TITLE: Find the maximum power of $24$ in $(48!)^2$?
QUESTION [9 upvotes]: Find the maximum power of $24$ in $(48!)^2$ ?

How to approach for such questions ?

REPLY [4 votes]: $24=\color\red{2^3}\cdot\color\green{3^1}$

The multiplicity of $\color\red{2}$ in $48!$ is $\sum\limits_{n=1}^{\log_{\color\red{2}}48}\Big\lfloor\frac{48}{\color\red{2}^n}\Big\rfloor=24+12+6+3+1=\color\red{46}$

The multiplicity of $\color\green{3}$ in $48!$ is $\sum\limits_{n=1}^{\log_{\color\green{3}}48}\Big\lfloor\frac{48}{\color\green{3}^n}\Big\rfloor=16+5+1=\color\green{22}$

Therefore:

The maximum value of $n$ such that $(\color\red{2^3})^n$ divides $48!$ is $\Big\lfloor\frac{\color\red{46}}{\color\red{3}}\Big\rfloor=\color\red{15}$
The maximum value of $n$ such that $(\color\green{3^1})^n$ divides $48!$ is $\Big\lfloor\frac{\color\green{22}}{\color\green{1}}\Big\rfloor=\color\green{22}$


Therefore:

The maximum value of $n$ such that $(\color\red{2^3})^n$ divides $(48!)^2$ is $\Big\lfloor\frac{\color\red{46}\cdot2}{\color\red{3}}\Big\rfloor=\color\red{30}$
The maximum value of $n$ such that $(\color\green{3^1})^n$ divides $(48!)^2$ is $\Big\lfloor\frac{\color\green{22}\cdot2}{\color\green{1}}\Big\rfloor=\color\green{44}$


Therefore, the maximum value of $n$ such that $24$ divides $(48!)^2$ is $\min(\color\red{30},\color\green{44})=30$.