Question

The sum of two numbers is 1215 and their HCF is 81. How many pairs of such numbers are possible?

Answer 1

Answer:

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.

Answer 2

Question:- The sum of two numbers is 1215 and their HCF is 81. How many pairs of such numbers are possible?⤵

Answer:-⤵

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.