What is the smallest n integer number such that 24^n does not divide 50! What is the smallest n integer number such that 24^n does not divide 50! (factorial) ? 
This is what I've done so far, but it seems a bit complicated 


$24=\color\red{2^3}\cdot\color\green{3^1}$

The multiplicity of $\color\red{2}$ in $50!$ is $\sum\limits_{n=1}^{\log_{\color\red{2}}50}\Big\lfloor\frac{50}{\color\red{2}^n}\Big\rfloor=25+12+6+3+1=\color\red{47}$

The multiplicity of $\color\green{3}$ in $50!$ is $\sum\limits_{n=1}^{\log_{\color\green{3}}50}\Big\lfloor\frac{50}{\color\green{3}^n}\Big\rfloor=16+5+1=\color\green{22}$

Therefore:


*

*The maximum value of $n$ such that $(\color\red{2^3})^n$ divides $50!$ is $\Big\lfloor\frac{\color\red{47}}{\color\red{3}}\Big\rfloor=15$

*The maximum value of $n$ such that $(\color\green{3^1})^n$ divides $50!$ is $\Big\lfloor\frac{\color\green{22}}{\color\green{1}}\Big\rfloor=22$



Therefore:


*

*The maximum value of $n$ such that $24^n$ divides $50!$ is $\min(15,22)=15$

*The minimum value of $n$ such that $24^n$ does not divide $50!$ is $15+1=16$

