The number of positive integers which are less than $mk$ and coprime to $m$ is $k\phi(m)$. 
Let $m,k$ be positive integers. Then the number of positive integers $\leq mk$ prime to $m$ is $k\phi(m)$.

My approach would be to use induction on $k$. If $k=1$, then by definition the result holds. If we assume it holds for some $k$. how can we  show it holds for $k + 1$ case?
 A: Note that $$\gcd(a,m) = 1 \iff \gcd(a+k \cdot m,m) = 1$$ Use this to prove the following.
If $S_0$ is the set of number relatively prime to $m$ and less than $m$, then $S_k$, the set of numbers relatively prime to $m$ and between $km$ & $(k+1)m$, is nothing but $$S_k = S_0 + km$$
Now conclude what you want.
A: Well you've got that there are $k\phi(m)$ numbers less than $mk$ which are coprime to $m$, so now what you need to do is show that there are exactly $\phi(m)$ numbers $r$ with $mk<r\leq m(k+1)$ which are coprime to $m$.  What can you say about the congruences of $mk+1, mk+2, \ldots , m(k+1)$ modulo $m$?  Can you make a counting argument using the definition of $\phi$? 
A: What divides or is prime to $k$ is usually completely unrelated to what divides or is prime to $k + 1$ so I don't think induction is a good approach.
I suggest you try and prove the following statement: A positive integer $n$ is prime to $m$ if and only $n + m$ is prime to $m$.  Then think about what $\phi(m)$ represents.
