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