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

[solakis1]

An easy one:

x+[2x]+[3x]=7

 

[HOI]

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.

 

[kaliprasad]

Spoiler: My Solution

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

 

[jonah1]

Beer soaked ramblings follow.

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

 

[solakis1]

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