Question

The number of real solutions of $\left[\frac{x}{2}\right]+\left[\frac{2x}{3}\right]=x$.

Answer 1

First it is clear that x can be only integer$(\because$ LHS will always give a integer value$)\to x=Z$

If we see for $x<0$ then we know that greatest integer function of negative number will always give minimum value $x(\because [x]\le x)$

$\therefore$ min $LHS$ value will be $\frac{x}{2}+\frac{2x}{3}=\frac{7x}{6}\ne x$ for $x<0$

So no solutions for $x<0$

For $x=0$ we have $0=0$

For positive integer values of x we need to check by hit and trail, so we start from $x=1$

Now for $x=2,3,4,5,7$ this equation satisfies but after that LHS always increases more then RHS so LHS>RHS always so no more solutions possible further

$\therefore$ Total real solutions$=x=0,2,3,4,5,7\to 6$ solutions.