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The Cattle Problem

This is a famous problem of intermediate analysis, also known as ‘Archimedes’ Cattle Problem Puzzle’, sent by Archimedes to Eratosthenes as a challenge to Alexandrian scholars. In it one is required to find the number of bulls and cows of each of four colors, the eight unknown quantities being connected by nine conditions. These conditions ultimately form a Pell equation which solution is necessary in case of finding the answer of the puzzle. The Greek puzzle is stated below with a little deviation. I have just tried to make the language simpler than the original, hope you’ll be able to grasp the puzzle easily.

O Stranger! If you are intelligent and wise, find the number of cattle of the Sun, who once upon a time grazed on the fields of an Island, divided into four groups (herds) of different colors, one white, another a black, a third yellow and the last dappled color.In each herd were bulls, mighty in number according to these proportions:

  • White bulls were equal to a half and a third of the black together with the whole of the yellow.
  • The black bulls were equal to the fourth part of the dappled and a fifth, together with, once more, the whole of the yellow.
  • The dappled bulls, were equal to a sixth part of the white and a seventh, together with all of the yellow.

So, these were the proportions of bulls, now the proportions of the cows were as following:

  • White cows were equal to the third part and a fourth of the whole herd of the black.
  • Black cows were equal to the fourth part once more of the dappled and with it a fifth part, when all cattle, including the bulls, went to pasture together. Now the dappled in four parts were equal in number to a fifth part and a sixth of the yellow herd.
  • Yellow cows were in number equal to a sixth part and a seventh of the white herd.

Keeping above conditions in focus, find the number of cattle of the Sun, giving separately the number of well-fed bulls and again the number of females according to each color. But come, this solution is not complete unless you understand  all these conditions regarding the cattle of the Sun:

  • When the white bulls mingled their number with the black, they stood firm, equal in depth and breadth. Number of bulls in a row were equal to the number of columns.
  • When the yellow and the dappled bulls were gathered into one herd they stood in such a manner that their number, beginning from one, grew slowly greater till it completed a triangular figure, there being no bulls of other colors in their midst nor none of them lacking.

Find the number of cows and bulls of each color separately.

Solution:

$ W$

= number of white bulls
$ B$ = number of black bulls
$ Y$ = number of yellow bulls
$ D$ = number of dappled bulls
$ w$ = number of white cows
$ b$ = number of black cows
$ y$ = number of yellow cows
$ d$ = number of dappled cows

The relations come as:

    •   $ W = (\frac{1}{2} + \frac{1}{3})B + Y$ The white bulls were equal to a half and a third of the black bulls together with the whole of the yellow bulls.
    • $ B = (\frac{1}{4} + \frac{1}{5})D + Y$ The black [bulls] were equal to the fourth part of the dappled bulls and a fifth, together with, once more, the whole of the yellow bulls
    •   $ D = (\frac{1}{6} + \frac{1}{7})W + Y$ The remaining bulls, the dappled, were equal to a sixth part of the white bulls and a seventh, together with all of the yellow bulls
    •   $ w = (\frac{1}{3} + \frac{1}{4})(B + b)$ The white cows were equal to the third part and a fourth of the whole herd of the black.
    •   $ b = (\frac{1}{4} + \frac{1}{5})(D + d)$ The black cows were equal to the fourth part once more of the dappled and with it a fifth part, when all, including the bulls, went to pasture together.
    •   $ d = (\frac{1}{5} + \frac{1}{6})(Y + y)$ the dappled cows in four parts [in totality] were equal in number to a fifth part and a sixth of the yellow herd.
    •  $ y = (\frac{1}{6} + \frac{1}{7})(W + w)$ the yellow cows were in number equal to a sixth part and a seventh of the white herd.

The arrangement on solving gives following relations in W,B,D,Y,w,b,d and y. which is a system of seven equations with eight unknowns. It is indeterminate, and has infinitely many solutions and form the following matrix:

6 -5 -6 0 0 0 0 0
0 20 -20 -9 0 0 0 0
-13 0 -42 42 0 0 0 0
0 -7 0 0 12 -7 0 0
0 0 0 -9 0 20 0 -9
0 0 -11 0 0 0 -11 30
-13 0 0 0 -13 0 42 0

Which yields the following solutions

W = 10,366,482k
B = 7,460,514k
Y = 4,149,387k
D = 7,358,060k
w = 7,206,360k
b = 4,893,246k
y = 5,439,213k
d = 3,515,820k

where $ k$ is an arbitrary constant, which can be equal to either 1 or 2 or 3 … etc. Again, from the second part of the problem:

