Question

What is the highest power of $6$ that divides $9!$

Answer 1

By the normal method $\frac{9}{6}=1and\frac{9}{6^2}=0$ Thus answers we get is $1$ which is wrong True there is just one multiple of $6$ from $1$ to $9$ but the product $2\times 3=6$ and also $4\times 9=36$ can further be divided by $6$ Thus when the divisor is a composite number find the highest power of its prime factors and then proceed In this case $9!$ can be divided by $2^7$ and $3^4$ and thus by $6^4$ (In this case we need not have checked power of $2$ as it would definitely be greater then that of $3$)