**Type : Highest power of p which divides the q! ,where p is not a prime number**

The approach for this type is same as that for calculating maximum power of prime in any factorial buthere first we will break p into product of primes.

Lets take an example to understand this

__Example 1__

**What will be the maximum power of 6 that divides the 9!**

In order to find maximum power of 6 we will first write as product of 2 and 3.

Maximum power of 2 in 9!

$ \displaystyle \begin{array}{l}\left[ \frac{9}{2} \right]+\left[ \frac{9}{{{2}^{2}}} \right]+\left[ \frac{9}{{{2}^{3}}} \right]+…..\\\begin{array}{*{35}{l}}

=\text{ }4\text{ }+\text{ }2\text{ }+\text{ }1\text{ }+\text{ }0 \\

=7 \\

Maximum\text{ }power\text{ }of\text{ }3\text{ }in\text{ }9! \\

\end{array}\\\left[ \frac{9}{3} \right]+\left[ \frac{9}{{{3}^{2}}} \right]+…..\\\begin{array}{*{35}{l}}

=\text{ }3+\text{ }1+0 \\

=4 \\

\end{array}\end{array}$

And there are seven 2â€™s and four 3â€™s are there, so we need to make the pairs of 2 and 3.

Now we know that 6 = 2 x3. So we need equal number of 2 and equal number of 3 in 9!

Hence there are maximum four pairs that can make 6.So the maximum power of 6 that

can divide 9! is 4.

**Other method**

Since the number 9! is not very big number in-fact we can write and check maximum power of 3

9! = 9x8x7x6x5x4x3x2x1 =**3×3**x**2x2x2**x7x**2×3**x5x**2×2**x**3**x**2**x1

So there are four pairs of 2 x 3, which can be formed

So the maximum power of 6 that can divide the 9! is 4 .

__Example 2__

**What will be the highest power of 12 that can exactly divide 32!**

We can write 12 = 2x2x3 i.e. we need pair of 2^{2} x 3

Maximum power of 2 in 32!

$ \displaystyle \begin{array}{l}\left[ \frac{32}{2} \right]+\left[ \frac{32}{{{2}^{2}}} \right]+\left[ \frac{32}{{{2}^{3}}} \right]+\left[ \frac{32}{{{2}^{4}}} \right]+\left[ \frac{32}{{{2}^{5}}} \right]+…..\\\begin{array}{*{35}{l}}

=\text{ }16\text{ }+\text{ }8\text{ }+\text{ }4\text{ }+\text{ }2\text{ }+1 \\

=31 \\

Maximum\text{ }power\text{ }of\text{ }3\text{ }in\text{ }32! \\

\end{array}\\\left[ \frac{32}{3} \right]+\left[ \frac{32}{{{3}^{2}}} \right]+\left[ \frac{32}{{{3}^{3}}} \right]+…..\\\begin{array}{*{35}{l}}

=\text{ }10+\text{ }3+\text{ }1 \\

=\text{ }14 \\

\end{array}\end{array}$

Now in 32! Number of 2 are 31

And the number of threeâ€™s = 14

So therefore if we take 14 threeâ€™s then we need 28 twoâ€™s because for each three we need two 2â€™s so therefore we need 28 twoâ€™s from 31 twoâ€™s to make the equal pairs of 2^{2} x3

So the maximum power of 12 that can divide the 32! is 14