Find the maximum power of $24$ in $(48!)^2$? Find the maximum power of $24$ in $(48!)^2$ ?

How to approach for such questions ?

Divide $48$ by $2$ repeatedly and add the quotient omitting any remainder,
$48/2=24, 24/2=12, 12/2=6, 6/2=3, 3/2=1$(with remainder as $1$). Therefore adding the quotients,i.e, $24+12+6+3+1=46$ $\Longrightarrow$ there are $46$ $2$s in $48!$ and $92$ $2$s in $(48!)^2$.
Similarly, divide 48 by 3 repeatedly and add the quotients omitting any remainder,
$48/3=16, 16/3=5$ (with remainder as $2$), $5/3=1$ (with remainder as $2$). Adding these quotients gives , $16+5+1=22$ $\Longrightarrow$ There are $22$ $3$s in $48!$ and hence $44$ $3$s in $(48!)^2$.
Now, the number of $24$s ($2^3\times3$)that can be made from $92$ $2$s and $44$ $3$s is $30$.
