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

We know that,
$33!=33\times 32\times 31\times \dots ..\times 8\times 7\times 6\dots ..2\times 1$
$33!=33\times (2{)}^5\times 31\times 30\times \dots .(2{)}^3\times 7\times 6\times 5\times (2{)}^2\times 3(2{)}^1\times 1$
Let us now consider all the 2’s form the above configuration, we get
$33!=2^5\times 2^4\times 2^3\times 2^2(33\times 30\dots \dots 1)$
$33!=2^{15}(33\times 31\times \dots \dots \times 1)\dots ..\left(1\right)$
$33!=2^{15}(33!)$
$2^{15}=\frac{33!}{33!}=1$
∴33! Is divisible by $2^{15}$
Now, let us find the value of $2^n$ having considered all the 2 terms in the configuration we get the value of $2^n=2^{16}$
So, from (1) we get,
$33!=2^{15}{.2}^{16}$
$33!=2^{31}$
We can say that 31 is the largest number such that 33! Is divisible by $2^n$
Hence proved.