Question

I sum of two munhers is 1215 and their HCF is 81, then the possible member of pairs of such numbers are (a) 2) 3 (c) 4 (d) 5 ​

Answer 1

Correct option is
C
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.

Answer 2

Answer:

______

Therefore, the numbers of such pairs are 4 .

Step-by-step explanation:

____________________

The answer is 4