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Each of the numbers x ₁, x ₂, . x ₁₀₁ is 1 . What is the smallest positive value of _ 1 i j 101 x_i x_j ?

Correct answer

10

Step-by-step solution

Let S= _ 1 i j 10 x_i x_j We have aligned & (x₁+x₂+x₃+ x₁₀₁ )^2=x₁^2+x₂^2+x₃^2+ . .+x₁₀₁^2+2 S & 2 S= ( _ i=1 ¹⁰¹ x_i )^2- _ i=1 ¹⁰¹ x_i^2 aligned Since we have x_i= 1 , So x_i^2=1 So 2 S= ( _ i=1 ¹⁰¹ x_i )^2-101 Since _ i =1 ¹⁰¹ x _ i will be an integer So ( _ i=1 ¹⁰¹ x_i )^2 will be a perfect square. For smallest positive S: ( _ i=1 ¹⁰¹ x_i )^2 must be the smallest perfect square greater than 101 . So ( _ i=1 ¹⁰¹ x_i )^2=121 _ i =1 ¹⁰¹ x _ i =11 or -11 We can verify that the desired sum can be achieved by putting

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