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If x, y, z, w ∈ R satisfy the following equations.

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in Matrices by (51.4k points)

If x, y, z, w ∈ R satisfy the following equations.

x + y + z + w = 1; x + 2y + 4z + 8w = 16, x + 3y + 9z + 27w = 81 and x + 4y + 16z + 64w = 256, then the pairs which have H.C.F as 2 is

a. \(|w|, |z|\)

b. \(|w|, |y|\)

c. \(|y|, |x|\)

d. \(|z|, |x|\)

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1 Answer

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by (51.2k points)

Correct option is c. \(|y|, |x|\)

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