Euler Phi Function and Divisibility

Question

Question: Suppose that the positive integer n is divisible by 7. Prove that $ 7φ(n) ≤ 6n $ When exactly does equality hold and why?

I have concluded that if n is divisible by 7, it obviously cannot be prime. This eliminates the use of the equation $ φ(p^k) = p^k − p^{k−1} $.

I have had a lot of practice with $n$ being prime, but have not learned a whole lot when $n$ is not prime.

Answer 1

By hypothesis, $n = 7^k m$ for some $k$, $m \in \mathbb{N}$, where $7 \nmid m$. Then $\gcd(7^k,m) = 1$, so $\varphi(7^k m) = \varphi(7^k) \varphi(m)$, so $$\begin{aligned}[t] 7 \varphi(n) = 7 \varphi(7^k m) = 7 \varphi(7^k) \varphi(m) &= 7 (7^k-7^{k-1}) \varphi(m) \\ &= 7\cdot 7^{k-1}(7-1) \varphi(m) = (7-1) 7^k \varphi(m) \leq 6 \cdot 7^k m = 6n.\end{aligned}$$

Answer 2

Because $7$ divides $n$, $$\varphi(n)=n\prod_{p|n}\left(1-\frac{1}{p}\right)\le n\left(1-\frac{1}{7}\right)$$ and$$7\varphi(n)\le7n\left(1-\frac{1}{7}\right)=6n$$