**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 2’s and four 3’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 three’s = 14

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

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