Before we learn how to find the Highest Common Factor (HCF) of numbers, let's first understand what factors are.
Factors of a number are the numbers that divide it exactly without leaving any remainder. For example, factors of 12 are 1, 2, 3, 4, 6, and 12 because each divides 12 completely.
When we look at two or more numbers, some factors may be common to all of them. These are called common factors. For example, the factors of 12 are 1, 2, 3, 4, 6, 12 and the factors of 18 are 1, 2, 3, 6, 9, 18. The common factors of 12 and 18 are 1, 2, 3, and 6.
The Highest Common Factor (HCF) is the greatest number among the common factors. In the example above, the HCF of 12 and 18 is 6.
Why is HCF important? It helps us simplify fractions, solve problems involving ratios, and divide things into equal parts without leftovers. For example, if you want to cut two ropes into equal pieces without any leftover, the length of each piece will be the HCF of the lengths of the ropes.
Imagine two sets of blocks representing the factors of two numbers. The blocks that appear in both sets are the common factors. The largest block in this overlap is the HCF.
There are three main methods to find the HCF of two or more numbers:
We will learn each method step-by-step with examples.
This method involves breaking down each number into its prime factors. A prime number is a number greater than 1 that has no factors other than 1 and itself (like 2, 3, 5, 7, 11, etc.).
Once we find the prime factors of each number, we look for the common prime factors. The HCF is the product of these common prime factors, each taken with the smallest power (or count) they appear in.
Let's see how to do this with an example.
In the factor trees above:
The common prime factors are two 2's and one 3.
So, the HCF is \( 2 \times 2 \times 3 = 12 \).
This method uses repeated division to find the HCF. Here's how it works:
graph TD A[Start with two numbers: a (larger), b (smaller)] B[Divide a by b] C{Is remainder zero?} D[Yes: HCF is b] E[No: Replace a with b, b with remainder] F[Repeat division] A --> B --> C C -->|Yes| D C -->|No| E --> F --> CThis method is straightforward and works well for numbers where division is easy.
The Euclidean algorithm is a faster and more efficient way to find the HCF, especially for large numbers. It is based on the principle:
HCF(a, b) = HCF(b, a mod b)
where \( a \mod b \) is the remainder when \( a \) is divided by \( b \).
You keep applying this formula repeatedly until the remainder becomes zero. The last non-zero divisor is the HCF.
graph TD A[Start with numbers a and b] B[Calculate remainder r = a mod b] C{Is r zero?} D[Yes: HCF is b] E[No: Set a = b, b = r] F[Repeat calculation] A --> B --> C C -->|Yes| D C -->|No| E --> F --> BThis method is very useful in exams because it reduces the problem quickly without full factorization.
Step 1: Find the prime factors of 36.
36 = 2 x 2 x 3 x 3
Step 2: Find the prime factors of 48.
48 = 2 x 2 x 2 x 3
Step 3: Identify the common prime factors.
Common prime factors are two 2's and one 3.
Step 4: Multiply the common prime factors to get the HCF.
HCF = 2 x 2 x 3 = 12
Answer: The HCF of 36 and 48 is 12.
Step 1: Divide the larger number 98 by the smaller number 56.
98 / 56 = 1 remainder 42
Step 2: Now divide 56 by the remainder 42.
56 / 42 = 1 remainder 14
Step 3: Divide 42 by the remainder 14.
42 / 14 = 3 remainder 0
Step 4: Since the remainder is 0, the divisor 14 is the HCF.
Answer: The HCF of 56 and 98 is 14.
Step 1: Calculate 270 mod 192.
270 / 192 = 1 remainder 78, so 270 mod 192 = 78
Step 2: Now find HCF(192, 78).
192 / 78 = 2 remainder 36, so 192 mod 78 = 36
Step 3: Find HCF(78, 36).
78 / 36 = 2 remainder 6, so 78 mod 36 = 6
Step 4: Find HCF(36, 6).
36 / 6 = 6 remainder 0, so 36 mod 6 = 0
Step 5: Since remainder is 0, HCF is 6.
Answer: The HCF of 270 and 192 is 6.
Step 1: Find the HCF of 84 and 126.
Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Factors of 126: 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126
Common factors: 1, 2, 3, 6, 7, 14, 21, 42
Highest common factor is 42.
Step 2: Divide numerator and denominator by 42.
\(\frac{84}{42} = 2\), \(\frac{126}{42} = 3\)
Answer: Simplified fraction is \(\frac{2}{3}\).
Step 1: Find the HCF of 150 and 210.
Prime factors of 150: 2 x 3 x 5 x 5
Prime factors of 210: 2 x 3 x 5 x 7
Common prime factors: 2, 3, 5
HCF = 2 x 3 x 5 = 30
Step 2: The maximum length of each piece is 30 cm.
Answer: Each piece should be 30 cm long.
When to use: When numbers are large or prime factorization is time-consuming.
When to use: To verify answers quickly in any method.
When to use: To avoid confusion and speed up calculations.
When to use: When under time pressure in competitive exams.
When to use: During prime factorization to avoid missing factors.
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