White bulls + black bulls = a square number, $ W+B=10366482k +7460514k$= a square number. or $ W+B=17,826,996k$ =a square number$ 2 \cdot 2\cdot 3 \cdot 11 \cdot 29 \cdot 4657 k = \textrm{a square number}$ . Thus $ k$ atleast be $ 3 \cdot 11 \cdot 29 \cdot 4657 $ or in general be $ 3\cdot 11\cdot 29 \cdot 4657 \cdot r^2=4456749r^2$ where $ r$ is any integer. Again, Dappled bulls + yellow bulls = a triangular number.or, $ Y + D = \textrm{a triangular number}$ where triangular numbers are numbers of the form $ 1 + 2 + 3 + 4 + 5 + \ldots + m =\frac{m(m+1)}{2}$ . where $ m$ is some positive integer. Thus $ 4,149,387k + 7,358,060k =\frac{m(m+1)}{2}$ or $ 11,507,447k =\frac{m(m+1)}{2}$ . Putting $ k=4456749 r^2$ we have $ 11,507,447 \times 4,456,749 r^2 = \frac{m(m+1)}{2}$ or $ 102,571,605,819,606 r^2 = m(m + 1)$ . The problem is now to find the values of $ r$ and $ m$ that we can find the value of $ k$ and thus the solution of the problem.  
The computer generated answers for smallest solutions are  at my Pastebin Account.
Recently, Ilan Vardi of Occidental College (Los Angeles, California, USA) developed simple explicit formulas to generate solutions to the cattle problem.Click here to read his paper on the cattle problem.

References and Further Readings: Weisstein, Eric W. “Archimedes’ Cattle Problem.”  From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/ArchimedesCattleProblem.html Archimedes’s Cattle Problem http://en.wikipedia.org/wiki/Archimedes%27_cattle_problem Archemedes’s Cattle Problem http://math.nyu.edu/~crorres/Archimedes/Cattle/Statement.html The Archemedes’s Cattle Problem http://www.maa.org/devlin/devlin_02_04.html

A Yes No Puzzle

This is not just math, but a very good test for linguistic reasoning. If you are serious about this test and think that you’ve a sharp [at least average] brain then read the statement (only) below –summarize it –find the conclusion and then answer that whether summary of the statement is Yes or No.
[And if you’re not serious about the test …then read the whole post to know what the stupid author was trying to tell you. :-) ]
STATEMENT: If the question you answered before you answered the question you answered after you answered the question you answered before you answered this one, was harder than the question you answered after you answered the question you answered before you answered this one, was the question you answered before you answered this one harder than this one? YES or NO?
 

Answer:

The answer is YES.

In other words, we could restate the statement as:

If the question you answered before this one was harder than THIS ONE, was the question you answered before this one harder than THIS ONE.

That makes the answer obvious. 

Gowers’ blog is Blog of the Month for October 2011

Reader’s brain is variable. It changes according to what it read. :) I have changed the pattern of selection and style of writing about BLOG OF THE MONTH. At the beginning of August, I planned that I will select some blogs from the education blog-o-sphere and will award to appreciate them for their excellent work. I know these awards will probably never make a difference but hope too that they’ll keep their good works on. So, here is the list of my 10 most favorite blogs, one of which, Gowers’ Weblog, is my Blog of The Month, for October 2011.

Previous Month’s Results:

Disclaimer: Please note that this selection is personal and I have no affiliation with any organization. Your views are invited in form of comments. I have a huge list of other blogs at My Blogs Page. Have a look. If you have a very good blog which I’ve not noticed yet ; or want to provide feedback about this selection, please feel free to comment below.

Cameron Counts – Blog of the Month for August 2011

I announced to choose a blog from the education blogosphere as Blog of the Month. To complete this task, I googled for days, read them, analyzed them and now I have the winner of ‘Blog of the Month’.
This is the first month of this series and discussing article is made in hurry, so one can feel an emptiness and lack of interest in it. But believe, Blog of the month was not selected in hurry. I took quick looks on about 500 blogs and thousands of posts. I created a list of all blogs I read and rated them on behalf of their qualities, visitors, content, language etc. From the list of 513 blogs, the shortlisted blogs were:

  1. What’s New (math)
  2. Gödel’s Lost Letter and P=NP(Math and Computer Science)
  3. Peter Cameron’s Blog(math)
  4. Let’s Play Math(math)
  5. Unapologetic Mathematician(math)
  6. Cock Tail Party Physics(Physics)
  7. WordPress Tips(Blogging)
  8. Honglang Wang’s Blog (Math and Programming)
  9. The GeomBlog(CS)
  10. Republic Of Mathematics (Math and Media)

I count a lot of things that there’s no need to count. Just because that’s the way I am. But I count all the things that need to be counted.

And Yes! The blog of the month is Peter Cameron’s Blog with useful content, interactive language and multidimensional approach to mathematics.
count1.jpg

About Peter Cameron’s Blog

Peter Cameron is a professor of mathematics in London and he writes about math, media and education at http://cameroncounts.wordpress.com. He mingles everything with math, like poetry – media – fun and internet. His blog is full of expositories, problems and results, Posts about doing – playing and learning mathematics, Poetry, Events Talks and Conferences, typesetting and Mathematics in Media. A list of categorized posts can be found here.

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