Explanation of the Problem and Solution

Exploring the problem of finding the smallest whole number that, when added to 9, is exactly divisible by 12, 15, 20, and 27, involves an understanding of least common multiples (LCM). Let's break down the solution step-by-step.

Understanding the Core of the Question

When a number is exactly divisible by 12, 15, 20, and 27, it should be a multiple of their least common multiple (LCM). The LCM of a set of numbers is the smallest number that is a multiple of each of the numbers in the set.

Prime Factorization and Finding LCM

Let's start by finding the prime factorization of each of the numbers:

12 2 x 2 x 3 15 3 x 5 20 2 x 2 x 5 27 3 x 3 x 3

From these factors, we determine the LCM as follows:

The LCM is the product of the highest powers of prime factors that appear in the factorization of each number. Therefore, the LCM of 12, 15, 20, and 27 is:

LCM 22 x 33 x 5 4 x 27 x 5 540

This means the smallest number that is exactly divisible by 12, 15, 20, and 27 is 540.

Solving the Given Problem

Given the problem of finding the smallest whole number that, when added to 9, is exactly divisible by 12, 15, 20, and 27, we need to find a number x such that:

x 9 540

Subtracting 9 from both sides:

x 540 - 9 531

Therefore, the smallest whole number that, when added to 9, is exactly divisible by 12, 15, 20, and 27 is 531.

Problem Solving

To solve this problem, we first find the least common multiple (LCM) of the divisors 12, 15, 20, and 27. The prime factorizations are as follows:

12 2 x 2 x 3 15 3 x 5 20 2 x 2 x 5 27 3 x 3 x 3

The LCM is calculated as:

LCM 22 x 33 x 5 4 x 27 x 5 540

Therefore, we need to find a number that, when added to 9, equals 540:

x 9 540

x 540 - 9 531

Thus, the smallest whole number that, when added to 9, is exactly divisible by 12, 15, 20, and 27 is 531.

Conclusion

In conclusion, the smallest whole number that, when added to 9, is exactly divisible by 12, 15, 20, and 27, is 531.