If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs
of such numbers are
(a) 2 (b) 3 (c) 4 (d) 5
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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.
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