Question

Sol. (2) Here, HCF of two numbers ( =13 ) let the numbers be ( 13 x ) and ( 13 y ), where ( x ) and ( y ) are prime to each other. Now, ( 13 x times 13 y=2028 )
( Rightarrow x y=frac{2028}{13 times 13}=12 )
The possible pairs are : (1,13) (3, 4) (2, 6) But the 2 and 6 are not co-prime.
( therefore ) The required number of pairs ( =2 )

# Ex.1. The product of the two ne product of the two numbers is 2028 and eir HCF is 13. The number of such pairs are. (1) 1 (2) 2 (3) 3 (4) 4

Solution