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Question
the number of ordered triplets(x,y,z) such that (x+y)(y+z)(z+x)=2013, where x,y,z are integers, is
the number of ordered triplets(x,y,z) such that (x+y)(y+z)(z+x)=2013, where x,y,z are integers, is
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Solution
If the question is to find "the number of ordered triplets (x, y, z) such that (x + y) (y + z) (z + x) = 2013, where x, y, z are real numbers", then here is the solution.
Prime factorisation of 2013 = 3 × 11 × 61
We want x + y = 3, y + z = 11 and z + x = 61 on solving three equation we get x = 26.5, y = – 23.5 and z = 34.5
∴ Required triplet is (26.5, – 23.5, 34.5)
If the question is to find "the number of ordered triplets (x, y, z) such that (x + y) (y + z) (z + x) = 2013, where x, y, z are real numbers", then here is the solution.
Prime factorisation of 2013 = 3 × 11 × 61
We want x + y = 3, y + z = 11 and z + x = 61 on solving three equation we get x = 26.5, y = – 23.5 and z = 34.5
∴ Required triplet is (26.5, – 23.5, 34.5)
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