Question

The largest integer $n$ such that $33!$ is divisible by $2^n$ is

• A. $31$
• B. $32$
• C. $33$
• D. $34$

Answer 1

The correct option is A $31$

To find the largest integer $n$ which divides $33!$,
First we find the multiples of $2$ which are less than $33!$.
$16$ multiples of $2$ are less than $33!$.
So in $33!$, all these $16$ numbers will contribute one power of $2$.
Similarly, all multiples of $4$ will contribute one extra power each.
Similarly for multiples of $2^3,2^4,2^5$.
Number of Multiples of $2^2$ less than $33!$ are $8$.
Number of Multiples of $2^3$ less than $33!$ are $4$.
Number of Multiples of $2^4$ less than $33!$ are $2$.
Number Multiple of $2^5$ less than $33!$ is $1$.
Next we add all these numbers to get the largetst power of $2$ in $33!$
$=16+8+4+2+1=31$
In terms of notation
Largest power of $2$ in $33!$
$=\left[\frac{33}{2}\right]+\left[\frac{33}{4}\right]+\left[\frac{33}{8}\right]+\left[\frac{33}{16}\right]+\left[\frac{33}{32}\right]$ (where $[.]$ denotes greatest integer function)
$=16+8+4+2+1$
$=31$