2023/009) Find n such that $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor + \lfloor \frac{n}{6} \rfloor = n$

We have $\lfloor x \rfloor \le x$ and equal ony of x is integer 

so    $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor  + \lfloor \frac{n}{6} \rfloor \le  \frac{n}{2}  + \frac{n}{3}+ \frac{n}{6} = n$ 

so  $\lfloor \frac{n}{2} \rfloor =  \frac{n}{2}$ 

 $\lfloor \frac{n}{3} \rfloor =  \frac{n}{3}$ 

 $\lfloor \frac{n}{6} \rfloor =  \frac{n}{6}$

the above is so because if any expression above is not true LHS is less that RHS in any of them shall make the sum < n

so  $\frac{n}{2}, \frac{n}{3} ,\frac{n}{6}$ all are ntegers of n is a multiple of 2,3 and 6 that is multiple of LCM(2,3,6) that s 6. so n is of the form 6k