Diving into the Relationship Between LCM and GCD: Solving for ( n ) where ( text{LCM}(20, n) 100 ) and ( text{LCM}(30, n) 120 )

Introduction to Least Common Multiple (LCM) and Greatest Common Divisor (GCD)

When working with numbers, the concepts of LCM and GCD play a crucial role in understanding relationships between them. The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both, while the greatest common divisor (GCD) is the largest positive integer that divides both numbers without leaving a remainder.

Solving for ( n ) where ( text{LCM}(20, n) 100 )

The problem is to find the possible values of ( n ) such that the least common multiple of 20 and ( n ) equals 100. We start by utilizing the relationship between LCM and GCD:

text{LCM}(a, b) frac{a times b}{text{GCD}(a, b)}

Substituting ( a 20 ) and ( b n ) and given that the LCM is 100, we get:

100 frac{20 times n}{text{GCD}(20, n)}

Rearranging, we have:

text{GCD}(20, n) frac{20 times n}{100} frac{n}{5}

We can denote the GCD as ( d ), so:

d frac{n}{5}

Naturally, ( d ) should be a divisor of both 20 and ( n ). The divisors of 20 are: 1, 2, 4, 5, 10, and 20. Given ( d frac{n}{5} ), we explore ( n ) for each divisor:

If ( d 1 ), ( n 5 times 1 5 ) If ( d 2 ), ( n 5 times 2 10 ) If ( d 4 ), ( n 5 times 4 20 ) If ( d 5 ), ( n 5 times 5 25 ) If ( d 10 ), ( n 5 times 10 50 ) If ( d 20 ), ( n 5 times 20 100 )

Next, we verify which of these ( n ) values satisfy the LCM(20, ( n )) 100 condition.

Verification:

For ( n 5 ): LCM(20, 5) 20 For ( n 10 ): LCM(20, 10) 20 For ( n 20 ): LCM(20, 20) 20 For ( n 25 ): LCM(20, 25) 100 For ( n 50 ): LCM(20, 50) 100 For ( n 100 ): LCM(20, 100) 100

The two valid values of ( n ) are:

( n 25 ) and ( n 50 )

Solving for ( n ) where ( text{LCM}(30, n) 120 )

Now, consider the problem where the LCM of 30 and ( n ) equals 120. Using the same GCD relationship and the fact that the GCD divides both numbers, we start by decomposing the numbers:

30 ( 2^1 times 3^1 times 5^1 ) 120 ( 2^3 times 3^1 times 5^1 )

This shows that ( n ) must be of the form ( 2^i times 3^j times 5^k ) where ( i 3 ), ( j leq 1 ), and ( k leq 1 ). Thus, the possible values of ( n ) are:

23 × 30 × 50 8 23 × 31 × 50 24 23 × 30 × 51 40 23 × 31 × 51 120

Hence, the possible values of ( n ) are 8, 24, 40, and 120.

Conclusion

By understanding the relationship between LCM and GCD and applying the concepts of prime factorization, we successfully solved for the values of ( n ) in both scenarios. These techniques are fundamental in number theory and provide a powerful tool for solving similar problems.