Wikipedia:Reference desk/Archives/Mathematics/2011 September 11

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September 11

Can't recall details of a math puzzle

I vaguely remember the gist of an old math puzzle, but I can't recall enough details to recreate it, and it's driving me nuts. Anyone know it?

There is a group of brothers who are dividing up of a herd of animals, according to some rules. The numbers don't work out. Another man comes along, and loans them one of his animals; then the numbers work out with exactly one animal left over, so the man gets his animal back.

If this rings a bell with anyone, I'd be grateful. --Floquenbeam (talk) 00:47, 11 September 2011 (UTC)

Three brothers have to divide up 11 animals such that one receives 1/2 of the animals, one receives 1/4 of the animals and the last one receives 1/6 of the animals. However, 11 is not divisible by 2, 4 or 6. So some man comes along and adds his animal, so now there are 12 animals. This means that the first brother gets 6 animals, the second gets 3 and the last gets 2. 12 - 6 - 3 - 2 = 1 left over, so the man takes his animal back. Or something like that. Widener (talk) 01:13, 11 September 2011 (UTC)

I have seen it as 17 animals with 1/2, 1/3 and 1/9 which when you all an animal allows the brothers to take 9, 6 and 2. Another case with three different would be 1/2, 1/3 and 1/8 for 24. The question is how to generate all of the answers to a,b,c,n (all different) where a|n, b|n, c|n and a+b+c+1=n. Any solution there gives n-1 animals with a/n, b/n and c/n shares. However the 17, 1/2,1/3 and 1/9 is the one I've seen most on the web.Naraht (talk) 01:36, 11 September 2011 (UTC)

According to http://www.jimloy.com/puzz/will.htm , there are *exactly* 7 solutions for 3 sons: (7,1/2,1/4,1/8);(11,1/2,1/4,1/6);(11,1/2,1/3,1/12);(17,1/2,1/3,1/9);(19,1/2,1/4,1/5);(23,1/2,1/3,1/8);(41,1/2,1/3,1/7) Naraht (talk) 01:36, 11 September 2011 (UTC)

Cool -- a meta-puzzle! (i.e., how many variants of this puzzle are possible?) Looie496 (talk) 02:25, 11 September 2011 (UTC)

• Thank you both very much. --Floquenbeam (talk) 02:18, 11 September 2011 (UTC)

The will specified that the last 1/12 of the animals be given to a 4th brother, who then sues the other brothers for stealing his inheritance. Rckrone (talk) 05:24, 11 September 2011 (UTC)

There are exactly 7 solutions if there are three sons and each fraction has a numerator of one and the neighbor loans exactly one animal. If you drop any of the restrictions then there are infinitely many solutions. For example: the sons are supposed to get 1/96, 1/379, and 1/940 apiece; they have 120,721 animals; and a neighbor loans them 8,429,519 more.

Really, the problem makes no sense (and was probably invented by a non-mathematician) because it's impossible to correctly divide property into fractions that don't add up to 1. No sane person would specify such a thing in a will, unless, I suppose, it was a typo or they made a mistake in the arithmetic. In that case, if the true intent couldn't be determined, the property would probably be divided in proportion to the fractions, e.g., in the proportion of ${\tfrac {1}{2}}:{\tfrac {1}{4}}:{\tfrac {1}{6}}$, or equivalently (multiplying by the least common denominator) 6:3:2, which, since they have 11=6+3+2 animals in this version of the problem, is easily satisfied without any extras. -- BenRG (talk) 19:39, 12 September 2011 (UTC)

Question about formula involving exponential functions

Is e¹×e^{[(−x÷2)+(x2÷3)+···]}=e[1−(x÷2)+(x²÷3)+···];where 'e' is the exponential function ?```` — Preceding unsigned comment added by Smprince055 (talk • contribs) 15:19, 11 September 2011 (UTC)

Note: I have supplied a section title for this question. Looie496 (talk) 15:40, 11 September 2011 (UTC)

Yes. And in general, $a^{b+c}=a^{b}a^{c}$. -- Meni Rosenfeld (talk) 16:40, 11 September 2011 (UTC)

Tensors over ℤ₂

I want have all possible 2x2x2-tensors over ℤ₂ listed with its ranks. --84.62.194.162 (talk) 17:44, 11 September 2011 (UTC)

I want a permanent position paying £200,000 per year, but hey. Try reading your notes, maybe open a book, and have a go. Tell us how you get on and we might be able to help you. — Fly by Night (talk) 18:47, 11 September 2011 (UTC)

There are three ways to pass to the quotient $(\mathbb {Z} _{2}^{2})^{\otimes 3}\to \mathbb {Z} _{2}^{2}\otimes \wedge ^{2}\mathbb {Z} _{2}^{2}\cong \mathbb {Z} _{2}^{2}.$ An element has rank zero, one, or two if and only if it is one of these kernels. It has rank zero or one if and only if it is in the intersection of any two (and hence all three). Sławomir Biały (talk) 19:45, 11 September 2011 (UTC)