Pleasee solve this ,.,.,.,.,.,.

 

It is given that the sum of two numbers is 1215 and their H.C.F is 81.
Let the two numbers be x and y.
Now,
81x+81y=1215     ...[sum of two numbers are 1215]
=>81(x+y)=1215
=>x+y=15
So,
For, x=1,y=14,the numbers are 1×81+14×81=81+1134=1215
For, x=7,y=8,the numbers are 7×81+8×81=567+648=1215
For, x=2,y=13,the numbers are 2×81+13×81=162+1053=1215
For, x=4,y=11,the numbers are 4×81+11×81=324+891=1215
Therefore, the numbers of such pairs are 4.

It is given that the sum of two numbers is?1215?and their H.C.F is?81.
Let the two numbers be?x?and?y.
Now,
81x+81y=1215? ? ?...[sum of two numbers are?1215]
=>81(x+y)=1215
=>x+y=15
So,
For,?x=1,y=14,the numbers are?1?81+14?81=81+1134=1215
For,?x=7,y=8,the numbers are?7?81+8?81=567+648=1215
For,?x=2,y=13,the numbers are?2?81+13?81=162+1053=1215
For,?x=4,y=11,the numbers are?4?81+11?81=324+891=1215
Therefore, the numbers of such pairs are?4.