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Q.
If $(201) !$ is divided by $24^{k}$ then the largest value of $k$ is
Permutations and Combinations
Solution:
In $201 !$ number of $2 ' s=100+50+25+12+6+3+1$
$=197$
$\therefore $ number of $8 's =\left[\frac{197}{3}\right]=65$
number of $3 's$ $=67+22+7+2=98$
$\therefore $ number of $24 's =65$