Question

Find the LCM of the following numbers (a)$9$ and $4$ (b)$12$ and $5$(c)$6$ and $5$(d)$15$ and $4$. Observe a common property in the obtained L.C.Ms. Is L.C.M the product of two numbers in each case?

Answer 1

Hint: We are given a few sets of numbers and we need to find the LCM of the numbers. We can find the LCM by using the prime factorization method. We need to factorize the given set of numbers using prime numbers and the LCM is given by the product of the prime numbers used for factorization. Then we need to check whether the LCM is equal to the product of the numbers.

Complete step-by-step answer:
We are given a few sets of numbers and asked to find the LCM of the numbers and whether there exists a common property between them.
Let’s start with the first set of numbers

(a)$9$ and $4$
The given two numbers are $9$ and $4$
In order to find the LCM, we need to factorize the numbers together using prime numbers.
It’s clear that both the numbers don’t have any common prime factors
So, let's factorize the numbers separately
Since $4$is an even number, it is divisible by $2$
So, let’s start our factorization using $2$ as it is the smallest positive prime number
$
  2\left| \!{\underline {\,
  {9,4} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {9,2} \,}} \right. \\
 $
Still, we have $2$, so we can once again factorize using the number $2$
$
  2\left| \!{\underline {\,
  {9,4} \,}} \right. \\
  2\left| \!{\underline {\,
  {9,2} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {9,1} \,}} \right. \\
 $
Now we know that the next prime number is $3$ and it is very well known that $9$ is divisible $3$. We need to factorize until we get $1$
$
  2\left| \!{\underline {\,
  {9,4} \,}} \right. \\
  2\left| \!{\underline {\,
  {9,2} \,}} \right. \\
  3\left| \!{\underline {\,
  {9,1} \,}} \right. \\
  3\left| \!{\underline {\,
  {3,1} \,}} \right. \\
  {\text{ }}1,1 \\
 $
Now the LCM is found by multiplying all the prime numbers used to factorize these two numbers
$LCM = 2*2*3*3 = 4*9 = 36$
The product of the two numbers is $9*4 = 36$
We can see that in this case, the LCM is equal to the product of the two numbers.

(b)$12$ and $5$
The given two numbers are $12$ and $5$
In order to find the LCM, we need to factorize the numbers together using prime numbers.
It’s clear that both the numbers don’t have any common prime factors
So, let's factorize the numbers separately
Since $12$is an even number, it is divisible by $2$
So, let’s start our factorization using $2$ as it is the smallest positive prime number
$
  2\left| \!{\underline {\,
  {12,5} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {6,5} \,}} \right. \\
 $
Still, we have $6$, so we can once again factorize using the number $2$
\[
  2\left| \!{\underline {\,
  {12,5} \,}} \right. \\
  2\left| \!{\underline {\,
  {6,5} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {3,5} \,}} \right. \\
 \]
Now we know that the next prime number is $3$. We need to factorize until we get $1$
$
  2\left| \!{\underline {\,
  {12,5} \,}} \right. \\
  2\left| \!{\underline {\,
  {6,5} \,}} \right. \\
  3\left| \!{\underline {\,
  {3,5} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {1,5} \,}} \right. \\
 $
Now we know that the next prime number is $5$. We need to factorize until we get $1$
$
  2\left| \!{\underline {\,
  {12,5} \,}} \right. \\
  2\left| \!{\underline {\,
  {6,5} \,}} \right. \\
  3\left| \!{\underline {\,
  {3,5} \,}} \right. \\
  5\left| \!{\underline {\,
  {1,5} \,}} \right. \\
  {\text{ }}1,1 \\
 $
Now the LCM is found by multiplying all the prime numbers used to factorize these two numbers
$LCM = 2*2*3*5 = 4*15 = 60$
The product of the two numbers is $12*5 = 60$
We can see that in this case, the LCM is equal to the product of the two numbers.

 (c)$6$ and $5$
The given two numbers are $6$ and $5$
In order to find the LCM, we need to factorize the numbers together using prime numbers.
It’s clear that both the numbers don’t have any common prime factors
So, let's factorize the numbers separately
Since $6$is an even number, it is divisible by $2$
So, let’s start our factorization using $2$ as it is the smallest positive prime number
$
  2\left| \!{\underline {\,
  {6,5} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {3,5} \,}} \right. \\
 $
Now we know that the next prime number is $3$ . We need to factorize until we get $1$
$
  2\left| \!{\underline {\,
  {6,5} \,}} \right. \\
  3\left| \!{\underline {\,
  {3,5} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {1,5} \,}} \right. \\
 $
Now we know that the next prime number is $5$ . We need to factorize until we get $1$
$
  2\left| \!{\underline {\,
  {6,5} \,}} \right. \\
  3\left| \!{\underline {\,
  {3,5} \,}} \right. \\
  5\left| \!{\underline {\,
  {1,5} \,}} \right. \\
  {\text{ 1,1}} \\
 $
Now the LCM is found by multiplying all the prime numbers used to factorize these two numbers
$LCM = 2*3*5 = 6*5 = 30$
The product of the two numbers is $6*5 = 30$
We can see that in this case, the LCM is equal to the product of the two numbers.

 (d)$15$ and $4$.
The given two numbers are $15$ and $4$
In order to find the LCM, we need to factorize the numbers together using prime numbers.
It’s clear that both the numbers don’t have any common prime factors
So, let's factorize the numbers separately
Since $4$is an even number, it is divisible by $2$
So, let’s start our factorization using $2$ as it is the smallest positive prime number
$
  2\left| \!{\underline {\,
  {15,4} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {15,2} \,}} \right. \\
 $
Still, we have $2$, so we can once again factorize using the number $2$
$
  2\left| \!{\underline {\,
  {15,4} \,}} \right. \\
  2\left| \!{\underline {\,
  {15,2} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {15,1} \,}} \right. \\
 $
Now we know that the next prime number is $3$ and it is very well known that $15$ is divisible $3$. We need to factorize until we get $1$
$
  2\left| \!{\underline {\,
  {15,4} \,}} \right. \\
  2\left| \!{\underline {\,
  {15,2} \,}} \right. \\
  3\left| \!{\underline {\,
  {15,1} \,}} \right. \\
  {\text{ }}\left| \!{\underline {\,
  {5,1} \,}} \right. \\
  {\text{ }} \\
 $
Now we know that the next prime number is $5$ . We need to factorize until we get $1$
$
  2\left| \!{\underline {\,
  {15,4} \,}} \right. \\
  2\left| \!{\underline {\,
  {15,2} \,}} \right. \\
  3\left| \!{\underline {\,
  {15,1} \,}} \right. \\
  5\left| \!{\underline {\,
  {5,1} \,}} \right. \\
  {\text{ 1,1}} \\
  {\text{ }} \\
 $
Now the LCM is found by multiplying all the prime numbers used to factorize these two numbers
$LCM = 2*2*3*5 = 4*15 = 60$
The product of the two numbers is $15*4 = 60$
We can see that in this case, the LCM is equal to the product of the two numbers.

Note: Generally, factorization can be done with composite numbers also but to find LCM prime factorization is always preferred. Herein all the problems the LCM was equal to the product of the two numbers, but it is not the case with all the numbers. This condition is satisfied only when the numbers don’t have any common prime factors.