Solution 63: Minimization with Floor and Ceiling

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  • Published on Oct 27, 2018
  • The discontinuities inherent in floor and ceiling functions make this nontrivial. Let's restrict the values of x with some inequalities to get rid of these pesky functions.
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Comments • 4

  • Sitanshu Chaudhary
    Sitanshu Chaudhary 8 days ago

    Ok please answer my easy question at which value of x ( x+1/x) this gives us 2
    0

  • Ashutosh Anand
    Ashutosh Anand 10 months ago +7

    {x} , [x] and x are in GP. Solve for x. [ x ] is greatest integer function and {x} is fractional function where {x} = x - [x].

    • The Gaming Nut
      The Gaming Nut 4 months ago

      Yup the ans is phi nasir has the solution. I solved it in INMO. Classic q from 2017 check out previous year papers

    • Nasir Khan
      Nasir Khan 10 months ago +1

      A beautiful question friend. If I am right the x= (phi) Golden ratio.
      Since these are in Geometric Progression :
      Squared([x])={x}x
      Substitute {x}=x-[x]
      [x]^2=x^2-x[x]
      Or
      x^2-x[x]-[x]^2=0
      And I hope u know u can complete the square method... You end up with
      x=[x](1+√5)/2
      And there u go a good eye can spot the golden ratio
      x=(phi) [x]
      And golden ratio approximates to 1.68(i hope it's right) anyways it's floor value is 1. Hence only phi satisfies ur problem.... A brilliant question indeed.