For $a,b,c,d\in \Bbb N$ prove that $(a-b)(c-a)(d-a)(c-b)(d-b)(d-c)$ is divisible by $12$ 
Given that $a,b,c,d\in \Bbb N$ and pairwise distinct, prove that the number
  $$(a-b)(c-a)(d-a)(c-b)(d-b)(d-c)\tag{1}$$
  is divisible by $12$.
Source: IME 1999/2000 entrance exam

My attempt:
Suppose WLG that $a<b<c<d$, and $p,q,r$ are defined as
$$p=|a-b|,q=|c-b|,t=|d-c|.$$ 
Therefore the original problem is equivalent (as signs are not relevant here), to finding that
$$P=p q r(p+q)(q+t)(p+q+t)\tag{2}$$
is divisible by 12, with $p,q,t\in \Bbb N$ and $p,q,t\ge 1$.
The problem is now showing that $P$ is divisible by 4 and by 3. To do that let us consider the possibilities for the parity (E=even, O=odd) for the numbers $p,q,t$ and see what happens to the parity of the other terms of the product $P$:
$$
\begin{array}{ l| c | c | c| c| c|c }
    case&p&q&t&(p+q)&(q+t)&(p+q+t)\\
  \hline 
       1&E&E&E&  E&    E&     E \\
       2&E&E&O&  E&    O&     O \\
       3&E&O&E&  O&    O&     O \\
       4&O&E&E&  O&    E&     O \\
       5&O&O&O&  E&    E&     O \\
       6&E&O&O&  O&    E&     E\\
       7&O&E&O&  O&    O&     E\\
       8&O&O&E&  E&    O&     E\\
  \hline  
\end{array}
$$
In all 8 cases, there are at least 2 even terms and, as each even term can be represented by $2k_i$, we will have $4k_i k_j$ as a factor in $P$ and therefore $P$ is divisible by 4.
Question: how can I show that $P$ is divisible by 3? It appears true but I'm not finding a simple formal argument to show that. There is another similar SE math question in here , but the argument in the answer is not sufficient clear for me.  I'm looking for an argument that is appealing for a 7th grader concerning the divisibility by 3, if that is possible.
Previous answer used, for instance, the notation "congruent in mod 3" that is not something appealing for a 7th grader. I'm looking for a simpler argument using terminology appealing for this level. 
Postscript: 
Given the helpful comment from Ronald below, I will try a complete explanation for the "divisibility by 3 part":
Now considering the same notation using $p,q,t$, it must be the case that for some $k_1,k_2,k_3\in \Bbb N$ it holds that
$$p=3 k_1+r_1, q=3 k_2+r_2, t=3 k_3+r_3$$
in which $r_1,r_2,r_3$ are the rests from the division of $p,q,t$ by 3. Therefore 
$r_1,r_2,r_3\in\{0,1,2\}$, in which $r_i=0$ denotes a case in which the number is divisible by 3. Let us now examine the possibilities for $r_i$ in $p,q,t$.
First, for at least one $r_i=0$, we are done because we have a factor in (2) divisible by 3. Now for the other possibilities. The table shows the rest for the division by 3 of the terms:
$$
\begin{array}{ l| c | c | c| c| c|c }
    case&p&q&t&(p+q)&(q+t)&(p+q+t)\\
  \hline 
       1&1&1&1&  2&    2&     0 \\
       2&1&1&2&  2&    3&     1 \\
       3&1&2&1&  0&    0&     1 \\
       4&2&1&1&  0&    2&     1 \\
       5&2&2&1&  1&    3&     2 \\
       6&2&1&2&  0&    0&     2 \\
       7&1&2&2&  0&    1&     2 \\
       8&2&2&2&  1&    1&     0 \\
  \hline  
\end{array}
$$
The rest for each of the terms $(p+q),(q+t),(p+q+t)$ can be found by adding the rests from the terms in each sum, and if this value is greater than 2, dividing this value again by 3, to find the rest. For example, in the first row, the rest for $(p+q+t)$ would be $(1+1+1)=3$, with rest 0 from the division by 3. 
Therefore, as each line in this last table has at least one term with rest 0 with respect to the division by 3, we have proved that $P$ is divisible by 3.
As Ronald suggested in the comments, this can be seen more easily with the original problem defined in terms of $a,b,c,d$. Because the rest of the division by 3 can be only 0,1,2, it must be the case that at least 2 of these numbers have the same rest with respect to 3, and if two numbers have the same rest with respect to $n$, their difference will have rest zero or will be divided by that number, as
$(k_1n+r)-(k_2n+r)=(k_1-k_2)n$ 
the rest with respect to $n$ is zero. Therefore in (1) we must have a term divisible by 3.
It is helpful to note, concerning math terminology that if two numbers have the same rest with respect to a number $n$, we can call these two numbers "congruent mod n". Therefore 0,3 and 6 are congruent mod 3 because all have rest 0 with respect with the division by 3. 1 and 4 are congruent mod 3 because 1 and 4 have rest 1 with respect to the division by 3. 
 A: Here's a proof accessible at $7$th grade level. Note that the product contains all possible differences of $a,b,c,d$ so to prove it is divisible by $3$ it suffices to prove that one such difference is divisible by $3$. We use a pigeonhole argument: we have $4$ integers $\,a,b,c,d\,$ but only $3$ possible remainders $\,0,1,2\,$ for division by $3$, therefore two integers have the same remainder $\color{#c00}r,\,$ hence their difference $\,3q\!+\!\color{#c00}r  - (3q'\!+\!\color{#c00}r) = 3(q\!-\!q')$ is divisible by $3,\,$  therefore the product $n$ has a factor of $3$.
Similarly two of them $\,x,y\,$ leave the same remainder $r$ after division by $2$ so $x-y$ is divisible by $2$. If there is another $\,z\,$ with remainder $r$ then also $y-z$ is divisible by $2;$ else the other two numbers have remainder $\neq r$ so they have equal remainders (since there are only $2$ possible remainders $0,1$), so their difference is divisible by $2$. Hence the product $n$ has two factors of $2,\,$ so it is divisible by $4$.
Finally $\ \color{#c00}3,\color{#0a0}4\mid n\,\Rightarrow\,12\mid n,\ $ by $\  n = \underbrace{4(n)}_{\Large 4(\color{#c00}3j)}\! -\! \underbrace{3(n)}_{\Large 3(\color{#0a0}4k)} = 12(j\!-\!k)$
