The sum of two numbers is 1224 and their HCF is 68. The number of pairs of numbers satisfying the above condition is:
Answer & explanation
Correct answer: option 1
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