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.
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