First slide

Permutations

Question

Let y be an element of the set A = {1, 2, 3, 5, 6, 10, 15, 30} and x₁, x₂, x₃ be integers such that x₁x₂x₃ = y, then the number of positive integral solutions of x₁x₂x₃ = y is

Moderate

A

64
B

27
C

81
D

None of these

Solution

Number of solutions of the given equation is the same as the number of solutions of the equation
$x_{1} x_{2} x_{3} x_{4} = 30 = 2 \times 3 \times 5$
Here, x₄ is there because if x₁x₂x₃ = 15, then x₄ = 2 and if x₁x₂x₃ = 5, then x₄ = 6, etc.
x₄ is in fact a dummy variable.
Each of 2, 3 and 5 will be a factor of exactly one of x₁, x₂, x₃, x₄ in 4 ways.
∴ Required number = 4³ = 64

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