In this post we will understand how to find possible numbers when the product of two numbers and their HCF is given. This is a type of application question from the chapter of HCF and LCM and is important enough to study for the entrance exams.

In the following post i have tried to explained the concept with the help of problem solving. My only request to the reader is that please don’t memorize the concept, try to understand the reason and concept behind the steps and if you have any doubt feel free to ask in the comment section

One formula that you need to remember for this topic is:**HCF * LCM = Product of two numbers**

## HCF and Product of two numbers

HCF of two numbers is 15

Since 15 is the common factor of the two numbers, the number can be written as 15x and 15y

Now, product of two numbers is 6300

So, 15x * 15y = 6300

=> 225 * x * y = 6300

=> x * y = 28 —-(1)

From eq(1), the possible **coprime values** of x and y are:

(Remember that we have to select number which do not have any common number left between them as we have already taken out common number in the form of HCF)

(x=1, y=28)

(x=4, y= 7)

Hence there are two possible pairs possible

(15 , 15*28) ===> **(15, 420)**

(15*4, 15*7) ===> **(60, 105)**

HCF of two numbers is 12

Since 12 is the common factor of the two numbers, the number can be written as 12x and 12y

Now, product of two numbers is 2160

So, 12x * 12y = 2160

=> 144 * x * y = 2160

=> x * y = 15 —-(1)

From eq(1), the possible **coprime values** of x and y are:

(Remember that we have to select number which do not have any common number left between them as we have already taken out common number in the form of HCF)

(x=1, y=15)

(x=3, y= 5)

Hence there are two possible pairs possible

(12 , 12*15) ===> **(12, 180)**

(12*3, 12*5) ===> **(36, 60)**

The product of two numbers is 1280

HCF of the number is 8

Using HCF-LCM formula

HCF * LCM = Product of two numbers

==> 8 * LCM = 1280

==> LCM = 1280 /8 =160

Hence the** LCM of the two numbers is 160**

The product of two numbers is 20736

HCF of the number is 54

Using HCF-LCM formula

HCF * LCM = Product of two numbers

==> 54 * LCM = 20736

==> LCM = 20736 /54 =384

Hence **384 is the LCM **of the two numbers

Let the three numbers are x, y and z

Since all the numbers are coprime, it means that there is no common factors between all three numbers

According to question

x * y =551 and —– eq(1)

z * y =1073 —– eq (2)

In both eq (1) and eq(2), the number y is common;

So we will do HCF (551, 1073) to find the value of y

Calculating individual factors

551 ==> 29 * 19

1072==> 29 * 37

From the above we can see that 29 is the HCF , hence it is also value of y

The other components are 19 and 37

Sum of numbers are **x + y + z** ==> 19 + 29 + 37 ==> **85**

**(06) The product of two numbers is 4107. The HCF of these number is 37. Find the greater number among the two**

HCF of two number is 37

Hence the first number can be written as 37x

and second number can be written as 37y

The product of two number is 4107

So, 37x * 37y =4107

==> x * y = 4107/1369

==> x * y =3 — eq(1)

From the eq (1), we can figure out that x=1 and y =3

So the numbers will be 37 and 37*3

Among 37 and 111, the greater number is 111

**(07)** **The product of two numbers is 2028. The HCF of these number is 13. Find the** **number of the possible pairs**

The purpose to provide you with solved examples is that you can understand the concept behind the topic. I have seen students who memorize the topic without any understanding and make error in the examination hall. With proper understanding of the concept you will solve question faster and ace any examination possible