# Solution 63: Minimization with Floor and Ceiling

Share
Embed
• 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.
Congratulations to Cobalt314, Jeremy Weissmann, Jaleb, Aswini Banerjee, Seth Harwood, Minh Cong Nguyen, Donovan B, Thomas Q, Bigg Barbarian, and adandap for successfully solving this math challenge question! Cobalt314 was the first person to solve the question.
Your support is truly a huge encouragement.
Every subscriber and every like are wholeheartedly appreciated.

• 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 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 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 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.