Explain me, please.
find all positive integers such that $\phi (n) = 6$

Question

Explain me, please. 
find all positive integers such that $\phi (n) = 6$

imqwerty · Answer

ϕ(n)=totient function?

loser66 · Answer

yes, this is what my prof said in class, but don't get the reason why

loser66 · Answer

If $p\geq 11$ we cannot have p|n , why?
since then $p_i -1 =10 \cancel |\phi (n)$ why do we need $p_i -1 |\phi (n)$?

imqwerty · Answer

$$n=p_1^{a_1} \times p_2^{a_2} \times p_3^{a_3}...$$p_i are all distinct primes and a_i>=1
$$\phi(n)=\phi(p_1^{a_1}) \times \phi(p_2^{a_2}) \times \phi(p_3^{a_3}).....$$

for any p>=11
phi(p) will be >6  and multiplying it with other terms will give phi(n)>6 
so we can pay p cannot be >=11

loser66 · Answer

$a_i$ can be 0

imqwerty · Answer

no i am assuming that $$n=p_1^{a_1} \times p_2^{a_2} \times p_3^{a_3}...$$^this is actual prime factorization of the number n 
so a_i can't be 0

loser66 · Answer

Got you. Thanks for it. I will work on it. If I don't understand somewhere, I will ask you then. :)

imqwerty · Answer

ok :) lol i too hav to wrk out the nxt step