What is the least number that must be multiplied with 69120 in order to make the product a perfect cube?

Introduction / Context:
This question is another example of working with perfect cubes using prime factorization. Instead of dividing, we now need to multiply the given number by the smallest possible integer so that the result becomes a perfect cube. This is a common type of problem in number systems that reinforces the idea that the exponents of prime factors in a perfect cube must all be multiples of 3.

Given Data / Assumptions:

• The given number is 69120.
• We must find the smallest positive integer k such that 69120 * k is a perfect cube.
• All calculations are done with exact prime factorizations.
• A perfect cube has prime exponents that are multiples of 3.

Concept / Approach:
The strategy is to write 69120 as a product of prime powers. We then examine the exponents of the primes and determine how many more of each prime are needed to make all the exponents multiples of 3. The smallest multiplier k will be formed by the extra prime factors required to adjust these exponents. This method avoids trial and error and leads directly to the minimal solution.

Step-by-Step Solution:
Step 1: Factorize 69120 into prime factors.Step 2: The prime factorization of 69120 is 2^9 * 3^3 * 5^1.Step 3: For a number to be a perfect cube, the exponent of each prime in its factorization must be a multiple of 3.Step 4: Check the exponent of each prime in 69120:- For prime 2, the exponent is 9, which is already a multiple of 3 (9 = 3 * 3).- For prime 3, the exponent is 3, also a multiple of 3.- For prime 5, the exponent is 1, which is not a multiple of 3.Step 5: We need to increase the exponent of 5 from 1 to the next multiple of 3, which is 3.Step 6: To raise the exponent from 1 to 3, we need 2 more factors of 5, that is multiply by 5^2 = 25.Step 7: Thus, k must at least contain 25, and no other extra prime factors are needed because the exponents of 2 and 3 are already multiples of 3.Step 8: Therefore, the least number that must be multiplied to 69120 is 25.

Verification / Alternative check:
Compute the product 69120 * 25. In prime factors, this is (2^9 * 3^3 * 5^1) * 5^2 = 2^9 * 3^3 * 5^3. Now each exponent is a multiple of 3: 9, 3, and 3. So 69120 * 25 is a perfect cube. Specifically, we can write the result as (2^3)^3 * (3^1)^3 * (5^1)^3, which equals (8 * 3 * 5)^3 = 120^3, confirming that the product is indeed a perfect cube.

Why Other Options Are Wrong:
Option 10: This equals 2 * 5; multiplying by this gives 2^10 * 3^3 * 5^2, where the exponents 10 and 2 are not multiples of 3.
Option 50: This is 2 * 5^2; multiplying gives 2^10 * 3^3 * 5^3, where the exponent of 2 is still not a multiple of 3.
Option 5: This only increases the exponent of 5 from 1 to 2, which is still not a multiple of 3.
Option 15: This is 3 * 5; multiplying gives 2^9 * 3^4 * 5^2, where both 4 and 2 are not multiples of 3.

Common Pitfalls:
A frequent mistake is to pick a number that makes one prime exponent a multiple of 3 but disrupts another exponent. Some students also forget to fully factor the given number, leading to incorrect exponent counts. Others try random multipliers instead of systematically adjusting exponents. The best practice is always to rely on prime factorization and exponent arithmetic when dealing with perfect powers.

Final Answer:
The least number that must be multiplied with 69120 to make a perfect cube is 25.