If the least common multiple of the numbers 1728 and K is 5184, how many possible positive integer values of K are there?

Introduction / Context:
This problem tests understanding of least common multiple and prime factorization. We are given that the least common multiple of 1728 and some positive integer K is 5184, and we must count how many possible K values can satisfy this condition. Such questions are standard in number theory sections of aptitude exams and help build confidence with factor exponents and lcm properties.

Given Data / Assumptions:

• Given numbers: 1728 and K.
• LCM(1728, K) = 5184.
• We consider only positive integer values of K.
• We need to count how many such K values exist.

Concept / Approach:
The least common multiple is determined by taking the maximum power of each prime that appears in the factorization of the given numbers. So we factor both 1728 and 5184 into primes. Then we determine the conditions on the prime powers of K so that the lcm of 1728 and K equals 5184. This gives us allowed exponent ranges for each prime in K, and we count all possible combinations of these exponents.

Step-by-Step Solution:
First factor 1728.1728 = 2^6 * 3^3.Next factor 5184.5184 = 2^6 * 3^4.Let K = 2^a * 3^b, because only primes 2 and 3 can appear; otherwise the lcm would contain additional primes not present in 5184.LCM(1728, K) takes the higher exponent of each prime from 1728 and K.So LCM(1728, K) = 2^(max(6, a)) * 3^(max(3, b)).We require 2^(max(6, a)) * 3^(max(3, b)) = 2^6 * 3^4.Therefore, max(6, a) = 6 and max(3, b) = 4.From max(6, a) = 6, we get a ≤ 6.From max(3, b) = 4, and since 1728 has exponent 3 for prime 3, K must supply exponent 4, so b must be 4.Thus b is fixed at 4, while a can be any integer from 0 to 6 inclusive.The possible values for a are 0, 1, 2, 3, 4, 5, and 6, which gives 7 choices.Each pair (a, b) with b = 4 and a in this range gives a distinct positive integer K.

Verification / Alternative check:
For example, if a = 0 and b = 4, then K = 3^4 = 81.LCM(1728, 81) = 2^6 * 3^4 = 5184, which satisfies the condition.If a = 6 and b = 4, then K = 2^6 * 3^4 = 5184 itself, and clearly LCM(1728, 5184) is 5184.Any choice of a greater than 6 would make max(6, a) larger than 6, causing the lcm to exceed 2^6 in the power of 2, which would violate the lcm condition.

Why Other Options Are Wrong:
Options 11, 8, and 6 correspond to counting more combinations than actually allowed or misreading the exponent restrictions.In reality, exponent b has only 1 valid choice (4), and exponent a has 7 valid choices (0 through 6), resulting in exactly 7 possible values for K.

Common Pitfalls:
Some learners incorrectly assume that K must also have exponent 6 for prime 2 or exponent 4 for prime 3 with no flexibility, which reduces the count too much.Others allow primes beyond 2 and 3 in K, not realizing that this would force the lcm to contain extra primes and no longer equal 5184.Miscomputing prime factorizations of 1728 or 5184 can also lead to incorrect exponent conditions.

Final Answer:
The number of possible positive integer values of K is 7.