An intuitive explanation of the multiplication identity of modular arithmetic Is there any intuitive explanation for the validity of this identity?

If $a\equiv b\pmod n$ and $c\equiv d\pmod n$, then
$$a\times c \equiv b\times d \pmod n$$

I want something which appeals to someone's who just beginning to learn number theory.
 A: This is not difficult to prove and understand. The idea is to use a simpler result two times.

If $a\equiv b\pmod{n} $ then we have $ka\equiv kb\pmod{n} $.

This should be obvious to understand as it is an immediate consequence of the definition of congruence. Just note that if $a-b$ is a multiple of $n$ then $k(a-b) $ is also a multiple of $n$.
Using this we can multiply first congruence with $c$ and second congruence with $b$ to get $$ac\equiv bc\pmod{n}, bc\equiv bd\pmod {n} $$ and these two give you $ac\equiv bd\pmod{n} $ (this is based on the fact that if two numbers are multiple of $n$ then so is their sum).

I don't think the above would count as a lot of hairy algebra and ideally should be accessible to anyone who has basic idea of factors and multiples (typically 6th standard math suited for children of 10-11 years in Indian curriculum). 
A: $a \equiv a'\pmod n$ means that $a$ and $b$ have the same remainders in terms of $n$. 
The this identity says that if you have two pairs of numbers $a \equiv a' \pmod n$ and $b\equiv b' \pmod n$ so $a$ and $a'$ have the same remainder and if $b\equiv b' \pmod n$ so $b$ and $b'$ have the same remainder, then the pair of products $ab$ and $a'b'$ will be so that $ab \equiv a'b'$ and the pair of products $ab$ and $a'b'$ will have the same remainder.
That should intuitively obvious. 
But if not:
$a$ and $a'$ having the same remainder means $a = kn + r$ for some integer $q$ and $r$ is the remainder and $a' = jn + r$ for a different integer $j$ but the same remainder $r$.  
And $b$ and $b'$ having the same remainder means $b = wn + s$ for some integer $w$ and remainder $s$.  And $b' = vn + s$ for a different integer $v$ but the same remainder $s$.
If you muliply $ab$ you get $kvn^2 + kns + vnr + sr=n\cdot (kvn+ks +vr) + sr$.  SO $ab$ will have whatever remainder that $sr$ will have. And if you multiply $a'b'$ you get $jwn^2 +jnr + wnr + sr = n\cdot(jwn+jr + wr) + sr$.  And $a'b'$ will also have whatever remainder that $sr$ will have.
So $ab$ and $a'b'$ will both have the same remainder in terms of $n$>
