Question

Find the HCF and LCM of 42 and 72 by prime factorization method i.e., by fundamental theorem of arithmetic.

Answer 1

Hint: First reduce both the numbers as the product of powers of their prime factors. This will be in the form of \[N = {p^a} \times {q^b} \times {r^c} \times ...\]. Then for HCF, take the prime numbers with lowest powers in both the numbers. And for LCM, take the prime numbers with the power that is highest in both the numbers.

Complete step-by-step answer:
According to the question, we have been given two numbers 42 and 72. We have to determine their HCF and LCM using a prime factorization method.
For this, first we will reduce the numbers as the product of powers of their prime factors. The reduced form of 42 is shown below:
$
   \Rightarrow 42 = 6 \times 7 \\
   \Rightarrow 42 = 2 \times 3 \times 7{\text{ }}.....{\text{(1)}}
 $
Similarly for 72, we have:
$
   \Rightarrow 72 = 8 \times 9 \\
   \Rightarrow 72 = {2^3} \times {3^2}{\text{ }}.....{\text{(2)}}
 $
Writing both the numbers together, we have:
$
   \Rightarrow 42 = 2 \times 3 \times 7{\text{ }}.....{\text{(1)}} \\
   \Rightarrow 72 = {2^3} \times {3^2}{\text{ }}.....{\text{(2)}}
 $
Now for HCF, we will take the prime numbers with lowest powers in both the numbers.
So for prime number 2, its power in 42 and 72 are 1 and 3 respectively. We will consider the lowest power i.e. ${2^1}$ will be taken.
For prime number 3, its power in 42 and 72 are 1 and 2 respectively. Again we will consider the lowest power i.e. ${3^1}$ will be taken.
At last, for prime number 7, its power in 42 is 1 and in 72 it is 0 (as 7 is not present so ${7^0}$). We will again consider the lowest power i.e. \[{7^0}\] will be taken.
Therefore the HCF of these numbers will be:
$
   \Rightarrow HCF = {2^1} \times {3^1} \times {7^0} \\
   \Rightarrow HCF = 6
 $
For HCF, we will take the prime numbers with highest powers in both the numbers.
Thus applying the same procedure for the highest power, we can see that the highest powers of 2, 3 and 7 in both the numbers are 3, 2 and 1 respectively (${2^3}$,${3^2}$ and \[{7^1}\] will be taken into account).
Therefore the LCM of these numbers will be:
$
   \Rightarrow LCM = {2^3} \times {3^2} \times {7^1} \\
   \Rightarrow LCM = 8 \times 9 \times 7 \\
   \Rightarrow LCM = 504
 $

Hence the HCF and LCM of these numbers are 6 and 504 respectively.

Note: For two numbers, the product of their HCF and LCM is equal to the product of the numbers. This result is widely used in factors and multiples problems. We can verify it for the above case.
The numbers are 42 and 72 and their HCF and LCM are 6 and 504 respectively. So we have:
$
   \Rightarrow 42 \times 72 = 6 \times 504 \\
   \Rightarrow 3024 = 3024
 $