This week ’s puzzle set an unmistakable paradox . give ten self - referential statements , your task is to set which of statements are truthful , and which are false .
Sunday Puzzle #23: Statements of Truth
This hebdomad ’s mystifier is presented at the top of the situation . I will duplicate it here . Above , there are ten numbered financial statement . Which of them is / are dependable ? What is your logical thinking ?
We ’ll be back next week with the solution – 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 trusted to include “ Sunday Puzzle ” in the subject line !
Solution to Sunday Puzzle #22: Archimedes’ Cattle Problem
Last week , I challenged you to take onone of Archimedes ’ most challenging mystifier . As many of you quickly deduct , Archimedes ’ Cattle Problem is , in fact , a math trouble . You might be of the same mind as commenter pyronius , who , upon realise the arithmetic nature of last calendar week ’s mystifier , observe :
“ Yo , archie . disguise math as cows does not a puzzler make . ”
You and pyronius would be improper . Many of the greatest puzzles are in fact maths problems in camouflage . That said , I understand the frustration over this particular mystifier . Partly because the language of the trouble itself can be hard to parse ( this screw thread , initiated by commenter hawkingdo , does a good caper of translate the teaser into some viable numbers ) ; but also because Archimedes ’ Cattle Problem – the second part , especially – is a real doozy . As with most multi - step problem , solving it requires a careful , unionized plan of attack .

The first somebody to provide not only a correct solution to the first part of the mystifier but an write up of his work was commenter zachparker , who wrote(solutions appear in bold ):
First thing first , I converted everything to equivalence , like so :
W = 5/6B + Y

bacillus = 9/20D + yttrium
D = 13/42W + Y
w = 7/12B + 7/12b

b = 9/20D + 9/20d
calciferol = 11/30Y + 11/30y
y = 13/42W + 13/42

This gives 7 equation for 8 variables , so I throw in an surplus one :
W = 1
which would grant all the other variable as a dimension of W.

catch as I really did n’t experience like doing all the algebra by helping hand , I recalculate each of the above equivalence to have a 0 to the rightfulness of each = , and built a dyad of matrices , an 8×8 for the left over side of meat and an 8×1 for the drive sides , and plugged the matrices into my trusty TI-89 ( I tried it in Excel first , and make a clustering of unusable decimals , and I wanted to keep my numbers nice and fractional so as to find an LCD ) .
Running M1 ^ -1 * M2 capture me the following results :
W= 1

Bel = 267/371
Y = 297/742
D = 790/1113

w = 171580/246821
b = 815541/1727747
y = 1813071/3455494

d = 83710/246821
Since I roll in the hay each of these numbers were as a fraction of W , I found the LCD of all the fractions ( 1036648 ) and multiplied it by each fraction , to get the following quantities :
double-u = 10,366,482

barn = 7,460,514
yttrium = 4,149,387
vitamin D = 7,358,060

w = 7,206,360
b = 4,893,246
y = 5,439,213

d = 3,515,820
As zachparker notes , this problem lend itself to analysis by what ’s known in linear algebra as a coefficient intercellular substance . Such matrices are easily resolved with a programme , or , in zachparker ’s grammatical case , a fancy figurer . But in the absence seizure of these dick , the trouble ’s seven equations and eight unknown quantity can also be solved for by hand . Commenter IrkedIndeed did exactly that , and came up with the same answer as zachparker . you could register IrkedIndeed ’s approach here , but a fair presentation of the necessary steps is ply inDavid Wells’Book of rummy and Interesting Puzzles :
Calculating the resolution to the 2d half of the problem is substantially more hard . As I mentioned last week , the first somebody to lick Archimedes ’ consummate problem in a satisfactory way was A. Amthor , who , in 1880 , showed that the small answer to the cattle problem take a full act of cattle give by a number of 206545 figure . But Amthor did not give all 206545 figure , andmathematician Hendrik Willem Lenstra , Jr. notesthat , of the four head digits Amthor provided ( 7766 ) , the fourth was in reality untimely , due to insufficiently accurate calculations

A full solution would come until 1965.WriteHarold Alkema and Kenneth McLaughlin , of the University of Waterloo :
In June of 1965 , Gus German , Robert Zarnke , and Hugh Williams solve the Archimedes Cattle Problem . No complete resolution to the problem had ever been devised by mathematicians in over 2000 class . The problem was first proposed by the Greek mathematician Archimedes of Alexandria in 200 BC . The UW team used a combination of the IBM 7040 and 1620 computers to produce the solution which had over 200,000 digits .
The research worker ’ findings were published inMathematics of Computation .

In 1981 , Harry L. Nelson of the Lawrence Livermore National Laboratory create the full , 206545 - finger answer on forty - seven electronic computer - printed pageboy , in The Journal of Recreational Mathematics :
In abbreviate bod , it read 77602714 … 237983357 … 55081800 , each of the six dots representing 34420 overlook digits .
A more detailed history of Archimedes ’ Cattle Problem and his continued influence in the twenty-first Century , include data link to several relevant publications on the matter , can be had at the website ofChris Rorres , Professor Emeritus of Mathematics at Drexel University .

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