Find the least positive integer by which 216 must be divided so that the quotient is a perfect square.

Introduction / Context:
This problem is the “division” analogue of making a perfect square. We need to remove minimal prime factors so that the remaining exponents are all even.

Given Data / Assumptions:

• Number: 216.
• Find smallest divisor d such that 216 / d is a perfect square.

Concept / Approach:
Prime factorize 216 and observe parity of exponents. Remove the smallest factor that turns all exponents even in the quotient.

Step-by-Step Solution:
216 = 2^3 * 3^3. To obtain a perfect square, exponents must be even. If we divide by 2 * 3 = 6, the quotient becomes 2^(3−1) * 3^(3−1) = 2^2 * 3^2 = 36. 36 is a perfect square (6^2). Therefore, the least such divisor is 6.

Verification / Alternative check:
Try smaller divisors: dividing by 3 leaves 2^3 * 3^2 (not all even exponents), dividing by 2 leaves 2^2 * 3^3 (still not all even). Dividing by 6 works.

Why Other Options Are Wrong:
3 or 4 alone do not make both prime exponents even. 9 removes too much from the 3-power but leaves 2^3 unchecked.

Common Pitfalls:
Confusing “multiply” versus “divide” adjustments. Ensure you reduce each odd exponent by one to make it even in the quotient.

Final Answer:
6