I couldn't figure out how to solve this one: $\phi(2^m\cdot3^n\cdot6^p)$
I'm assuming there is a connection between the the three numbers since $6=2\cdot3$,$\space$ and $2$ and $3$ are prime numbers.
I was thinking, maybe: $$\phi(n)=\Big(2^m\cdot3^n\cdot6^p\Big)\cdot\Bigr(1-\frac12\Bigr)\cdot\Bigr(1-\frac13\Bigr)$$ But it doesn't goes anywhere..
Indeed, this is true. For this, you may simply look at the formula : $$ N = \prod_i p_i^{k_i} \implies \phi(n) = N \prod_i\frac{p_i-1}{p_i} $$
(This is an integer since each $p_i$ is a divisor of $n$) where $\{p_i\}$ is the set of prime numbers which are divisors of $N$, and $k_i \geq 1$ are then decided by uniqueness of prime factorization.
Now, simply note that if any two of $m,n,p \geq 1$ and the other is non-negative, we have that the prime factors of $N = 2^m3^n6^p$ is exactly the set $\{2,3\}$(if the conditions are not fulfilled, it may end up being a smaller set). Therefore, by the formula, we have that $\phi(N) =N \times \frac{1}{2} \times \frac 23 = \frac N3$. This is very much an integer since $n \geq 1$ so $N$ contains $3$ in its prime factorization.
