Solving the Equation: x+[2x]+[3x]=7

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Discussion Overview

The discussion revolves around solving the equation x + [2x] + [3x] = 7, where participants explore different approaches to classify the values of x and analyze the equation's behavior. The scope includes mathematical reasoning and exploratory analysis of the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant suggests classifying real numbers into different sets based on whether they are integers or non-integers and how they relate to the terms [2x] and [3x].
  • Another participant mentions that Desmos indicates the equation behaves similarly to the line x = 4/3, implying a potential solution.
  • A later reply challenges the previous claim by stating that 4/3 is not a solution, as it results in a left-hand side value of 8, which does not equal the right-hand side of 7.

Areas of Agreement / Disagreement

Participants express differing views on the potential solutions to the equation, with no consensus on the correct approach or solution at this time.

Contextual Notes

Participants have not fully resolved the implications of the classifications proposed, nor have they clarified the conditions under which certain values of x yield integer results for [2x] and [3x].

solakis1
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An easy one:

x+[2x]+[3x]=7
 
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Yes, that is an easy problem! Why did you post it?

I would start by dividing the real numbers into classes:
1) The set of integers.
2) The set of non-integer, x, such that 2x is an integer
3) The set of non-integers, x, such that 2x is not an integer but 3x is.
4) The set of non-integers, x, such that neither 2x nor 3x is an integer but 6x is.
5) The set all other real numbers.
 
Because RHS is integer so LHS is integer

As $\lfloor 2x \rfloor$ and $\lfloor 3x \rfloor$ are integers so x is integer so $\lfloor 2x \rfloor = 2x $ and $\lfloor 3x \rfloor = 3x$

so x + 2x + 3x = 6x = 7 so $x = \frac{7}{6}$ which is not integer so NO solution
 
Beer soaked ramblings follow.
solakis said:
An easy one:

x+[2x]+[3x]=7
Desmos somehow gives the impression that $x + \lfloor 2x \rfloor + \lfloor 3x \rfloor = 7$ is basically the line $x = \frac{4}{3}$.
https://www.desmos.com/calculator/stz3o2wn2h
 
The answer given by Kaliprasad is the right one 7\6 is not an integer
4/3 is not a solution of the above equation LHS is 8 RHS Is 7
 
Last edited:

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