How many natural numbers $x\leqslant 21 !$ there are such that $\gcd(x,20!)=1$

Question

How many natural numbers $x\leqslant 21 !$ there are such that $\gcd(x,20!)=1$

Attempt:

I used this methode and I have found that:

$21!$$=2^{18}\times3^{9}\times5^{4}\times7^{3}\times11\times13\times17\times19$

$20!$$=2^{18}\times3^{\color{red}8}\times5^{4}\times7^{\color{red}2}\times11\times13\times17\times19$

It's easy to see that the prime factorization contains the same prime numbers, but how can I know how many numbers $x\leqslant 21 !$ there are such that $\gcd(x,20!)=1$

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If I look at $3$ and $3^2$, so there are $4$ numbers between $3$ and $9$ such that $\gcd(3,x)=1$ $3,\color{blue}4,\color{blue}5,6,\color{blue}7,\color{blue}8,9$

Answer 1

You can easily notice that for a natural number $n$, gcd$(n,20!)=1$ if and only if gcd$(n,21!)=1$. Thus you can ask the question as follows :

How many natural numbers $x\leq 21!$ such that gcd($x,21!)=1$. Then the answer is $\varphi(21!)$ which is easy to calculate.