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Food For Thought. - Balinares

Nov. 12th, 2005

04:27 pm - Food For Thought.

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(Aside: yeah, I've been cranky and unapproachable all week. Yeah, for good reason. Serenity is a perpetual conquest. Sometimes too much is too much, and it takes a few days to find your center again. That's all I'll say about that.)

It occurred to me just a moment ago that while some mathematical riddles required a certain amount of background (in particular, I was trying to remember if proving there are infinitely many prime numbers is one of those), some don't, and thus make a good brainteaser for the non-mathematician.

Here's a good one. Comments are screened. Impress me!

Let p be a prime number other than 2.
Prove that (p-1).p.(p+1) must be a multiple of 24.

Comments:

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From:unciaa
Date:November 12th, 2005 04:23 pm (UTC)
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*walks up the counter, holding a Beozar stone. Hey. It worked for Harry Potter*
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From:grey_wolf_xvii
Date:November 12th, 2005 05:03 pm (UTC)
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à vue de pif :

(p-1)p(p+1)

2 étant exclu, p est impaire (les autres paires étant divisibles par 2). D'où p-1 paire et p+1 paire.

(2xk)p(2xk') = 4kk'p

k et k' sont eux-mêmes soit paires, soit impaires. Mais bon, en partant d'un nombre impaire et en ajoutant et enlevant 1, on retombe forcément à la fois sur un k ou k' paire et l'autre impaire (puisque en considérant les terminaisons 0, 2, 4, 6 et 8, il y a alternance de k paires et impaires).

Prenons k' paire = 2m

D'où 8kmp

Les multiples de trois sont disposés de trois en trois (sans blague!). De ce fait, si p est impair, p+1 ou p-1 sera multiple de 3. A moins ce soit p lui-même qui soit égal à trois, et dans ce cas on est tranquille les mains dans les poches.

Alors bon, disons que (p-1) soit multiple de trois, par exemple : 2xk = 2x3xd

Au total, on a 24 x d x m x p.
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From:thenightwolf
Date:November 12th, 2005 06:01 pm (UTC)
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p-1, p et p+1 sont trois nombres qui se suivent, donc l'un au moins est un multiple de 3, donc (p-1)p(p+1) est multiple de 3.

De plus, p étant premier, p-1 et p+1 sont pairs; donc p-1 est multiple de 2 et p+1 de 4 (nombre pair suivant), donc (p-1)(p+1) est multiple de 8.

Donc (p-1)p(p+1) est multiple de 3*8 = 24.

Par contre, pour avoir aidé des personnes ayant des problèmes de maths, je ne suis pas d'accord pour dire que c'est faisable pour les gens qui ne manipulent pas aisément les maths, car les histoires de multiples sont à la base conceptuellement compliquées (d'ailleurs en terminale S, ils ne sont plus abordés, sauf en spécialité Maths), même si l'idée de base (trois nombres qui se suivent : au moins 1 multiple de 3) est de nature logique.
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From:jallora
Date:November 12th, 2005 06:14 pm (UTC)
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I don't trust my brain to try and resolve your little riddle, plus I fear it would make me feel dumb and very blond, but I just wanted to say I liked your icon.
I knew you weren't fine, I'm sorry I couldn't help. *Bisous*
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From:msondo
Date:November 12th, 2005 07:45 pm (UTC)
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Hmm.. well if p = 3 then the obvious answer is 24. If you can add something equally to one side of any equation you can take it away equally?
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From:timduru
Date:November 12th, 2005 09:22 pm (UTC)
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42 !
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From:issarlk
Date:November 12th, 2005 11:19 pm (UTC)
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Je ne trouve pas, ca me rend fou, j'ai envie de casser quelque chose.
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From:kemonotsukai
Date:November 13th, 2005 12:04 am (UTC)
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So, at the first sight it seems very difficult, but in fact it's really easy ^^
What is 24? 24 = 4 * 2 * 3
So we have to proof that (p-1).p.(p+1)=X*(2*3*4)

One of the term must be divisible by 3 since they are consecutive
A prime number superior than 2 is mandatoty an odd number. So (p-1) and (p+1) are par numbers. They are both divisible by 2, and more since they are two consecutive par nimbers, one of the both must be divisible by 4.
So we have the three multiplicators.

Exemples :
3=> 2 * 3 * 4 = 24
5=> 4 * 5 * 6 = (4)* 5 (2 * 3) = 5*24
7=> 6 * 7 * 8 = (2 * 3) * 7 * (4 * 2 )= (2*7)*24 = 14*24
and so on....
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From:reveille_d
Date:November 13th, 2005 12:52 am (UTC)
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Just glancing at that makes my brain cramp in unreasonable psychosomatic response.
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From:kemonotsukai
Date:November 13th, 2005 03:08 am (UTC)
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Just to know : it's normal that my comment have been erased ?
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From:balinares
Date:November 13th, 2005 03:12 pm (UTC)
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Not erased, hon, just screened. So as to give other people a chance to look for the answer. :)
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From:kefen
Date:November 14th, 2005 12:19 am (UTC)
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You make me hope. I'll give it a try and hope I can figure it out later, need rest right now.

(Maybe I'll just manage to prove how helpless I am at such things;)
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From:thenightwolf
Date:November 16th, 2005 07:14 pm (UTC)
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Tiens, puisque tu aimes ça (TU AIMES CA, HEIN ??!), en voilà un joli (mais qui n'est pas du tout non-mathématique).

Soit P un polynôme de C de degré n tel que P(i) = 1/i pour i=1,...,n+1.
Trouver P(n+2).
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From:janetraeness
Date:November 17th, 2005 02:42 pm (UTC)
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prime numbers must be odd, barring 2. even * odd = even. even * even = even

even * odd * even = even.


Every third number is a multiple of 3 (for obvious reasons). Every even number is a multiple of 2. So one of the factors of the product (p-1)p(p+1) must be 6 (3*2) .. Since (p-1) (p+1) must both be even, there are 2 multiples of 2. This means that factors must be 3*2*2 (or 12).

Since 3 is the smallest prime (other than 2), p+1 has 2*2 as prime factors. This means that for EVERY case, 3*2*2*2 (one 2 from (p-1) 2 from (p+1), and a 3 from one of those numbers), ... there is a 24 as a factor.

:) That good enough?
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