There was a trick to it, of course. Mandating that

*p*be prime was not necessary. All it took was that

*p*be odd. As is the case for all primes other than two!

Henceforth, given

*p*any odd number:

*(p-1)*and

*(p+1)*are both even.

Furthermore,

*(p-1)*and

*(p+1)*are two consecutive even numbers, meaning that one of the two is a multiple of four.

Hence, of

*(p-1)*and

*(p+1)*, one is a multiple of

*2*and the other is a multiple of

*4*.

Hence,

*(p-1).(p+1)*is a multiple of

*8*.

Additionally,

*(p-1)*,

*p*and

*(p+1)*being three consecutive numbers, one of them must be a multiple of

*3*.

In conclusion,

*(p-1).p.(p+1)*is a multiple of

*8*3*=

*24*.

Kudos to

**grey_wolf_xvii**,

**krdbuni**,

**thenightwolf**,

**kemonotsukai**and

**janetraeness**for figuring it out! And honorable mention to

**unciaa**and

**timduru**for the witty answers. :) And thank you to the others for trying!

I've unscreened the comments now.

**thenightwolf**also makes a good point that it was not that easy for the non-mathematician, because even though there is no advanced math concept involved, it does require a manner of mathematical thinking. I think he's right. So, uh, apologies for my lack of empathy in this.

The same woofiebutt also offers another problem, a lot more mathy:

Let

*P*be a polynomial of degree

*n*such that

*P(i) = 1/i*for

*i=1,...,n+1*.

Calculate

*P(n+2)*.

And as for the rest, perhaps I'll post later on. We'll see.

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