divisibility

[defeated_soldier]

Every integer of the form (n^3-n)*(n-2) for n=3,4,5....is

a) divisible by 6 but not always divisible by 12
b)divisible by 12 but not always divisible by 24
c)divisible by 24 but not always divisible by 48
d) divisible by 9


which one is answer ?


I tried this way but failed

I put n=3 expression becomes =24*1 ===>NOT divisible by 9 so d) is ruled out.


but for othervalues of n i.e 4,5,6.......etc (a) (b) and (c) remains true .

How do i find out then which is the answer ?

 

[soroban]

Hello, defeated_soldier!

Every integer of the form \(\displaystyle (n^3\,-\,n)*(n\,-\,2)\;\) for \(\displaystyle n\,=\,3,4,5,....\) is:

a) divisible by 6, but not always divisible by 12
b) divisible by 12, but not always divisible by 24
c) divisible by 24, but not always divisible by 48
d) divisible by 9

Factor that integer: \(\displaystyle \,(n^3\,-\,n)(n\,-\,2)\;=\;n(n^2\,-\,1)(n\,-\,3)\;=\;n(n\,-\,1)(n\,+\,1)(n\,-\,2)\)

We have: \(\displaystyle (n\,-\,2)(n\,-\,1)n(n\,+\,1),\,\) the product of four consecutive integers.

The product of four consecutive integers is divisible by \(\displaystyle \,1\cdot2\cdot3\cdot4\,=\,24\) . . . answer (c)

 

[defeated_soldier]

Thank you.

its perfect