Question

The sum of a pair of positive integers is 336 and their h.c.f. is 21. the number of such possible pairs is

Answer 1

HCF of two numbers = 21
Hence the numbers can be 21*a , 21* b

Given sum = 336
21a + 21b = 336
21(a+b) = 336
a+b = 16
Possibilities: (a,b) == (9,7) , (8,8) 
So the numbers can be:
[21*9 , 21*7 ]&& [21*8 , 21*8]
But [21*8 , 21*8] isn't possible because it doesn't satisfy the HCF condition which says that HCF is only 21.In this case 21*8 has to be the HCF.

So there is only 1 possible pair [21*9 , 21*7]
Hope it helps.