November 29th, 2005


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.