Factors of a Number Using Prime Factorisation:
Let us recall
- A factor of a number divides it exactly. For example, \(5\) is a factor of \(30\).
- A prime is a number greater than \(1\) that has only \(1\) and itself as factors.
- The prime factorisation of a number is that number written as a product of primes. For example, \(180 = 2 × 2 × 3 × 3 × 5\).
Every "subpart" of a prime factorisation is a factor
Take the prime factorisation of \(180\):
\(180 = 2 × 2 × 3 × 3 × 5\)
Take some of the primes, for example \(2 × 3 × 3 = 18\). This is a subpart of the factorisation. The primes we did not take are \(2\) and \(5\), so,
\(180 = (2 × 3 × 3) × (2 × 5) = 18 × 10\)
So \(18\) is a factor of \(180\), and \(10\) is its partner factor. Reordering factors does not change the product, so every subpart gives a factor.
Important!
Is every product of primes a factor?
No, only if it is a subpart. We can only use a prime as many times as it appears in the factorisation.
- \(2 × 2 × 3\) is a subpart of \(2 × 2 × 3 × 3 × 5\), so \(12\) is a factor of \(180\).
- \(2 × 2 × 2\) is not a subpart, because \(180\) has only two \(2\)s. So \(8\) is not a factor of \(180\).
- \(7\) is not a factor, because \(7\) does not appear at all.
How to list all the factors of a number:
Example: Find all the factors of \(150\).
Step 1: Find the prime factorisation using the division method.
\(\begin{array}{r|l}
2&150 \\ \hline
3&75 \\ \hline
5&25\\ \hline
5&5 \\ \hline
& 1
\end{array}\)
2&150 \\ \hline
3&75 \\ \hline
5&25\\ \hline
5&5 \\ \hline
& 1
\end{array}\)
So \(150 = 2 × 3 × 5 × 5\).
Step 2: Form the subparts systematically.
- Single primes: \(2, 3, 5\)
- Combinations of two primes: \(2 × 3 = 6\), \(2 × 5 = 10\), \(3 × 5 = 15\), \(5 × 5 = 25\)
- Combinations of three primes: \(2 × 3 × 5 = 30\), \(2 × 5 × 5 = 50\), \(3 × 5 × 5 = 75\)
- Combination of all four primes: \(2 × 3 × 5 × 5 = 150\)
Step 3: Add \(1\), which is a factor of every number.
Factors of \(150\): \(1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150\) (12 factors).
Important!
- If two different subparts give the same number, write that factor only once. For example, \(2 × 5\) and \(5 × 2\) are the same.
- Always include \(1\) and the number itself.
Finding the HCF of Numbers Using Prime Factorisation:
Common Factor:
A common factor is a factor that divides each of the given numbers exactly.
For example, consider \(18\) and \(30\).
Factors of \(18\): \(1,2,3,6,9,18 \)
Factors of \(30\): \( 1,2,3,5,6,10,15,30 \)
The common factors are \( 1,2,3,6 \)
Therefore, \(1,2,3,\) and \(6\) are common factors of \(18\) and \(30\).
Highest Common Factor (HCF):
The Highest Common Factor (HCF) of two or more numbers is the highest (or greatest) of their common factors. It is also known as the Greatest Common Divisor (GCD).
In the previous example, the common factors of \(18\) and \(30\) are \(1,2,3,\) and \(6\).
The highest among them is \(6\).
Therefore, \(\boxed{\operatorname{HCF}(18,30)=6} \)
Steps to find the HCF Using Prime factorisation:
Find the HCF of \(48\) and \(80\)
Step 1: Find the prime factorisation.
\(48 = 2 \times 2 \times 2 \times 2 \times 3 \)
\(80 = 2 \times 2 \times 2 \times 2 \times 5 \)
Step 2: Identify the common prime factors.
The only common prime factor is \(2\).
Step 3: Select the minimum number of occurrences.
Both numbers contain four occurrences of \(2\). Therefore, we select four occurrences.
Step 4: Multiply the common prime factors.
\( \begin{aligned} \operatorname{HCF}(48,80)& = 2 \times 2 \times 2 \times 2\\ & =16 \end{aligned} \)
Therefore, \( \boxed{\operatorname{HCF}(48,80)=16} \)
Example:
A gardener has \(42\) rose plants and \(56\) jasmine plants. She wants to plant them in rows so that each row has plants of only one kind, every row has the same number of plants, and the number of rows is as small as possible. How many plants should be in each row?
Solution:
Each row’s size must divide both \(42\) and \(56\), so it is a common factor. We want the fewest rows, so we want the largest row size, which is the HCF.
\(42 = 2 × 3 × 7\)
\(56 = 2 × 2 × 2 × 7\)
\(\text{HCF} = 2 × 7 = 14\).
Each row has \(14\)plants.
There will be \(42 ÷ 14 = 3\) rows of roses and \(56 ÷ 14 = 4\) rows of jasmine, so \(7\) rows in all.
Important!
- To find the HCF, we must choose only the prime factors common to all the given numbers.
- For the HCF, select the minimum number of occurrences of each common prime factor
- Thinking the HCF is \(0\) when there are no common primes. It is always \(1\).