Question

Prove the following property by using the example.
\[LCM\times HCF=\text{Product of two numbers}\]

Answer 1

Hint: We solve this problem by taking the example of two numbers and check the property that is given is correct or not.
We find the LCM and HCF by using the prime factorization method.
If two numbers are represented in the prime factorization method then the LCM is given by taking the product of common primes along with remaining primes and the HCF is given by taking the product of common primes from two numbers.

Complete step by step answer:
We are asked to prove the property
\[LCM\times HCF=\text{Product of two numbers}\]
Let us assume that the two numbers are 12 and 30.
Let us use the prime factorisation method for the number 12
We know that prime factorisation is the method of representing the given number as the product of prime numbers.
Now, by representing the number 12 with first prime number that is 2 we get
\[\Rightarrow 12=2\times 6\]
Here, we can see that the number 6 can be further represented as the product of other primes as follows
\[\Rightarrow 12={{2}^{2}}\times 3\]
Now, let us use the prime factorisation method for 30
By representing the number 30 with the product of first prime number 2 we get
\[\Rightarrow 30=2\times 15\]
Here we can see that the number 15 can be represented further as the product of primes as follows
\[\Rightarrow 30=2\times 3\times 5\]
Now, let us take the prime factorisation for both the numbers
\[\Rightarrow 12={{2}^{2}}\times 3\]
\[\Rightarrow 30=2\times 3\times 5\]
We know that the LCM of two numbers that are represented in the prime factorisation method is given by taking the product common primes along with remaining primes.
By using the above condition we get the LCM of 12 and 30 as
\[\begin{align}
  & \Rightarrow LCM=\left( \text{common primes} \right)\times \left( \text{remaining primes} \right) \\
 & \Rightarrow LCM=\left( 2\times 3 \right)\times \left( 2\times 5 \right) \\
 & \Rightarrow LCM=60 \\
\end{align}\]
We know that the HCF of two numbers that are represented in the prime factorisation method is given by taking the product of common primes.
By using the above condition we get the HCF of 12 and 30 as
\[\begin{align}
  & \Rightarrow HCF=\left( \text{common primes} \right) \\
 & \Rightarrow HCF=2\times 3 \\
 & \Rightarrow HCF=6 \\
\end{align}\]
Now, let us multiply LCM and HCF then we get
\[\begin{align}
  & \Rightarrow LCM\times HCF=60\times 6 \\
 & \Rightarrow LCM\times HCF=360........equation(i) \\
\end{align}\]
Now, let us multiply the numbers 12 and 30 then we get
\[\Rightarrow 12\times 30=360........equation(ii)\]
Now, by comparing equation (i) and equation (ii) we get
\[\begin{align}
  & \Rightarrow LCM\times HCF=12\times 30 \\
 & \Rightarrow LCM\times HCF=\text{Product of two numbers} \\
\end{align}\]
Hence the required result has been proved that is
\[\therefore LCM\times HCF=\text{Product of two numbers}\]

Note:
We can find the LCM and HCF in other methods also.
Let us consider the number 12, 30
Let us find the LCM by using the division method.
We take the numbers 12 and 30 and use the prime numbers that divides in both 12 and 30 and divide them accordingly,
Here we can see that first prime number 2 divides both 12 and 30 then by using the division method we get
 \[\begin{align}
  & 2\left| \!{\underline {\,
  12,30 \,}} \right. \\
 & =6,15 \\
\end{align}\]
Here, the numbers 6, 15 both cannot be divided by 2
By using the next prime number 3 then we get
\[\begin{align}
  & 2\left| \!{\underline {\,
  12,30 \,}} \right. \\
 & 3\left| \!{\underline {\,
  6,15 \,}} \right. \\
 & =2,5 \\
\end{align}\]
Here, we can see that the number 2, 5 are prime numbers so, we can stop the division and take the LCM as the product of all primes that is
\[\begin{align}
  & \Rightarrow LCM=2\times 3\times 2\times 5 \\
 & \Rightarrow LCM=60 \\
\end{align}\]
Now, let us find the HCF of 12 and 30 by using the Euclid’s division.
That is let us divide the large number30 with smaller number that is 12 then we get
\[12\overset{2}{\overline{\left){\begin{align}
  & 30 \\
 & -\left( 24 \right) \\
 & =6 \\
\end{align}}\right.}}\]
Now, let us divide the divisor with the remainder then we get
\[6\overset{2}{\overline{\left){\begin{align}
  & 12 \\
 & -\left( 12 \right) \\
 & =0 \\
\end{align}}\right.}}\]
Here we can see that the remainder is 0 so, we can stop the division and take the divisor as HCF as
\[\Rightarrow HCF=6\]
Now, let us multiply LCM and HCF then we get
\[\begin{align}
  & \Rightarrow LCM\times HCF=60\times 6 \\
 & \Rightarrow LCM\times HCF=360........equation(i) \\
\end{align}\]
Now, let us multiply the numbers 12 and 30 then we get
\[\Rightarrow 12\times 30=360........equation(ii)\]
Now, by comparing equation (i) and equation (ii) we get
\[\begin{align}
  & \Rightarrow LCM\times HCF=12\times 30 \\
 & \Rightarrow LCM\times HCF=\text{Product of two numbers} \\
\end{align}\]
Hence the required result has been proved that is
\[\therefore LCM\times HCF=\text{Product of two numbers}\]