Question

If the H.C.F of 336 and 54 is 6 then find the L.C.M of 336 and 54

Answer 1

Hint: We solve this problem by using the prime factorisation method. We use the prime factorisation for both the given numbers then the L.C.M is given by product taking the common primes from both the numbers once and remaining time factors from both numbers. That is for example if the prime factorisation of two numbers is in the form
$$\begin{array}{rl}& \Rightarrow a=2^2\times 3\times 5\\ & \Rightarrow b=2\times 5^2\times 7\end{array}$$
Then the L.C.M is given as
$$\Rightarrow LCM=(2\times 5)\times (2\times 3\times 5\times 7)$$
Here the numbers in the first bracket are common primes and the numbers in the second bracket are remaining primes.

Complete step-by-step solution
We are asked to find the L.C.M of 336 and 54
Let us use the prime factorisation method of 336 that is by using the first prime 2 then we get
$$\Rightarrow 336=2\times 168$$
Now again dividing the above equation furthermore with 2 we get
$$\begin{array}{rl}& \Rightarrow 336=2^2\times 84\\ & \Rightarrow 336=2^3\times 42\\ & \Rightarrow 336=2^4\times 21\end{array}$$
Here we can see that we cannot divide the above equation using the number 2
Now let us go for next prime number 3 then we get
$$\Rightarrow 336=2^4\times 3\times 7$$
Here, we can see that the product of numbers on RHS include only prime numbers
Therefore the prime factorisation of 336 gives
$$\Rightarrow 336=2^4\times 3\times 7$$
Now, let us use the prime factorisation for 54
By dividing the number 54 with first prime number 2 we get
$$\Rightarrow 54=2\times 27$$
Here we can see that we cannot divide the above equation using the number 2
Now let us go for next prime number 3 then we get
$$\begin{array}{rl}& \Rightarrow 54=2\times 3\times 9\\ & \Rightarrow 54=2\times 3^3\end{array}$$
Here, we can see that the product of numbers on RHS include only prime numbers
Therefore the prime factorisation of 54 gives
$$\Rightarrow 54=2\times 3^3$$
We know that the L.C.M is given by product taking the common primes from both the numbers once and remaining time factors from both numbers. That is for example if the prime factorisation of two numbers is in the form
$$\begin{array}{rl}& \Rightarrow a=2^2\times 3\times 5\\ & \Rightarrow b=2\times 5^2\times 7\end{array}$$
Then the L.C.M is given as
$$\Rightarrow LCM=(2\times 5)\times (2\times 3\times 5\times 7)$$
Here the numbers in the first bracket are common primes and the numbers in the second bracket are remaining primes.
By using the above definition we get the LCM of 336 and 54 as
$$\Rightarrow LCM(336,54)=(2\times 3)\times (2^3\times 7\times 3^2)$$
Now, by finding the product of above equation we get
$$\Rightarrow LCM(336,54)=3024$$
Therefore, the LCM of 336 and 54 is 3024.

Note: We have a shortcut for solving this problem.
We are given that HCF of 336 and 54 is 6
For any two numbers $$a,b$$ we have a standard result that is
$$\Rightarrow HCF\times LCM=a\times b$$
By using the above formula we get the LCM of 336 and 54 as
$$\begin{array}{rl}& \Rightarrow 6\times LCM=336\times 54\\ & \Rightarrow LCM={\displaystyle \frac{18144}{6}}\\ & \Rightarrow LCM=3024\end{array}$$
Therefore, the LCM of 336 and 54 is 3024.