A. 3 B. 4 C. 2 D. 1
The answer is c)2
Given:
The product of two numbers is 2028.
i.e. XY=2028
The prime factorisation of 2028 are 2,2,3,13,13.
The hcf of two number is 13.
X=13(A) Y=13(B)
From the prime factorisation ,the combinations formed are
1)X=13(1) Y=13(2*2*3)
2)X=13(2) Y=13(2*3)
3)X=13(2*2) Y=13(3)
1) AND 3) have the HCF 13.2)have HCF 26.
Hence there are totally 2 pairs available.(13,2028),(52,39)