Find the least positive integer which, when multiplied with 74088, makes the product a perfect square. Show the reasoning clearly and retain the numerical value 74088 without approximation.

Introduction / Context:
This question tests prime factorization and the condition for a number to be a perfect square. A product is a perfect square if every prime factor appears with an even exponent in its prime factorization.

Given Data / Assumptions:

• Number given: 74088.
• We seek the smallest positive integer k such that 74088 * k is a perfect square.
• No approximation is needed; work with exact prime powers.

Concept / Approach:
Factorize 74088 into primes. For a perfect square, every prime's exponent must be even. If an exponent is odd, multiply by that prime once to make it even. Combine all such missing primes for the least multiplier.

Step-by-Step Solution:
74088 = 2^3 * 3^3 * 7^3 (by prime factorization). Each exponent (3) is odd for 2, 3, and 7. To make exponents even, multiply by one 2, one 3, and one 7. Least multiplier k = 2 * 3 * 7 = 42. Now 74088 * 42 = 2^(3+1) * 3^(3+1) * 7^(3+1) = 2^4 * 3^4 * 7^4, a perfect square.

Verification / Alternative check:
A quick check: any odd exponent becomes even by multiplying once with that prime. All three primes required exactly one copy, confirming k = 42 is minimal.

Why Other Options Are Wrong:
44 = 2^2 * 11 introduces 11 and leaves 3 and 7 odd. 46 = 2 * 23 introduces 23 and leaves 3 and 7 odd. 48 = 2^4 * 3 leaves 7 odd; also adds unnecessary powers.

Common Pitfalls:
Students sometimes add large multipliers or multiply by existing odd exponents directly without ensuring minimality. Always fix each odd exponent by multiplying once by that prime only.

Final Answer:
42