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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 =3×3x2x2x2x7x2×3x5x2×2x3x2x1
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 22 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 22 x3
So the maximum power of 12 that can divide the 32! is 14

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