Problem 77 - Elementary Algebra

Question

For exercises \(55-86\), use prime factorization to find the least common multiple. $$ 120 n^{2} p^{2} ; 180 n^{5} p^{2} $$

Step-by-Step Solution

Answer

The least common multiple is \(360 n^5 p^2\).

Step 1: Find the Prime Factorization of 120

Break down 120 into its prime factors: \[ 120 = 2^3 \times 3 \times 5 \]

Step 2: Find the Prime Factorization of 180

Break down 180 into its prime factors: \[ 180 = 2^2 \times 3^2 \times 5 \]

Step 3: Determine the Highest Power of Each Prime

For the least common multiple, take the highest power of each prime number appearing in the factorizations: \[ 2^3, 3^2, 5 \]

Step 4: Consider the Variables with Their Highest Powers

In the expressions provided, the highest powers are:\[ n^5 \text{ and } p^2 \]

Step 5: Calculate the Least Common Multiple

Combine the highest powers of all primes and variables to find the least common multiple: \[ \text{LCM} = 2^3 \times 3^2 \times 5 \times n^5 \times p^2 \]

Step 6: Simplify the Expression

Multiply the constants to simplify the expression: \[ 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360 \] So, \[ \text{LCM} = 360 n^5 p^2 \]

Key Concepts

Prime Factorization

Prime factorization is breaking down a number into the set of prime numbers that multiply together to give the original number.
Primes are numbers greater than 1 that have no divisors other than 1 and themselves.
Examples of prime numbers include 2, 3, 5, 7, and 11.
To perform prime factorization for 120, we would:

• Start with the smallest prime, 2, and divide: \(120 \div 2 = 60\)
• Continue dividing by 2: \(60 \div 2 = 30\), and again: \(30 \div 2 = 15\)
• 15 is not divisible by 2, so we move to the next prime, 3: \(15 \div 3 = 5\)
• Lastly, we divide by 5: \(5 \div 5 = 1\)

Each of these primes is a factor, making the prime factorization for 120: \(2^3 \times 3 \times 5\).
This breakdown helps us when finding the least common multiple (LCM) later on.

Highest Powers

When calculating the LCM, we must consider the highest powers of each prime factor found in any of the factorizations.
The highest power of a number is the greatest exponent of that prime found in the factorization steps.
In our example factorization:

• \(2^3\) (from 120)
• \(3^2\) (from 180)
• \(5\) (common in both)

We use these highest powers to avoid missing any multiples.
We do the same with variables (e.g., n or p) by taking the highest exponent for each:

• \(n^5\)\ from \(120 n^2 p^2\) and \(180 n^5 p^2\)\lags highest power is \(n^5\)
• \(p^2\) common to both\

Using the highest powers ensures our LCM includes all necessary factors.

Variable Exponents

Variables in the expressions contribute to the LCM by taking the highest exponent of each variable.
Exponents indicate how many times a variable is multiplied by itself.
In our example:

• 120 has exponents for n and p: \(n^2\) and \(p^2\)
• 180 has \(n^5\) and \(p^2\)

The highest exponent of 'n' between the two is 5 (\(n^5\)), and for 'p', it is 2 (\(p^2\)).
Thus, combining these, we include \(n^5\) and \(p^2\) in the LCM.
This method ensures that the LCM is the smallest expression that holds all the necessary factors and their respective powers.

LCM Calculation

To calculate the LCM using prime factorization:

• Find the prime factors of each number involved.
• Identify the highest powers of each prime.

For 120 and 180, combining their factorizations:

• 120 = \(2^3 \times 3 \times 5\)
• 180 = \(2^2 \times 3^2 \times 5\)

The highest powers are \(2^3\), \(3^2\), and 5, as well as the highest exponents for any variables.
Then, combine:

• Multiply the highest power of each prime: \(2^3 \times 3^2 \times 5\)
• Include variable exponents: \(n^5 \times p^2\)

The resulting expression is: \(LCM = 2^3 \times 3^2 \times 5 \times n^5 \times p^2 = 360 n^5 p^2\).
This gives us the least common multiple, combining all necessary factors into one expression.