Problem From Hungerford (Abstract Algebra) (Ch 1 Section 1) The Question:  


  Let $a,b,c,$ and $q$ be as in Exercise 5. Suppose that when $q$ is divided by $c$, the quotient is $k$. Prove that when $a$ is divided by $bc$, the quotient is also $k$.  


Exercise 5 Question:  


  Let $a$ be any integer and let $b$ and $c$ be positive integers. Suppose that when $a$ is divided by $b$, the quotient is $q$ and the remainder is $r$, such that
  $$a = bq + r \text{ and } 0 \leq r &lt; b.$$
  If $ac$ is divided by $bc$, the quotient is $q$ and the remainder is $rc$.


What I have done so far:
Let $a=bq+r, 0 \leq r &lt; b$, and $q=ck+z, 0 \leq z &lt; c$, where $a,b,c,q,k,z$ are integers. Then it follows that:
$$a = bq + r \implies a = b(ck+z)+r \implies a = bck + bz + r, 0 \leq bz+r &lt; bc.$$
What I am stuck on is how to show the last inequality is true. I have tried manipulating the inequalities given:
$$ 0 \leq r &lt; b, 0 \leq z &lt; c \implies 0 \leq rc+bz &lt; bc \implies 0 \leq bz + r &lt; bz + rc &lt; 2bc $$
$$ 0 \leq bz + r &lt; 2bc $$
I think this is to no avail. Can anyone give me tips on how to proceed?
$$
0\le bz+r \le b(c-1)+(b-1)=bc-1 &lt; bc.
$$
I know I'm super late I'm sorry. but the problem only want you to show k. so set r = to zero and you are done. you don't need z. when you substitute you will get 2r, then set are to zero.