This week's puzzle poses an apparent paradox. Given ten self-referential statements, your task is to determine which of statements are true, and which are false.

### Sunday Puzzle #23: Statements of Truth

This week's puzzle is presented at the top of the post. I will repeat it here. Above, there are ten numbered statements. Which of them is/are true? What is your reasoning?

We'll be back next week with the solution â€“ 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!

### Solution to Sunday Puzzle #22: Archimedes' Cattle Problem

Last week, I challenged you to take on one of Archimedes' most challenging puzzles. As many of you quickly deduced, Archimedes' Cattle Problem is, in fact, a math problem. You might be of the same mind as commenter pyronius, who, upon recognizing the arithmetic nature of last week's puzzle, remarked:

**"Yo, archie. disguising math as cows does not a puzzle make."**

You and pyronius would be wrong. Many of the greatest puzzles are in fact math problems in disguise. That said, I understand the frustration over this particular puzzle. Partly because the language of the problem itself can be difficult to parse (this thread, initiated by commenter hawkingdo, does a good job of translating the puzzle into some workable numbers); but also because Archimedes' Cattle Problem â€“ the second part, especially â€“ is a real doozy. As with most multi-step problems, solving it requires a careful, organized approach.

#### PART I SOLUTION

The first person to provide not only a correct solution to the first part of the puzzle but an account of his work was commenter zachparker, who wrote (solutions appear in bold):

First things first, I converted everything to equations, like so:

W = 5/6B + Y

B = 9/20D + Y

D = 13/42W + Y

w = 7/12B + 7/12b

b = 9/20D + 9/20d

d = 11/30Y + 11/30y

y = 13/42W + 13/42

This gives 7 equations for 8 variables, so I threw in an extra one:

W = 1

which would yield all the other variables as a proportion of W.

Seeing as I really didn't feel like doing all the algebra by hand, I recalculated each of the above equations to have a 0 to the right of each =, and built a couple of matrices, an 8x8 for the left sides and an 8x1 for the ride sides, and plugged the matrices into my trusty TI-89 (I tried it in Excel first, and got a bunch of unusable decimals, and I wanted to keep my numbers nice and fractional in order to find an LCD).

Running M1^-1 * M2 got me the following results:

W= 1

B = 267/371

Y = 297/742

D = 790/1113

w = 171580/246821

b = 815541/1727747

y = 1813071/3455494

d = 83710/246821

Since I knew 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:

W = 10,366,482

B = 7,460,514

Y = 4,149,387

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 lends itself to analysis by what's known in linear algebra as a coefficient matrix. Such matrices are easily resolved with a program, or, in zachparker's case, a fancy calculator. But in the absence of these tools, the problem's seven equations and eight unknowns can also be solved for by hand. Commenter IrkedIndeed did exactly that, and came up with the same solution as zachparker. You can read IrkedIndeed's approach here, but a cleaner presentation of the necessary steps is provided in David Wells' *Book of Curious and Interesting Puzzles*:

#### PART II SOLUTION

Calculating the solution to the second half of the problem is considerably more difficult. As I mentioned last week, the first person to solve Archimedes' complete problem in a satisfactory way was A. Amthor, who, in 1880, showed that the smallest solution to the cattle problem requires a total number of cattle given by a number of 206545 digits. But Amthor did not give all 206545 digits, and mathematician Hendrik Willem Lenstra, Jr. notes that**, **of the four leading digits Amthor provided (7766), the fourth was actually wrong, due to insufficiently precise calculations

A full solution would come until 1965. Write Harold Alkema and Kenneth McLaughlin, of the University of Waterloo:

In June of 1965, Gus German, Robert Zarnke, and Hugh Williams solved the Archimedes Cattle Problem. No complete solution to the problem had ever been devised by mathematicians in over 2000 years. 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 researchers' findings were published in *Mathematics of Computation**.*

In 1981, Harry L. Nelson of the Lawrence Livermore National Laboratory produced the full, 206545-digit answer on forty-seven computer-printed pages, in *The* *Journal of Recreational Mathematics:*

In abbreviated form, it reads 77602714... 237983357... 55081800, each of the six dots representing 34420 omitted digits.

A more detailed history of Archimedes' Cattle Problem and his continued influence in the 21st Century, including links to several relevant publications on the subject, can be had at the website of Chris Rorres, Professor Emeritus of Mathematics at Drexel University.

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