Find the Prime Factorization of $\varphi(11!)$

Question

Find the Prime Factorization of $\varphi(11!)$

What I did:

$\varphi(11!)=\varphi(11)\cdot\varphi(10)...\varphi(1)$

$$\varphi(11)=2\cdot 5\\ \varphi(10)=2^2\\ \varphi(9)=3\cdot 2\\ \varphi(8)=2^2\\ \varphi(7)=2\cdot 3\\ \varphi(6)=2\\ \varphi(5)=2^2\\ \varphi(4)=2\\ \varphi(3)=2\\ $$

$$\Longrightarrow\text{ the answer is } 2^{12}\cdot 5\cdot 3^2$$

$11!$

$\varphi(11!)=\varphi(39916800)=8294400$

$8294400=2^{12}3^45^2$

Where am I wrong?

Is there a simpler way to solve this?

Answer 1

\begin{align} & \bigg\lfloor \frac{11}{2} \bigg\rfloor+\bigg\lfloor \frac{11}{4} \bigg\rfloor+\bigg\lfloor \frac{11}{8} \bigg\rfloor=8 \\ & \bigg\lfloor \frac{11}{3} \bigg\rfloor+\bigg\lfloor \frac{11}{9} \bigg\rfloor=4 \\ & \bigg\lfloor \frac{11}{5} \bigg\rfloor=2 \\ & \bigg\lfloor \frac{11}{7} \bigg\rfloor=1 \\ &\bigg\lfloor \frac{11}{11} \bigg\rfloor=1 \\ \end{align} $$\varphi(11!)=\varphi ({{2}^{8}}\times {{3}^{4}}\times {{5}^{2}}\times {{7}^{1}}\times {{11}^{1}})=\varphi ({{2}^{8}})\varphi ({{3}^{4}})\varphi ({{5}^{2}})\varphi ({{7}^{1}})\varphi ({{11}^{1}})$$ Note if $P$ be prime then $$\varphi(P^n)=P^n-P^{n-1}$$

Answer 2

While it is true that you should exploit the fact that $\varphi$ is multiplicative, what you do is not quite correct.

Recall that $\varphi(nm)= \varphi(n) \varphi(m)$ needs that $m$ and $n$ are co-prime.

Thus, first write $11!$ as a product of distinct prime powers. Only, then do what you did.

Answer 3

$\varphi(11!)=(11!) (1-\frac{1}{2})(1-\frac{1}{3}) (1-\frac{1}{5}) (1-\frac{1}{7}) (1-\frac{1}{11})$. I do not really know if this will help.