Question

Prove that $33!$ is divisible by ${{2}^{15}}$. What is the largest integer n such that $33!$ is divisible by ${{2}^{n}}$ ?

Answer 1

Hint: One can prove ${{2}^{15}}$ divides $33!$ by expanding $33!$ and then separating factors of the for ${{2}^{l}}$ such as 2, 4, 8, 16, 32 and then multiply them. For ${{2}^{nd}}$ part we can use formula
$\left[ \dfrac{n}{P} \right]+\left[ \dfrac{n}{{{P}^{2}}} \right]+\left[ \dfrac{n}{{{P}^{3}}} \right]+\left[ \dfrac{n}{{{P}^{4}}} \right]+......$
Where n is $\left( n \right)!$ and P is the prime number whose highest power is to be found out.

Complete step-by-step answer:

In the question we have to prove that $33!$ is divisible by ${{2}^{15}}$.
Now, we will try to expand $33!$ using the formula $n!=n\times \left( n-1 \right)\times \left( n-2 \right)....\times 3\times 2\times 1$.
So, we get
$33!=33\times \left( 32 \right)\times \left( 31 \right)\times \left( 30 \right)....\times 3\times 2\times 1$
Let’s consider the factor which can be purely written in 2’s form which is,
$33!=33\times \left( 32 \right)\times 31\times 30....\times \left( 16 \right)\times .....\times \left( 8 \right)\times ....\times \left( 4 \right)\times 3\times 2\times 1$
Which can written as,
$33!=33\times {{2}^{6}}\times 31\times 30....\times {{2}^{4}}\times .....\times {{2}^{3}}\times ....\times {{2}^{2}}\times 3\times {{2}^{1}}\times 1$
Hence, we can write it as
$33!={{2}^{5}}\times {{2}^{4}}\times {{2}^{3}}\times {{2}^{2}}\times {{2}^{1}}\times \left( 33\times 31\times 30\times .....\times 17\times 15\times ....\times 9\times 7....\times 3\times 1 \right)$
So,
$33!={{2}^{15}}\times \left( 33\times 31\times 30\times .....\times 17\times .....\times 3\times 1 \right)$
So we can say that $33!$ factorial can be written as ${{2}^{15}}\times $ (any integer value).
So $33!$ is divisible by ${{2}^{15}}$.
If there is a number given $'n!'$ in which it contains prime factor P o we can find the coefficient of power of ‘P’ using formula,
$\left[ \dfrac{n}{P} \right]+\left[ \dfrac{n}{{{P}^{2}}} \right]+\left[ \dfrac{n}{{{P}^{3}}} \right]+.....$
Here $\left[ x \right]$ means the integral value of number x.
So, the coefficient of power of number 2 contain in $33!$ is,
$\left[ \dfrac{33}{2} \right]+\left[ \dfrac{33}{{{2}^{2}}} \right]+\left[ \dfrac{33}{{{2}^{3}}} \right]+\left[ \dfrac{33}{{{2}^{4}}} \right]+\left[ \dfrac{33}{{{2}^{5}}} \right]$
No more terms will be there as only till ${{2}^{5}}=32<33$ after that all the terms will be ‘0’.
So, the value is
$\left[ 16.5 \right]+\left[ 8.25 \right]+\left[ 4.125 \right]+\left[ 2.0625 \right]+\left[ 1.03125 \right]$
Which is equal to,
\[16+8+4+2+1=31\]
So, we can say that \[{{2}^{31}}\] is the maximum number in the form of \[{{2}^{l}}\] can divide \[33!\].
The value of n is 31.

Note: Instead of doing the first part they can do directly ${{2}^{nd}}$ part using the formula and from that they can directly say without proving as ${{2}^{15}}$ divides ${{2}^{31}}$.