# November 29th, 2005 - Balinares — LiveJournal

## Nov. 29th, 2005

### 10:35 am - *Math Riddle Answer.*

I'm way overdue in posting the answer to the little math riddle I offered you guys a while back; I'm sorry, I've not been very much online.

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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