Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
When m and n are positive integers, is m!*n! an integer squared?
1) m = n + 1
2) m is an integer squared
\[m,n\,\, \geqslant \,\,1\,\,\,{\text{ints}}\,\,\,\left( * \right)\]
\[m!\,\,n!\,\,\,\mathop = \limits^? \,\,{K^{\,2}}\,\,\,,\,\,\,K\mathop \geqslant \limits^{\left( * \right)} 1\,\,\,\operatorname{int} \]
\[\left( 1 \right)\,\,\,m = n + 1\,\,\,\,\,\, \Rightarrow \,\,\,\,m! = \left( {n + 1} \right)n!\,\,\,\,\,\, \Rightarrow \,\,\,\,m! \,n! = \left( {n + 1} \right){\left( {n!} \right)^2}\]
\[\left\{ \begin{gathered}
\,{\text{Take}}\,\,n = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\left[ {\,2 \cdot 1\,\,{\text{is}}\,\,{\text{not}}\,\,{\text{a}}\,\,{\text{perfect}}\,\,{\text{square}}\,} \right]\,\, \hfill \\
\,{\text{Take}}\,\,n = 3\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\left[ {\,4 \cdot {6^2}\,\, = \,\,{{12}^2}\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{perfect}}\,\,{\text{square}}\,} \right]\,\, \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,m = {J^2}\,\,\,,\,\,\,J\mathop \geqslant \limits^{\left( * \right)} 1\,\,\,\operatorname{int} \]
\[\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {m,n} \right) = \left( {1,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,n + 1 = {J^2}\,\,\, \Rightarrow \,\,\,\,m!\,n! = {J^2} \cdot {\left( {n!} \right)^2}\,\, = {\left( {J \cdot n!} \right)^2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\left[ {\,K = \,J \cdot n!\,} \right]\,\,\,\,\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
