Can You Solve One Of Archimedes’ Most Challenging Puzzles?

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 gift you with two interlingual rendition of what is arguably Archimedes ’ most challenging puzzle ever .

I am speaking of Archimedes ’ Cattle Problem . Said puzzler appear below , in two character . Archimedes in the beginning bewilder the complete , two - part version to his Quaker Eratosthenes , a majuscule mathematician , poet , geographer , and astronomer who at one time served as master bibliothec at the Library of Alexandria . Eratosthenes was a brilliant man , but it is indecipherable whether he ever solved the puzzle . What we do jazz is that when amanuscript containing the problem was strike in 1773 , the solution was not immediately detect . In fact , it was not until more than a century later , in 1880 , that mathematician A. Amthor would invent a general resolution .

uncalled-for to say , it is a very difficult puzzle .

For this ground we have show the puzzle in two parts . Make no error , the first half of the puzzler is still quite intriguing . It is , however , markedly more approachable . The complete variation of the brain-teaser includes the conditions outlined in the 2nd and third paragraphs . It involves some ( probably designed ) ambiguity , and also much , much bountiful routine .

Sunday Puzzle #22: Archimedes’ Cattle Problem

PART I

If thou prowess diligent and wise to , O alien , cypher the number of cows of the Sun , who once upon a time graze on the arena of the Thrinacian islet of Sicily , divide into four herds of dissimilar colors , one Milk River snowy , another glossy Shirley Temple , the third yellow and the last dappled . In each herd were bulls , mighty in number grant to these proportions : Understand , unknown , that the clean bulls were equal to a one-half and a third of the black together with the whole of the chicken , while the black were equal to the fourth part of the dappled and a fifth , together with , once more , the whole of the yellowish . take note further that the remaining bulls , the cloud , were equal to a sixth part of the white and a 7th , together with all the yellow . These were the proportions of the cows : The white were precisely adequate to the third part and a fourth of the whole ruck of the black ; while the black were equal to the fourth part once more of the dappled and with it a 5th part , when all , including the bulls , go to crop together . Now the mottled in four parts were equal in telephone number to a fifth part and a sixth of the yellow-bellied herd . lastly the chickenhearted were in bit equal to a sixth part and 7th of the livid herd . If thou canst accurately tell , O alien , the identification number of Bos taurus of the Sun , have on an individual basis the numeral of well - feast bulls and again the identification number of females consort to each color , thou wouldst not be called amateurish or illiterate of numbers , but not yet shalt thou be come among the wise .

PART II

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But come , understand also all these conditions regarding the cattle of the Sun . When the white bulls jumble their figure with the black , they stood firm , equal in depth and breadth , and the plain of Thrinacia , extend far in all ways , were filled with their hoi polloi . Again , when the yellow and the mottled bulls were gathered into one herd they stood in such a manner that their number , beginning from one , grow slowly great till it completed a triangular figure , there being no bulls of other colours in their midst nor none of them lacking .

If thou art able , O stranger , to find out all these things and gather them together in your intellect , giving all the recounting , thou shalt depart crowned with glory and be intimate that thou hast been adjudge perfect in this species of wisdom .

We ’ll be back next hebdomad with the answer – and a new teaser ! Got a great brainteaser , original or otherwise , that you ’d like to see featured?E - post me with your recommendation . As always , be sure to let in “ Sunday Puzzle ” in the subject line !

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SOLUTION To Sunday Puzzle #21: Slicing Pie

Last hebdomad , I asked you to specify the maximal number of pieces you could producewith six unbent cuts through a Proto-Indo European .

https://gizmodo.com/this-weeks-puzzle-is-as-easy-as-pie-1687313362

Several of you cater a correct solution , butI consider the first was commenter Tauromachy . ( Reader Stephen M. was the first to reply correctly via e - ring armour , albeit five minutes after bullfighting ) . I ’ve sum their solution , with additional explanations , below .

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One cut down the midsection give you two pieces . A 2nd cut hybridise 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 drafting can get pretty messy . ( Perhaps I spoke too soon ; a reader by the name of BJ manage to puff the solution rather clearly – and on an iPhone , no less ! ) Already , however , we can see a pattern issue : Each gash adds to the total identification number of opus the number of the baseball swing . Martin Gardner explains why , inEntertaining Mathematical Puzzles :

deliberate , for representative , the third cutting . It get across two previous transmission line . Those two lines will divide the third line into three sections . Each of those three sections cut a bit of pie into two parts , so each segment will append one extra piece and the three sections course add three pieces . The same is true of the 4th line . It can be cast so it crosses the other three lines . These three lines will divide the fourth line into four sections . Each section add an supererogatory objet d’art so the four section will add four more musical composition . And the same is true of the fifth line , sixth parentage , and so on for as many lines as we care to tally .

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

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On the 2d stinger , you add 2 to the total number of existing slice . 2 + 2 = 4

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

Fourth cut : 7 + 4 = 11

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Fifth cut : 11 + 5 = 16

Sixth track : 16 + 6 = 22

So the answer is 22 ! It ’s a classic case of elicitation .

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But what if you want to roll in the hay the number of pieces that could be created by a really big turn of cuts ? We could brute force it according to our little prescript – but the solution can actually be evince mathematically . For that , seethis pleasant exchange between commenters hawkingdo and EndlessMe .