Question

The sum of two numbers is 1212 and their HCF is 101. Find the pairs

Answer 1

Answer:

Step-by-step explanation:

Sum of two numbers = 1212

HCF of the numbers = 101

Let H be the HCF of two numbers such as N1 and N2

Then, N1 = H×a and N2 =H×b

where a and b are coprime numbers

Sum of the two numbers N1+N2 = H×(a+b)

According to the question, we have H= 101 and N1+N2= 1212

Thus, 101×(a+b) = 1212

= a+b=12

Thus, we have to find all possible (a,b) such that a and b are the coprime numbers satisfying a+b = 12

= (a,b)= (6,6)

Thus, there is only one such possibilities, and one pair