Introduction

In this article, we will explore the problem of finding the smallest possible value of a natural number b given that the least common multiple (LCM) of a and b is 2520, where a is a specific natural number (120 in this case). We will delve into the mathematical principles behind the solution, including the relationship between LCM, GCD, and prime factors, and apply these concepts to reach a rigorous solution.

Multiplicative Relationship between LCM and GCD

The relationship between the LCM and GCD of two numbers can be expressed as:

LCM(a, b) times; GCD(a, b) a times; b

To find the smallest possible value of b, let's use the given information and follow a step-by-step approach.

Given Information and Initial Setup

We are given:

LCM(a, b) 2520 a 120

Using the relationship:

2520 times; GCD(120, b) 120 times; b

Rearranging for b:

120 times; b / GCD(120, b) 2520

Hence,

b 2520 times; GCD(120, b) / 120

To minimize b, we need to determine the GCD(120, b) and apply it in the equation.

Prime Factorization and Prime Factors

Let's first perform the prime factorization of both 120 and 2520:

120 2^3 times; 3 times; 5

2520 2^3 times; 3^2 times; 5 times; 7

We can see that the LCM(120, 2520) is indeed 2520, as it includes all the prime factors from both numbers with the highest exponent in each prime factor.

Since 2520 is the LCM, the prime factors of b should include 3^2 and 7 (since 120 already includes 2^3, 3^1, and 5).

Calculating the GCD and Solving for b

The GCD(120, b) should be a common divisor of both 120 and b. From the prime factorization of 120, we see that the smallest GCD is 3 (since 3 is a common factor and is the smallest prime factor of 120 that can also be a factor of 2520).

Using the derived formula:

b 2520 times; GCD(120, b) / 120

Since the GCD(120, b) must be a multiple of 3, let's calculate for the smallest possible GCD, which is 3:

b 2520 times; 3 / 120

b 63

Verification

Let's verify the solution:

Prime factorization of 120: 2^3 times; 3 times; 5 Prime factorization of 63: 3^2 times; 7 LCM(120, 63) 2^3 times; 3^2 times; 5 times; 7 2520 GCD(120, 63) 3

This satisfies the equation LCM(120, 63) times; GCD(120, 63) 120 times; 63, as 2520 times; 3 120 times; 63.

Conclusion

The smallest possible value of b given that LCM(120, b) 2520 is b 63.

This solution is based on the principles of LCM and GCD, and the prime factorization of the given numbers. Understanding these concepts is crucial for solving similar problems in number theory and discrete mathematics.