Solve the equation: $\lfloor a/2\rfloor+\lfloor a/3\rfloor+\lfloor a/5\rfloor=a$, where $a> 0$. Let $a\in\mathbb{Z}^+$, then $a$ we can write it as $a = 30k + i$, where $i\in\{0,1,2,\ldots, 29\}$ and $k\in\mathbb{Z}_0^+$. Thus substituting we have:
\begin{align*}
     \lfloor(30k + i)/2\rfloor + \lfloor(30k + i)/3\rfloor + \lfloor(30k + i)/5\rfloor &= 30k + i\\
     \lfloor 15k + (i/2)\rfloor + \lfloor 10k + (i/3)\rfloor + \lfloor 6k + (i/5)\rfloor &= 30k + i\\
     15k + \lfloor i/2\rfloor + 10k + \lfloor i/3\rfloor + 6k + \lfloor i/5\rfloor &= 30k + i\\
     31k + \lfloor i / 2 \rfloor + \lfloor i / 3 \rfloor + \lfloor i / 5 \rfloor & = 30k + i \\
     k &= i - (\lfloor i/2\rfloor + \lfloor i/3\rfloor + \lfloor i/5\rfloor)
\end{align*}
Thus we have for each $i$ we will have a value of $k$. Also note that if $i=0$,
\begin{align*}
     k &= 0 -(\lfloor 0/2\rfloor + \lfloor 0/3\rfloor + \lfloor 0/5 \rfloor) \Rightarrow k=0.
\end{align*}
So $a=0$, but $a>0$. So $i=0$ is not taken, for all others we have a solution.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.

Yes, this is an excellent solution.
It would perhaps be a good idea, after obtaining
\begin{align*}
     k &= i -(\lfloor i/2\rfloor + \lfloor i/3\rfloor + \lfloor i/5 \rfloor) 
\end{align*}
to show that $k$ is non-negative. This is not totally trivial and, since$$\frac{1}{2}+\frac{1}{3}+\frac{1}{5}>1,$$ it is only just true.
