number theory

**deleted**

What is the largest integer $n$ such that $3^n$ is a factor of

$1 * 3 * 5 * \ldots * 97 * 99 * 101 * 103 *\ldots * 197 * 199$?

$\begin{array}{|rcll|} \hline && 1*3*5*7*9*11 \ldots *195*197*199 \\\\ &=& \dfrac{ 1*2*3*4*5* \ldots *196*197*198*199} {2*4*6*8*10* \ldots *194*196*198 } \\\\ &=& \frac{ 1*2*3*4*5* \ldots *196*197*198*199} {(2*1)*(2*2)*(2*3)*(2*4)*(2*5)* \ldots *(2*97)*(2*98)*(2*99) } \\\\ &=& \dfrac{ 1*2*3*4*5* \ldots *196*197*198*199} {2^{99}*1*2*3*4*5* \ldots *97*98*99 } \\\\ &=& \dfrac{ 199!} {2^{99}*99! } \\\\ && \boxed{ 199! = \ldots * \left.3^{ \lfloor\frac{199}{3}\rfloor + \lfloor\frac{199}{9}\rfloor + \lfloor\frac{199}{27}\rfloor + \lfloor\frac{199}{81}\rfloor } \right. * \ldots \\ 199! = \ldots * \left.3^{66 +22 +7 +2 }\right. * \ldots \\ 199! = \ldots * \left.3^{97}\right. * \ldots \\\\ 99! = \ldots * \left.3^{ \lfloor\frac{99}{3}\rfloor + \lfloor\frac{99}{9}\rfloor + \lfloor\frac{99}{27}\rfloor + \lfloor\frac{99}{81}\rfloor } \right. * \ldots \\ 99! = \ldots * \left.3^{33 +11 +3+1 }\right. * \ldots \\ 99! = \ldots * \left.3^{48}\right. * \ldots \\ } \\\\ &=& \dfrac{ \ldots * \left.3^{97}\right. * \ldots} {2^{99}*\ldots * \left.3^{48}\right. * \ldots } \\\\ &=& \dfrac{ \ldots * \left.3^{97-48}\right. * \ldots} {2^{99}*\ldots} \\\\ &=& \dfrac{ \ldots * \left.3^{\color{red}49}\right. * \ldots} {2^{99}*\ldots} \\ \hline \end{array}$

$\mathbf{n=49}$