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Math Riddle Answer. - Balinares — LiveJournal

Nov. 29th, 2005

10:35 am - Math Riddle Answer.

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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.

Comments:

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From:grey_wolf_xvii
Date:November 29th, 2005 12:13 pm (UTC)
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Hé bien, si p est un polynôme de degrée n, P(x) = somme de r=0 à n de a (indice r) * x^r.

Ce que je suis franchement en train de me demander, c'est ce que veut dire le "P(i) = 1/i for i=1,...,n+1.". Est-ce la somme des 1/i de i=1 jusqu'à i = n+1? Et si c'est le cas, en quoi ça qualifie de polynôme?
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From:balinares
Date:November 30th, 2005 12:26 am (UTC)
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Ca veut dire que P(1)=1, P(2)=1/2... P(n+1)=1/(n+1).

Comme tu as n+1 valeurs connues d'un polynôme de degré n, ce polynôme est défini de manière unique.

Le but de l'exercice est donc de calculer P(n+2) en fonction de ces données. Sous forme d'une fonction de n, donc.

J'ai calculé à la main, pour voir, que pour n=1, P(3)=0, et pour n=2, P(4)=1/2.

Il y a une astuce, mais je ne la trouve pas encore.
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From:thenightwolf
Date:November 30th, 2005 04:44 pm (UTC)
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P(3) ne peut pas valoir 0, puisque P(3) = 1/3 par définition (c'est chiant les définitions, hein ? :p), de même que P(4) = 1/4.
On connaît P(1), P(2)..., P(n+1), reste plus qu'à trouver P(n+2).

Astuce, utiliser un polynôme auxilliaire Q bien choisi, qui nous facilitera la tâche.

(mais c'est pas évident)
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From:grey_wolf_xvii
Date:November 30th, 2005 06:04 pm (UTC)
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Euh... P(n+2) = 1/(n+2)?

Quel genre de réponse est attendue, exactement?
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From:thenightwolf
Date:November 30th, 2005 06:09 pm (UTC)
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Non, les P(i) = 1/i pour i allant de 1 à n *donné* sont une définition. Comme le disait Sun, ceci définit le polynôme de façon unique.

Connaissant P(1)... P(n+1), il faut trouver P(n+2), qui est une fonction de n.

C'est pas tant la forme de P(n+2) qui est intéressante, mais la manière d'y parvenir.
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From:balinares
Date:November 30th, 2005 10:34 pm (UTC)
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On s'est peut-être mal compris...? Pour n=1 et n=2 j'ai calculé P puis P(n+2) pour voir la tronche que ça avait, histoire de voir s'il y a pas moyen de faire ça par récurrence sur le degré de P (et puis parce que ça m'occupe à peu de frais), mais j'ai peut-être mal compris le problème initial?
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From:thenightwolf
Date:December 1st, 2005 04:04 pm (UTC)
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Aaah, mes plus plates z'excuses, j'ai vraiment pas le cerveau en face des sphincters cérébraux, en ce moment.
Si si tu as parfaitement compris le problème ^^
Mais ce n'est pas une récurrence :)
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From:thenightwolf
Date:December 1st, 2005 06:11 pm (UTC)
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Un petit indice, il faut former un polynôme Q sympathique.
Et ô miracle, d'ailleurs, P(i)=1/i (ou réécrit autrement, i*P(i)-1 = 0).

La prochaine étape est l'invocation d'un mathématicien dont le nom commence par D'Al et finit par Lambert :p
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From:grey_wolf_xvii
Date:December 1st, 2005 06:24 pm (UTC)
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Alfred Lambert?

(Bon, d'accord, je retourne me pendre)
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From:timduru
Date:November 29th, 2005 02:31 pm (UTC)
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<== old meerkat
it's too old memories for me. ;)
head ache now,
/me runs to pharmacy

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From:balinares
Date:November 30th, 2005 12:27 am (UTC)
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Ack! Sorry! *hands over the aspirin*
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From:weasely
Date:November 29th, 2005 02:56 pm (UTC)
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ive kept -well- away from the last one like this, but i happened gto glance something along this post. ... nnnnumbers are fun. they make brainmeat bleed.
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From:balinares
Date:November 30th, 2005 12:29 am (UTC)
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No they don't! >O.o< Numbers are cool! Numbers want you happy, just like dragons do!
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From:kefen
Date:November 29th, 2005 05:11 pm (UTC)
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Yeah, I had a little hope to solve the riddle, being somehow allergic to math, but no luck with this one :P

*goes sulk in a corner*
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From:balinares
Date:November 30th, 2005 12:29 am (UTC)
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Just wait till we discuss cardinalities while curled up on my couch with beer at hand. All will seem clear then. :)
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From:thenightwolf
Date:November 30th, 2005 04:38 pm (UTC)
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THAT'S how mathematics should be !
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From:balinares
Date:November 30th, 2005 10:35 pm (UTC)
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I have a mind to set up a couch-beer-math party. You're game? =)
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From:thenightwolf
Date:December 1st, 2005 04:55 pm (UTC)
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OMG ! I'm in !
I'd even make a strip-tease if it could help proving P=NP (or not).
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From:grey_wolf_xvii
Date:December 1st, 2005 06:33 pm (UTC)
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Je veux bien, hein, mais je bois de la grenadine moi
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From:unblue
Date:December 2nd, 2005 09:09 am (UTC)
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Damn man wait till I'm back and count me in ! I suck at maths, but a beer and Nightwolf performing a strip-tease is enough to make it interesting !
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From:kefen
Date:December 4th, 2005 09:05 pm (UTC)
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Fair enough, count me in! :)
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From:grey_wolf_xvii
Date:December 1st, 2005 10:40 am (UTC)
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En 20 symboles finaux, pas mieux.

http://membres.lycos.fr/vixem/Truc/Pas%20mieux.jpg

http://membres.lycos.fr/vixem/Truc/Pas%20mieux%202.jpg

(J'ignore si je peux injecter le 2^k comme ça, vlan, mais autant essayer)
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From:grey_wolf_xvii
Date:December 1st, 2005 01:46 pm (UTC)
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(Le coup du 2^k me semble bien peu évident en fait. donc, plutôt la réponse d'au-dessus)
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From:thenightwolf
Date:December 1st, 2005 04:52 pm (UTC)
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La méthode cherchée n'est pas du tout aussi calculatoire, sans sommes et sans Cnk :) (d'ailleurs j'ai même pas eu envie de lire tout ça, désolé, mais ça fatigue le neurone les somme ;)
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From:grey_wolf_xvii
Date:December 1st, 2005 04:56 pm (UTC)
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Zut.
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