Question

If the HCF of 2 numbers is 43 and sum 430 then total number of distinct pairs of such numbers is

Answer 1

Answer:

5

Step-by-step explanation:

Let H be the HCF of two numbers N1 and N2.

Then, we get, N1 = H×a and N2 = H×b

where a and b are co-primes.

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

From given question, we have H = 43 and N1 + N2 = 430

Thus, 43 × (a+b) = 430

⇒  a+b = 10

So, we have to find all possible (a,b) such that a & b are coprime numbers satisfying a+b=10

(a,b)= (1,9), (2,8), (3,7), (4,6), (5,5)

Thus, we get 5 such possibilities.