Question

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

Answer 1

Let 13a & 13b be the two numbers

13a*13b =2028

ab =12

(1,12) & (3,4) are the co prime no with product 12

Hence required numbers are (13*1, 13*12) & (13*3 , 13*4) So there are 2 such pairs

Answer 2

Step-by-step explanation:

which is the required solution