What is the least number by which 37044 must be divided so that the resulting quotient is a perfect cube?

Introduction / Context:
This question is about perfect cubes and prime factorization. It asks for the smallest divisor that, when used to divide a given number, produces a perfect cube. Problems like this are common in number theory sections of aptitude tests and help learners practice expressing numbers in terms of their prime factors and analysing exponents to achieve required properties such as being a perfect square or perfect cube.

Given Data / Assumptions:

• The given number is 37044.
• We are looking for the least positive integer k such that 37044 / k is a perfect cube.
• All calculations are done with exact integer arithmetic.
• A perfect cube is a number of the form n^3 where n is an integer.

Concept / Approach:
The main idea is to factorize 37044 into its prime factors. For a number to be a perfect cube, the exponent of every prime in its factorization must be a multiple of 3. We need to find the smallest divisor k that we can remove from 37044 so that the remaining factorization has all exponents divisible by 3. This turns into an exponent adjustment problem, where we choose exponents for the divisor in such a way that the remaining exponents are multiples of 3.

Step-by-Step Solution:
Step 1: Factorize 37044 into prime factors.Step 2: One finds that 37044 = 2^2 * 3^3 * 7^3.Step 3: For a perfect cube, each prime exponent in the quotient must be a multiple of 3.Step 4: Suppose we divide 37044 by k = 2^a * 3^b * 7^c, where 0 ≤ a ≤ 2, 0 ≤ b ≤ 3, 0 ≤ c ≤ 3.Step 5: The quotient will then be 2^(2 - a) * 3^(3 - b) * 7^(3 - c).Step 6: For this quotient to be a perfect cube, 2 - a, 3 - b, and 3 - c must each be multiples of 3.Step 7: For the prime 2, exponents available are 2, 1, 0. Only 0 is a multiple of 3, so 2 - a = 0 which gives a = 2.Step 8: For the prime 3, we have 3 - b. Values 3 and 0 are multiples of 3, so b can be 0 or 3. To keep k minimal, choose b = 0.Step 9: For the prime 7, similarly 3 - c must be 3 or 0, so c = 0 or 3. To keep k minimal, choose c = 0.Step 10: Therefore the least such k is 2^2 = 4.

Verification / Alternative check:
If we divide 37044 by 4, we get 37044 / 4 = 9261. Now check if 9261 is a perfect cube. Note that 21^3 = 21 * 21 * 21 = 441 * 21 = 9261. Therefore 9261 is indeed a perfect cube, confirming that 4 is valid. If we tried k = 2, the quotient would be 2 * 3^3 * 7^3, which has exponent 1 for prime 2, not a multiple of 3, so it is not a perfect cube. Thus 4 is the least valid divisor.

Why Other Options Are Wrong:
Option 2: Leaves a factor of 2^1 in the quotient, so the exponent of 2 is not a multiple of 3.
Option 14: Includes unnecessary factors of 7, making k larger than needed without improving the cube condition beyond what 4 already achieves.
Option 21: Similar to 14, it includes extra factors which are not required to adjust exponents to multiples of 3.
Option 6: Equal to 2 * 3 and again leaves the exponent of 2 as 1 in the quotient, which fails the cube requirement.

Common Pitfalls:
Some learners try to guess divisors instead of systematically using prime factorization, which often leads to incorrect answers. Another mistake is to confuse the condition for a perfect square (exponents must be even) with that for a perfect cube (exponents must be multiples of 3). It is also easy to forget that we want the least divisor, so including unnecessary prime factors in k leads to a larger number than required.

Final Answer:
The least number by which 37044 must be divided to obtain a perfect cube is 4.