Answer :
### Solution:
Let's start by addressing each of the parts of the problem step-by-step.
#### (i) Express each given number as the products of their prime factors.
1. Prime factorization of 12:
- 12 can be divided by 2: [tex]\(12 \div 2 = 6\)[/tex]
- 6 can be divided by 2: [tex]\(6 \div 2 = 3\)[/tex]
- 3 is a prime number, so it remains as is.
- Therefore, the prime factors of 12 are [tex]\(2, 2, 3\)[/tex].
3. Prime factorization of 16:
- 16 can be divided by 2: [tex]\(16 \div 2 = 8\)[/tex]
- 8 can be divided by 2: [tex]\(8 \div 2 = 4\)[/tex]
- 4 can be divided by 2: [tex]\(4 \div 2 = 2\)[/tex]
- 2 is a prime number, so it remains as is.
- Therefore, the prime factors of 16 are [tex]\(2, 2, 2, 2\)[/tex].
3. Prime factorization of 24:
- 24 can be divided by 2: [tex]\(24 \div 2 = 12\)[/tex]
- 12 can be divided by 2: [tex]\(12 \div 2 = 6\)[/tex]
- 6 can be divided by 2: [tex]\(6 \div 2 = 3\)[/tex]
- 3 is a prime number, so it remains as is.
- Therefore, the prime factors of 24 are [tex]\(2, 2, 2, 3\)[/tex].
Prime factorization results:
- [tex]\(12 = 2 \cdot 2 \cdot 3\)[/tex]
- [tex]\(16 = 2 \cdot 2 \cdot 2 \cdot 2\)[/tex]
- [tex]\(24 = 2 \cdot 2 \cdot 2 \cdot 3\)[/tex]
#### (ii) Find the product of the common prime factors and the factors which are not common.
1. Common Prime Factors:
- By examining the prime factors of 12, 16, and 24, we can identify the common prime factor.
- The common prime factor is [tex]\(2\)[/tex].
2. All Prime Factors (considering each unique factor included at least once across all numbers):
- From 12, we use [tex]\(2, 3\)[/tex]
- From 16, we use [tex]\(2\)[/tex]
- From 24, we use [tex]\(2, 3\)[/tex]
- The unique factors here are [tex]\(2, 3\)[/tex].
3. Products:
- Product of common prime factors: [tex]\(2\)[/tex]
- Product of all prime factors: [tex]\(2 \cdot 3 = 6\)[/tex]
Therefore, the common prime factor product is [tex]\(2\)[/tex] and all unique prime factors product is [tex]\(6\)[/tex].
#### (iii) Is the product the L.C.M. of the given numbers? Give reason.
To find the Least Common Multiple (LCM), consider:
- The LCM is the smallest positive integer that is evenly divisible by all the given numbers (12, 16, and 24).
- We include the highest powers of all prime factors found in any of the numbers.
Prime factors in highest powers are:
- From 12: [tex]\(2^2 \cdot 3\)[/tex]
- From 16: [tex]\(2^4\)[/tex]
- From 24: [tex]\(2^3 \cdot 3\)[/tex]
Combining the highest powers:
- Highest power of 2: [tex]\(2^4\)[/tex] (from 16)
- Highest power of 3: [tex]\(3^1\)[/tex] (from 12 and 24)
Therefore:
[tex]\[ \text{LCM} = 2^4 \cdot 3^1 = 16 \cdot 3 = 48 \][/tex]
Thus, the calculated LCM (48) is the product of the highest powers of all factors and satisfies the condition that it's the smallest number divisible by all of 12, 16, and 24.
Let's start by addressing each of the parts of the problem step-by-step.
#### (i) Express each given number as the products of their prime factors.
1. Prime factorization of 12:
- 12 can be divided by 2: [tex]\(12 \div 2 = 6\)[/tex]
- 6 can be divided by 2: [tex]\(6 \div 2 = 3\)[/tex]
- 3 is a prime number, so it remains as is.
- Therefore, the prime factors of 12 are [tex]\(2, 2, 3\)[/tex].
3. Prime factorization of 16:
- 16 can be divided by 2: [tex]\(16 \div 2 = 8\)[/tex]
- 8 can be divided by 2: [tex]\(8 \div 2 = 4\)[/tex]
- 4 can be divided by 2: [tex]\(4 \div 2 = 2\)[/tex]
- 2 is a prime number, so it remains as is.
- Therefore, the prime factors of 16 are [tex]\(2, 2, 2, 2\)[/tex].
3. Prime factorization of 24:
- 24 can be divided by 2: [tex]\(24 \div 2 = 12\)[/tex]
- 12 can be divided by 2: [tex]\(12 \div 2 = 6\)[/tex]
- 6 can be divided by 2: [tex]\(6 \div 2 = 3\)[/tex]
- 3 is a prime number, so it remains as is.
- Therefore, the prime factors of 24 are [tex]\(2, 2, 2, 3\)[/tex].
Prime factorization results:
- [tex]\(12 = 2 \cdot 2 \cdot 3\)[/tex]
- [tex]\(16 = 2 \cdot 2 \cdot 2 \cdot 2\)[/tex]
- [tex]\(24 = 2 \cdot 2 \cdot 2 \cdot 3\)[/tex]
#### (ii) Find the product of the common prime factors and the factors which are not common.
1. Common Prime Factors:
- By examining the prime factors of 12, 16, and 24, we can identify the common prime factor.
- The common prime factor is [tex]\(2\)[/tex].
2. All Prime Factors (considering each unique factor included at least once across all numbers):
- From 12, we use [tex]\(2, 3\)[/tex]
- From 16, we use [tex]\(2\)[/tex]
- From 24, we use [tex]\(2, 3\)[/tex]
- The unique factors here are [tex]\(2, 3\)[/tex].
3. Products:
- Product of common prime factors: [tex]\(2\)[/tex]
- Product of all prime factors: [tex]\(2 \cdot 3 = 6\)[/tex]
Therefore, the common prime factor product is [tex]\(2\)[/tex] and all unique prime factors product is [tex]\(6\)[/tex].
#### (iii) Is the product the L.C.M. of the given numbers? Give reason.
To find the Least Common Multiple (LCM), consider:
- The LCM is the smallest positive integer that is evenly divisible by all the given numbers (12, 16, and 24).
- We include the highest powers of all prime factors found in any of the numbers.
Prime factors in highest powers are:
- From 12: [tex]\(2^2 \cdot 3\)[/tex]
- From 16: [tex]\(2^4\)[/tex]
- From 24: [tex]\(2^3 \cdot 3\)[/tex]
Combining the highest powers:
- Highest power of 2: [tex]\(2^4\)[/tex] (from 16)
- Highest power of 3: [tex]\(3^1\)[/tex] (from 12 and 24)
Therefore:
[tex]\[ \text{LCM} = 2^4 \cdot 3^1 = 16 \cdot 3 = 48 \][/tex]
Thus, the calculated LCM (48) is the product of the highest powers of all factors and satisfies the condition that it's the smallest number divisible by all of 12, 16, and 24.
