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# problem

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Given that $\displaystyle {{\left((3!)!\right)!}\over{3!}}= k\cdot n!$, where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k+n$.

Jun 25, 2018

#1
+20831
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Given that $\displaystyle {{\left((3!)!\right)!}\over{3!}}= k\cdot n!$, where $k$ and $n$ are positive integers and

$n$ is as large as possible,

find $k+n$.

$$\begin{array}{|rcll|} \hline \displaystyle {{\left((3!)!\right)!}\over{3!}} &=& k\cdot n! \quad & | \quad 3! = 6 \\ \displaystyle {{\left(6!\right)!}\over{6}} &=& k\cdot n! \quad & | \quad 6! = 720 \\ \displaystyle {{720!}\over{6}} &=& k\cdot n! \\ 720! &=& 6k\cdot n! \quad & | \quad 720! = \underbrace{720}_{=6k}\cdot \underbrace{719}_{=n}! \\\\ 6k &=& 720 \quad & | \quad : 6 \\ \mathbf{k} & \mathbf{=} & \mathbf{120} \\ \mathbf{n} & \mathbf{=} & \mathbf{719} \\ \mathbf{k+n} & \mathbf{=}&\mathbf{ 839 } \\\\ \boxed { \dfrac{ \left((3!)!\right)!}{3!} = 120\cdot 719! } \\ \hline \end{array}$$

Jun 26, 2018