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