mohankumarbd wrote:
I have a question on the below problem.
If x^3- x = p, and x is odd, is p divisible by 24?
And the answer is yes. It is divisible by 24.
the reason being the above can be simplified into (x-1)(x)(x+1) which are consecutive integers. so (x-1) & (x+1) are even integers. and so the the total product should have factors 2*3*4.
Now, if the problem is exactly as given above, should we also not consider the below scenarios.
X-1 could be zero which is also an even integer. So p = 0. But again considering that zero is also divisible by 24, is this why the answer is correct. How are such questions to be Handled. Any inputs on how to consider the last 'zero' scenario please.
\(x^3-x=(x-1)*x*(x+1)\).
Since \(x=odd\) then \(x-1\) and \(x+1\) are consecutive even integers. Now, the product of two consecutive even integers is always divisible by 8 (since one of them is divisible by 4 and another by 2).
Next, \((x-1)*x*(x+1)\) is also the product of three consecutive integers. Out of three consecutive integers one is always divisible by 3, so \((x-1)*x*(x+1)\) is divisible by 3 too.
Which means that \((x-1)(x)(x+1)\) is divisible by both 3 and 8, so by 3*8=24.
As for zero:
zero is a divisible by every integer, except zero itself. So, if \(p=0\)then it's divisible by 24 as well as by all other integers but zero itself.
Hope it's clear.