Is (a/b)/c equal to a/(b*c) for integer division? Let $\div$ denote a binary operator that returns the integer quotient of two integers, i.e. (assuming that both integers are positive) $a \div b = \left\lfloor \frac ab \right\rfloor$.  This corresponds to the integer division operator in many programming languages (e.g. the // operator in Python).
I observed that, when $a$, $b$ and $c$ are positive integers, the values of $(a \div b) \div c$ and $a \div (b \times c)$ are equal.
I have tried to find a counter-example by using the following Python code, but wasn't able to find one:
from random import randint

while True:

    a = randint(1, 10000000000)
    b = randint(1, 10000)
    c = randint(1, 10000)

    lhs = a//b
    lhs = lhs//c

    rhs = a//(b *c)

    if lhs != rhs:
        print a, b, c
        break

Could anyone please provide a counter example if the assertion that I made is not true or a proof which shows that it is always true?
Additional Information:


*

*Please note that all the division operators used above correspond to integer division.

*The version of Python is 2.7.12.

*I asked this question on StackOverflow and it was suggested there, that I ask it here.

*I was not able to find a tag which says integer-division, so I didn't use it and any suggestions are welcome.

 A: Write $a=qb+r$, with $0 \le r \lt b$, so that $a \div b=q$.
Then write $q=sc+t$, with $0 \le t \lt c$, so that $(a \div b) \div c=s$.
We now have $a=b(sc+t)+r=bcs+bt+r$. As $$\begin{aligned} bt+r &\le b(c-1)+(b-1) \\ &=bc-b+b-1 \\ &=bc-1, \end{aligned}$$ we have $a \div (bc)=s$.
A: It's slightly easier to do the other way than Ross.
Let $q,r$ be integers such that $a = b·c·q+r$ and $0 \le r < b·c$. Then $r \div b \le r / b < c$.
Then $( a \div b ) \div c = ( c·q + ( r \div b ) ) \div c = q + ( r \div b ) \div c = q = a \div ( b·c )$.
(We simply twice used the easy fact that $(d·x+y) \div d = x + y \div d$ for integers $x,y,d$ with $d > 0$.)
A: Another way to think about this is, say you have $N$ candies that you want to distribute among $a * b$ kids. What is the maximum number of candies a kid can get, given that you distribute the candies equally.
The answer is $\Big\lfloor\frac{N}{a*b}\Big\rfloor$
Note that you can divide the $a * b$ kids into $a$ classes. Since each student gets an equal number of candies, each class also gets an equal number of candies, which is $\le \left\lfloor\frac{N}{a}\right\rfloor$. Now for each class, you distribute these many candies among $b$ kids, so each kid gets
$$\left\lfloor\frac{\left\lfloor\frac{N}{a}\right\rfloor}{b}\right\rfloor$$
Hence,
$$\left\lfloor\frac{\left\lfloor\frac{N}{a}\right\rfloor}{b}\right\rfloor = \left\lfloor\frac{N}{a*b}\right\rfloor$$

I didn't prove that we can give $\Big\lfloor\frac{N}{a}\Big\rfloor$ candies to each class and still ensure that each student will get the optimal number of candies. This can be proven easily as well. Let's say each class gets $Y$ candies in the optimal solution. Then, we have:
$$
Y * a \le N
\implies Y \le \frac{N}{a}
$$
Hence we can give each class $\left\lfloor\frac{N}{a}\right\rfloor$ candies for optimal distribution.
Note that each class can also reject some candies.
A: $$a \text{\\} b  \text{\\} c = d$$
$$\exists x.~~ a\text{\\}b = x ~~\land~~ x\text{\\}c = d$$
$$\exists x.~~ xb \in [a - b + 1 \dots a]  ~~\land~~ cd \in  [x - c + 1 \dots x]$$
$$\exists x.~~ xb \in [a - b + 1 \dots a]  ~~\land~~ bcd \in  [xb - bc + b \dots xb]$$
$$\exists x.~~ xb \in [a - b + 1 \dots a]  ~~\land~~ bcd \in  [(a - b + 1) - bc + b \dots a]$$
$$bcd \in [a - bc + 1 \dots a]$$
$$a\text{\\}(bc) = d$$
A: This seems to work always, because it is mathematical tautology $-$ not an obvious one though. Let's denote integer division by $\div$, and say we have $(a\div b)\div c=s$. This means
$$a\div b=sc+r_c\quad\implies\quad a=(sc+r_c)b+r_b=sbc+\underbrace{r_cb+r_b}_R$$
where $r_c\in\{0,...,c-1\}$ and $r_b\in\{0,...,b-1\}$ denote the remainder after division by $c$ and $b$ respectively. We therefore have 
$$R=r_cb+r_b\le b(c-1)+b-1=bc-b+b-1=bc-1<bc.$$
This suffices to conclude that
$$a\div(bc)=(sbc+R)\div(bc)=s$$
since $R$ is to small to make a difference after the division.
A: Let $\div$ denote integer division, and $/$ normal division. Let $a,b,c$ be integers and $x$ an abitrary number.
First observation: 
Because we're dividing an iteger, the division $(a+x)\div b$ can only be greater than $(a\div b)$ when x is at least 1, i.e. $x\ge 1$. 
Furtrher we already know that $(a/b)/c = a/(b\cdot c)$ for regular division holds.
Therefore, the only reason integer division $\div$ wouldn't hold for $(a\div b)\div c$ would be when the first division cuts off so much, the second division has an altered result.
We can now split $a$ in a divisble part $p\cdot b$ and a remainder $q$. Note that the remainder is smaller than $b$.
However, because it's a remainder, $q/b$ is smaller than 1. And because it's smaller than one, and $a\div b$ is an integer, 
$(a\div b)\div c = p\div c= (p + q/b)\div c $
And therefore, the equality $(a\div b)\div c) = a\div (b\cdot c)$ holds.
