# 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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Ashutosh Anand7 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 Nut24 days agoYup the ans is phi nasir has the solution. I solved it in INMO. Classic q from 2017 check out previous year papers

Nasir Khan7 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.