2026/052 ) Let $a, b, c, d$ be four integers. Prove that $(b−a)(c−a)(d−a)(d−c)(d−b)(c−b) $ is divisible by 12.

If is is divisible by 12 then it is divisible by 3 and also by 4. this is so because 3 and 4 are co-primes

First Let us prove that is divisible by 3

Let us take $(b-a) , (c-a), (d-a)$ mod 3

As taking mod 3 there are 3 different remainders there are 2 possibilities

1. One of them is zero so it is divisible by 3
2. Non of them is zero so at least 2 remainder are same say $b-a$ and $c-a$. then the difference $(c-a) - (b-a) or (c-b) gives remainder 0 so divisible by 3

Having proved divisible by 3 Now let us prove that is is divisible by 4

Let  us take $(b-a) , (c-a), (d-a)$

If at least two of them are even then it is divisible by 4

        1. If one is even say $d-a$ then $b-a$ and $c-a$ are odd and $(c-a) - (b-a)$ that is c-a is even . As                there are 2 even numbers so product is divisible by 4

        2. If none is even the $b-a$ and$ c-a$ being odd $c-b$ is even and $b-a$ and $d-a $being odd $d-b$             is even . As $c-b$  and  $d-b$ are even so product is divisible by 4

In all cases it is divisible by 4

Hence it is divisible by 12