Archimedes was the greatest mathematician of antiquity. He was also a lover of puzzles, which he would devise and pose to his contemporaries. This week, we present you with two versions of what is arguably Archimedes' most challenging puzzle ever.

#### Art by Jim Cooke

I am speaking of Archimedes' Cattle Problem. Said puzzle appears below, in two parts. Archimedes originally posed the complete, two-part version to his friend Eratosthenes, a great mathematician, poet, geographer, and astronomer who at one time served as chief librarian at the Library of Alexandria. Eratosthenes was a brilliant man, but it is unclear whether he ever solved the puzzle. What we do know is that when a manuscript containing the problem was discovered in 1773, the solution was not immediately found. In fact, it was not until more than a century later, in 1880, that mathematician A. Amthor would devise a general solution.

Needless to say, it is a very difficult puzzle.

For this reason we have presented the puzzle in two parts. Make no mistake, the first half of the puzzle is still quite challenging. It is, however, markedly more accessible. The complete version of the riddle includes the conditions outlined in the second and third paragraphs. It involves some (probably intentional) ambiguity, and also much, *much* bigger numbers.

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## Sunday Puzzle #22: Archimedes' Cattle Problem

*PART I*

*If thou art diligent and wise, O stranger, compute the number of cattle of the Sun, who once upon a time grazed on the fields of the Thrinacian isle of Sicily, divided into four herds of different colors, one milk white, another glossy black, the third yellow and the last dappled. In each herd were bulls, mighty in number according to these proportions: Understand, stranger, that the white bulls were equal to a half and a third of the black together with the whole of the yellow, while the black were equal to the fourth part of the dappled and a fifth, together with, once more, the whole of the yellow. Observe further that the remaining bulls, the dappled, were equal to a sixth part of the white and a seventh, together with all the yellow. These were the proportions of the cows: The white were precisely equal to the third part and a fourth of the whole herd of the black; while the black 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. Now the dappled in four parts were equal in number to a fifth part and a sixth of the yellow herd. Finally the yellow were in number equal to a sixth part and seventh of the white herd. If thou canst accurately tell, O stranger, 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, thou wouldst not be called unskilled or ignorant of numbers, but not yet shalt thou be numbered among the wise.*

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*PART II*

*But come, understand also 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, and the plains of Thrinacia, stretching far in all ways, were filled with their multitude. Again, 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 colours in their midst nor none of them lacking. *

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*If thou art able, O stranger, to find out all these things and gather them together in your mind, giving all the relations, thou shalt depart crowned with glory and knowing that thou hast been adjudged perfect in this species of wisdom.*

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*We'll be back next week with the solution β and a new puzzle! Got a great brainteaser, original or otherwise, that you'd like to see featured? E-mail me with your recommendations. As always, be sure to include "Sunday Puzzle" in the subject line!

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## SOLUTION To Sunday Puzzle #21: Slicing Pie

Last week, I asked you to determine the maximum number of pieces you could produce with six straight cuts through a pie.

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Several of you supplied a correct solution, but I believe the first was commenter Tauromachy. (Reader Stephen M. was the first to reply correctly via e-mail, albeit five minutes after Tauromachy). I've summarized their solutions, with additional explanations, below.

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One cut down the middle gives you two pieces. A second cut crossing the first gives you four. But a third will give you as many as seven slices. With pen and paper, it's pretty easy to come up with the maximum number of pieces achievable by four cuts (11), but any more than that and the drawing can get pretty messy. (Perhaps I spoke too soon; a reader by the name of BJ manage to draw the solution rather clearly β and on an iPhone, no less!) Already, however, we can see a pattern emerging: **Each cut adds to the total number of pieces the number of the cut. **Martin Gardner explains why, in *Entertaining Mathematical Puzzles*:

Consider, for example, the third cut. It crosses two previous lines. Those two lines will divide the third line into three sections. Each of those three sections cuts a piece of pie into two parts, so each section will add one extra piece and the three sections naturally add three pieces. The same is true of the fourth line. It can be drawn so it crosses the other three lines. These three lines will divide the fourth line into four sections. Each section adds an extra piece so the four sections will add four more pieces. And the same is true of the fifth line, sixth line, and so on for as many lines as we care to add.

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So, for instance: An uncut pie is one piece. On cut number one, you add 1 to the total number of pieces. 1+1 = 2.

On the second cut, you add 2 to the total number of existing pieces. 2+2 = 4

On the third cut, you add 3 to the total number of existing pieces. 4+3 = 7

Fourth cut: 7+4 = 11

Fifth cut: 11+5 = 16

Sixth cut: 16+6 = 22

So the answer is 22! It's a classic case of induction.

But what if you wanted to know the number of pieces that could be created by a really big number of cuts? We could brute force it according to our little rule β but the solution can actually be expressed mathematically. For that, see this pleasant exchange between commenters hawkingdo and EndlessMe.

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