Question

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

Answer 1

Sol :- Given,

To prove that 33! is divisible by $2^{15}$

as we know that,

$33!=33\times 32\times 31\times 30\times ---\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$

$33!=33\times (2{)}^5\times 31\times 30\times ----(2{)}^3\times 7\times 6\times 5\times (2{)}^2\times 3(2{)}^1\times 1$

let us now consider all the 2's form the

above configuration, we get

$33!=2^5\times 2^4\times 2^3\times 2^2\times 2^1(33\times 31\times 30\times ---\times 6\times 5\times 3\times 1)$

$33!=2^{5+4+3+2+1}(33\times 31\times 30\times ---\times 6\times 5\times 3\times 1)$

$33!=2^{15}(33\times 31\times 30\times ---\times 6\times 5\times 3\times 1)$ $\to$ eq(1)

$33!=2^{15}(33!)$

$2^{15}=\frac{33!}{33!}=1$

$\therefore$ 33! is divisible by $2^{15}$

Hence Proved $\to$ proved.

Now, let us find the value of $2^n$

having considered all the 2 terms

in the configuration we get the

value of $2^n=2^{16}$

Let us substitute this in eq(1) we get,

$33!=2^{15}{.2}^{16}$

$33!=2^{31}$

we can now say that 31 is the largest

number such that 33! is divisible by $2^n$

Hence Proved