Question

$(n^3-n)$ is divisible by:

• A. 2
• B. 3
• C. 6
• D. 4

Answer 1

Answer: (A), (B), (C)

The correct options are
A 2
B 3
C 6

Here,
$(n^3-n)$ can be written as
$(n^3-n)$ = $(n-1)\times (n+1)\times n$

If we divide any number, let's say P by 3, we get the remainder as 0, 1 or 2.
Example:
$6÷3$ = $3\times 2$ + 0
$7÷3$ = $3\times 2$ + 1
$8÷3$ = $3\times 2$ + 2
So, we can write:
$P÷3$ = $3\times z$ + (0 or 1 or 2); where z is the multiple be any real number.
Now,
For 3z,
$(n-1)\times (n+1)\times n$
= $(3z-1)\times (3z+1)\times 3z$
This is divisible by 3 as it is a multiple of 3.

For 3z + 1, we can rewrite the equation
$=\left(\right(3z+1)-1)\times \left(\right(3z+1)+1)\times (3z+1)$
$=\left(3z\right)\times (3z+2)\times (3z+1)$
This is also divisible by 3, as it is a multiple of 3.

For 3z + 2, we can rewrite the equation as
$=\left(\right(3z+2)-1)\times \left(\right(3z+2)+1)\times (3z+2)$
$=(3z+1)\times (3z+3)\times (3z+2)$
$=(3z+1)\times \left(3\right(z+1\left)\right)\times (3z+2)$
$=3(3z+1)\times (z+1)\times (3z+2)$
Taking 3 as common, we can see that this equation is also divisible by 3.

Divisibility by 2
Here,
$(n-1)\times (n+1)\times n$ ,
For any number n, there are two cases.

Case I: n is odd
Then, n + 1 = even number
Even number is always divisible by 2 so, $(n-1)\times (n+1)\times n$ is divisible by 2.

Case II: n is even
$(n-1)\times (n+1)\times n$ is divisible by 2 as it is an even multiple of the whole term.

Now,
$(n-1)\times (n+1)\times n$ is divisible by both 3 and 2.
Therefore, it is divisible by 6 as well.

Hence, $(n-1)\times (n+1)\times n$ is divisible by 2, 3 and 6.