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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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    See https://math.stackexchange.com/questions/141196/highest-power-of-a-prime-p-dividing-n – lab bhattacharjee Jul 02 '17 at 11:32
  • I actually took a look but i couldn't see that. Seems like it gives an explanation, but i don't know the notation he used in answer. @SahibaArora –  Jul 02 '17 at 11:40
  • @AbdullahUYU There is a link in the second answer which tells you what the notation is. If you have already gone through the same question on the site you didn't understand then your question should have provided a link to that and asked for the specific doubt. Or you could have just commented on the answer there to ask about the notation. – Sahiba Arora Jul 02 '17 at 11:44
  • Thanks i got it. –  Jul 02 '17 at 11:51

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$$61!=\\ 1.2.\underbrace{3}_{3^1K}.4.5.\underbrace{6}_{3^1K}.7.8.\underbrace{9}_{3^2K}..\underbrace{12}_{3^1K}...\underbrace{15}_{3^1K}...\underbrace{18}_{3^2K}...\underbrace{21}_{3^1K}...\underbrace{24}_{3^1K}...\underbrace{27}_{3^3K}...$$ SO YOU CAN FIND $$\lfloor\frac{n}{3}\rfloor+\lfloor\frac{n}{9}\rfloor+\lfloor\frac{n}{27}\rfloor+\lfloor\frac{n}{81}\rfloor+\cdots$$ in this case $$\lfloor\frac{61}{3}\rfloor+\lfloor\frac{61}{9}\rfloor+\lfloor\frac{61}{27}\rfloor+\lfloor\frac{61}{81}\rfloor=\\20+6+2=28$$so max $ k$ in $$3^k|61!$$is $28$

Prakhar Dwivedi
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Khosrotash
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