elementary number theory - Dividing $61!$ by $3$ as we can.

Find maximum possible $n$ in the equation $61!=3^n\cdot m$.

Some textbooks gives a solution to this question like this:
$$\begin{align*} 61&=\textbf{20}\cdot 3+1 \\ 20&=\textbf{6}\cdot 3+2 \\ 6&=\textbf{2}\cdot3+0 \end{align*}$$
$n_{max} = 20+6+2$. But this comes me unintuitive. And seems like it's working. Why this solution is true. Can you give an explanation?

I don't think the answer in Highest power of a prime $p$ dividing $N!$ gives an understanding directly to solution i wrote since the algorithm continues dividing with the last full part.