Question

The most challenging mathematical puzzle of all time: Archimedes' Riddle...
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...

Answer 1

Complete text of the riddle:


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.

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.

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.

Answer 2

Have fun and see you in 100 years! LOL

Answer 3

this is in Diophantine analysis....

Answer 4

@Seratul Can you figure it out? :D I know it's \[7.76*10^{something}\]
it stumped thousands of people
this is so exciting I love problems that have practically never been solved. it's supposed to be a trick: it's got 2 answers conceived 2 different ways but only 1 of those is correct...

Answer 5

That's too much to read :'(

Answer 6

LOL!!!!!

Answer 7

I've ALMOST got it....

Answer 8

White bulls=\[\left( \frac{ 1 }{ 2 }+ \frac{ 1 }{ 3 }\right)\]black bulls+yellow bulls

Answer 9

Black Bulls= \[\left( \frac{ 1 }{ 4 }+\frac{ 1 }{ 5 } \right)\]
dappled bulls +yellow bulls

Answer 10

Dappled Bulls=\[\left( \frac{ 1 }{ 6 }+\frac{ 1 }{ 7 } \right)\] white +yellow bulls

Answer 11

White Cows=\[\left( \frac{ 1}{ 3 } +\frac{ 1 }{ 4 }\right)\] black herd

Answer 12

Black Cows=\[\left( \frac{ 1 }{ 4 } +\frac{ 1 }{ 5 }\right)\] dappled herd

Answer 13

Dappled Cows=\[\left( \frac{ 1 }{ 5 }+\frac{ 1 }{ 6 } \right)\] yellow herd

Answer 14

Yellow Cows=\[\left( \frac{ 1 }{ 6 }+\frac{ 1 }{ 7 } \right)\] white herd

Answer 15

yay that's over next step..

Answer 16

White bulls + black bulls = a square number,
Dappled bulls + yellow bulls = a triangular number.

Answer 17

let's write the number of white, black, dappled, and yellow bulls as W,B,D, and Y and the number of white, black, dappled, and yellow cows as w , b , d , and y...

Answer 18

???it creates a system of seven equations with eight unknowns!!! :(

Answer 19

50,389,082 is the total but it's not the answer! :D

Answer 20

the solution given above for the first part of the problem should be multiplied by
\[n=\frac{ \left( w ^{4658j}-w ^{-4658j} \right)^{2} }{ \left( 4657 \right)\left( 79072 \right) }\]

Answer 21

where\[w=300426607914281173363\sqrt{609} + 84129507677858393258\sqrt{7766}\]
and j is any positive integer

Answer 22

Equivalently, squaring w makes:
\[w ^{2}=u+v \sqrt{(609)(7766)}\] where (u, v) are fundamental solutions of the Pell Equation (which takes forever to explain, pardon me :D) which is \[u ^{2}-(609)(766)v ^{2}=1\]

Answer 23

If we put 1 in the place of j, this will make the smallest herd satisfying both equations. Let's skip this part....

Answer 24

*and satisfies both parts of the problem..

Answer 25

So the size of the smallest herd is.....



\[7.76*10^{206544}\]


There you go! :) tell me if I'm wrong :D

Answer 26

Don't try solving the exponent, you'll get an "error" on your calculator and the number won't fit in your house, much less on your tiny computer screen!!!

Answer 27

the answer is 42

Answer 28

???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????? 42 COWS???????

Answer 29

The life, universe and everything does include cows last time I checked.

Answer 30

there's no way its 42. Any calculator will tell u it's a number too big. u try it!!!

Answer 31

Too big with decimal, or too big whole number?

Answer 32

You could probably round if its a decimal number

Answer 33

10^206544= INVALID INPUT./thank you very much.

Answer 34

and no its an exponent.

Answer 35

oh just saw it. Geez thats a huge number

Answer 36

they've only calculated it once and i'm not even going to TRY posting it on here because you'd be reading #s for the rest of your life.

Answer 37

Haha I bet! I mean thats a huge exponent!

Answer 38

yeah and then multiply 7.76 to it!!!!! have fun. totally! I mean it.
XD

Answer 39

I think all my brilliant calculations are slowing down the page's loading time LOL

Answer 40

Nope not fun... :p

Answer 41

^lol

Answer 42

thanks for the medal bro I need those :D

Answer 43

haha same

Answer 44

lol this problem is soooo confusing.. It is too huge!

Answer 45

lol I took 8 days solving it bfor I posted it on here

Answer 46

lol yea it would probably take me 8 years...

Answer 47

Check it out:

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


I am perfect???? O.O And crowned with glory??? :O

Answer 48

give me a minute, I have to pick my jaw up off the floor XD

Answer 49

Lol bro you are a comedian haha

Answer 50

Thanks. everybody needs a good joke every now and then :D

Answer 51

Haha especially one dealing with math. :D

Answer 52

LOL