The product of two numbers is 2028 and their HCF is 13. The number of such of pair is = ?

Solution (By Examveda Team)

HCF = 13 (given)
Let number are 13x & 13y respectively
$$\eqalign{ & {\text{Also given}} \cr & 13x \times 13y = 2028 \cr & 13 \times 13 \times xy = 2028 \cr & xy = \frac{{2028}}{{13 \times 13}} = 12 \cr & \therefore {\text{Possible pairs are}} \cr & {\text{ = }}\left( {1,12} \right)\left( {3,4} \right) \cr} $$
Only two pair are possible