Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Let us take the composite number \(N\).
Decompose the number \(N\) into the product of primes.
Here, the number \(N = x_1 \times x_2\). But, both \(x_1\) and \(x_2\) are again a composite number. So, factorise it further to obtain a prime number.
The prime factors of \(x_1 = p_1 \times p_2\).
The prime factors of \(x_2 = p_3 \times p_4\).
We get, \(N = p_1 \times p_2 \times p_3 \times p_4\) where \(p_1\), \(p_2\), \(p_3\) and \(p_4\) are all prime numbers.
If we have repeated primes in a product, then we can write it as powers.
In general, given a composite number \(N\), we factorise it uniquely in the form N=p1q1×p2q2×p3q3×...×pnqn where p1,p2,p3, are prime numbers, and q1,q2,q3,...qn are natural numbers.
Thus, every composite number can be expressed as a product of primes apart from the order.
Consider a composite number \(26950\).
Let us factor this number using the factor tree method.
The prime factor of \(26950\) \(=\) \(2 \times 5 \times 5 \times 7 \times 7 \times 11\).
That is, \(26950 = 2 \times 5^2 \times 7^2 \times 11\).
Here, a composite number \(26950\) is written as a product of prime numbers.
If we change the order of the prime numbers, then also the answer will be the same composite number.
We can write \(26950 = 2 \times 7^2 \times 5^2 \times 11\) or \(26950 = 11 \times 7^2 \times 5^2 \times 2\).
Thus, the prime factorisation of a natural number is unique, except for the order of its factors.
HCF \(=\) Product of the smallest power of each common prime factor in the numbers.
LCM \(=\) Product of the greatest power of each prime factor involved in the numbers.