The sum of two numbers is 1224 and their HCF is 68. The number of pairs of numbers satisfying the above condition is:

3

The sum of two numbers is 1224 and their HCF is 68 

Let the two numbers be x and y respectively

x + y = 1224

HCF is 68 

So, let x = 68a and y = 68b , ( m + n) then 

68m + 68n = 1224 

⇒ 68 ( m + n ) = 1224 

⇒ ( m + n) = 18

The required numbers are ( 1,17), ( 5,13), ( 7,11)  = 3