Question

H.C.F. of two numbers is always a factor of their L.C.M. ( True/False) $?$

Answer 1

Hint: Basics of H.C.F and L.C.M. : $\left( 1 \right)$ H.C.F. ( highest common factor ) : H.C.F. is also called the greatest common factor ( G.C.F.) and it is defined as the highest or largest number which divides the given two or more numbers completely means there must be no remainder. $\left( 2 \right)$ L.C.M. ( lowest common multiple ) : It is defined as the lowest or smallest common multiple of the given two or more numbers. There are two methods which are used to find the H.C.F and L.C.M. commonly: $\left( {\text{i}} \right)$ Prime factorization method and $\left( {{\text{ii}}} \right)$ Long division method.

Complete step-by-step solution:
Let us discuss the prime factorization method to find H.C.F. and L.C.M. :
$\left( 1 \right)$ H.C.F. by prime factorization method: Let us understand the concept with an example ;
Find the H.C.F. of $24{\text{ and 12}}$ $?$
The steps are as follows :-
Step $1$ : Find the prime factors of the given numbers.
${\text{The prime factors of 24}} = 2 \times 2 \times 2 \times 3 \times 1 = {2^3} \times 3 \times 1$
${\text{The prime factors of 12}} = 2 \times 2 \times 3 \times 1 = {2^2} \times 3 \times 1$
Step $2$ : List the common factors and the product of least powers of common prime factors gives the H.C.F.
$ \Rightarrow 24 = {2^3} \times {3^1} \times 1$
$ \Rightarrow 12 = {2^2} \times {3^1} \times 1$
Since $2$ is common in both and it’s least power is ${2^2}$ and $3$ is also common with least power as ${3^1}$ ;
H.C.F. $\left( {24,12} \right) \Rightarrow {2^2} \times {3^1} \times 1 = 12$
$\left( 2 \right)$ L.C.M. by prime factorization method: Let us understand the concept with the same example ;
Step $1$ : Find the prime factors of the given numbers.
${\text{The prime factors of 24}} = 2 \times 2 \times 2 \times 3 \times 1 = {2^3} \times 3 \times 1$
${\text{The prime factors of 12}} = 2 \times 2 \times 3 \times 1 = {2^2} \times 3 \times 1$
Step $2$ : List the common factors and the product of highest powers of common prime factors and the remaining prime factors gives the L.C.M.
$ \Rightarrow 24 = {2^3} \times {3^1} \times 1$
$ \Rightarrow 12 = {2^2} \times {3^1} \times 1$
Since $2$ is common in both and it’s highest power is ${2^3}$ and $3$ is also common with highest power as ${3^1}$ ;
L.C.M. $\left( {24,12} \right) \Rightarrow {2^3} \times {3^1} \times 1 = 24$
Now, let us come to our question;
Statement: H.C.F. of two numbers is always a factor of their L.C.M. ( True/False) $?$
Let us take some examples and try to analyze this statement ;
$\left( 1 \right)$ Find L.C.M. and H.C.F. of $\left( {26,91} \right)$ :
Step $1$ : ${\text{The prime factors of 26}} = 2 \times 13 \times 1$
${\text{The prime factors of 91}} = 7 \times 13 \times 1$
Step $2$ : $26 = {2^1} \times {13^1} \times 1$
\[ \Rightarrow 91 = {7^1} \times {13^1} \times 1\]
L.C.M. $ \Rightarrow {13^1} \times 2 \times 7 \times 1 = 182$
H.C.F. $ \Rightarrow {13^1} \times 1 = 13$
Conclusion: Here, we notice that L.C.M. $ = 182$ and H.C.F. $ = 13$ means $182$ is completely divisible by $13$ . Hence we can say that H.C.F. is a factor of the L.C.M.
$\left( 2 \right)$ Find L.C.M. and H.C.F. of $\left( {24,32,54} \right)$ :
Step $1$ : ${\text{The prime factors of 24}} = 2 \times 2 \times 2 \times 3 \times 1 = {2^3} \times {3^1} \times 1$
${\text{The prime factors of 32}} = 2 \times 2 \times 2 \times 2 \times 2 \times 1 = {2^5} \times 1$
${\text{The prime factors of 54}} = 2 \times 3 \times 3 \times 3 \times 1 = {2^1} \times {3^3} \times 1$
Step $2$ : $24 = {2^3} \times {3^1} \times 1$
\[ \Rightarrow 32 = {2^5} \times 1\]
\[ \Rightarrow 54 = {2^1} \times {3^3} \times 1\]
L.C.M. $ \Rightarrow {2^5} \times {3^3} \times 1 = 864$
H.C.F. $ \Rightarrow 2 \times 1 = 2$
Conclusion: Here, we notice that L.C.M. $ = 864$ and H.C.F. $ = 2$ means $864$ is completely divisible by $2$ . Hence we can say that H.C.F. is a factor of the L.C.M.
So, with the help of above examples we have proved that the given statement “H.C.F. of two numbers is always a factor of their L.C.M.” is true.

Note: The important relation between H.C.F. and L.C.M. is : The product of H.C.F. and L.C.M. of two numbers is equivalent to the product of the given two numbers and this relation is very useful while solving questions. Example : For two numbers $\left( {26,91} \right);{\text{ H}}{\text{.C}}{\text{.F}}{\text{. = 13 and L}}{\text{.C}}{\text{.M}}{\text{. = 182 }}$ ; then $26 \times 91 = 13 \times 182 = 2366$ . However, this relation fails in the case of three numbers.