers.

Common Pitfalls: Confusing LCM with the final answer without making it a square. Always ensure even exponents for primes in the factorization.

Final Answer: 3600