How to find a number $n$ such that $$\frac{n}{\phi(n)} > 10,$$ where $\phi(n)$ denotes the Euler's phi function?
I was trying to find the smallest one, so was keeping each prime once.
I tried with the number $$n = 2\times3\times5\times7\times11\times13\times17\times19$$ and some more numbers, but not working. Need some help!
Note that $$\phi(n) = n \times \prod_{p} \left(1-\frac1p \right) = n \times \prod_{p} \left(\frac{p-1}{p} \right)$$ hence
$$\frac{n}{\phi(n)} = \prod_{p} \left(\frac{p}{p-1} \right).$$
Best Answer
The solution in the comments, done by several people, is saying that, with $f(n)=\frac{n}{\phi(n)}$ we have $$ f(p_{55}\#)=10.003719732091010383325131639818913973 $$ where $$ p_{55}\#=16516447045902521732188973253623425320896207954043566485360902980990824644545340710198976591011245999110 $$ and that this is the smallest such integer with $f(n)>10$.
More information is found in the following post, by considering $1/f(n)$:
$\frac{\phi(m)}{m}$ is dense in $[0,1]$