Mathpuzzle.com

I've made a few corrections, and posted the answer and solution for 57|x³ - 3y³| < x. Answer. You can also see the answer at the Mathematica Information Center. I extended the problem to 13874|x³ - 3y³| < x. The solution values have 1781 digits. I'd like to see other interesting cases of Diophantine Inequalities, if anyone can find any.

Theo Gray has some aluminum foil, a blender, a bucket of rust, and a sparkler, so we're going to try to set up a thermite reaction. Later: It worked out well, but we needed a bit of a aluminum dust to start up the blender-ground aluminum. In a pinch, we could have used an Etch-a-Sketch for the aluminum starter. You can see Theo's Thermite video.

R. William Gosper: Last month I had the pleasure to visit the Microsoft Theory Group. Mike Sinclair and Gary Starkweather in a neighboring group generously helped me laser-cut a puzzle consisting of a 8" diameter circular tray and twelve unequal circular disks with diameters {1.14662, 1.24041, 1.33278, 1.52265, 1.65921, 1.74812,
1.83679, 2.36627, 2.54162, 2.63057, 2.73347, 3.03309 ©2003 Bill Gosper}. To my amazement, *no one* has yet succeeded in fitting all twelve disks back into the tray, although Christian Borgs, a Microsoft physicist, solved two slightly easier versions. This puzzle is fairly resistant to computer solution--I'm not even sure how to quantify the combinatorics of the search.

The Catalan sequence comes out of the Mersenne Primes. C0 = 2, C1 = 2^C0-1 = 3, C2 = 2^C1-1 = 7, C3 = 2^C2-1 = 127, C4 = 2^C3-1 = 170141183460469231731687303715884105727. So far, all are prime. Is C5 = 2^C4-1 a prime number? Landon Curt Noll, co-discoverer of the 25th and 26th Mersenne primes, has proved that any factors of C5 have at least 50 digits. Harry Nelson, co-discoverer of the 27th Mersenne, is now a puzzle designer. Incidently, M21-M23 were all discovered here in Urbana IL. Bryant Tuckerman, who found M24, also developed flexagons with Richard Feynman, and was the subject of Martin Gardner's first column.

Some very striking new visual illusions have been found by Akiyoshi Kitaoka. One particularly effective illusion is his rabbit circles - a non-moving picture appears strongly animated.

Cihan Altay -- Odd Event: Locate non-zero digits into an 6x6 grid so that each even digit tells the number of odd digits in the neighbouring squares, and each odd digit tells the number of even digits in the neighbouring squares. Two squares are neighbours if they share an edge or a corner. Same digits can not be neighbours. a) Maximize the number of digits on the grid. b) Maximize the total of the digits on the grid. Any further anlayses from your visitors would be appreciated. Send answers. Without the same digits restriction, the best answer is in PQRST 4.

Jorge Luis Mireles has made many fascinating pictures of how different polyforms can be covered with the same shape. Patrick Hamlyn: Arrange 81 Y-pentominoes to make a side-9 Y-pentomino. Answer. Alternately, arrange 64 solid Y-pentominoes to make a side-4 solid Y-pentomino. Answer.

How do you brace a unit-side pentagon with non-overlapping unit rods? Martin Gardner mentioned that an answer had been found, and Gred Frederickson tracked the answer down for Hinged Dissections: Swinging and Twisting. T H O'Beirne's pentagon can now be seen at Mathworld.

Michael Trott brought to my attention two beautiful new proofs for the infinitude of primes. Proof #1: Pi<4. Proof #2: Pi^2 is not an integer. Further details on why these are proofs can be seen in Super-regularization of Infinite Products, by Garcia Munoz and Marco Perez.

At work, I've lately been using the Bayes Junk Tool, Mozilla Mail, and Mathematica to analyze spam. Just a small project , but fun, using mathematics as a flamethrower against annoying spammage. The basic outline can be seen in A Plan for Spam. If curious, feel free to write me there at edp@wolfram.com. If you already use Mozilla's spam filter, I'm especially interested in seeing false positives -- valid messages that accidently get flagged as spam.

I had a wonderful time at the 23rd International Puzzle Party in Chicago. Part of it was the Puzzle Design Competition, which had many delightful entries. The Binary Burr mechanically adapts the Towers of Hanoi, which you can order from Bill Cutler. I presented Patrick Hamlyn's 30-60-90 Partridge Puzzle, and you can order that (and other puzzles) from craftsman Walter Hoppe. I picked up a few great puzzles by woodmaster Tom Lensch, miniature master Allan Boardman, LiveWire Puzzles, PuzzleCraft, Kadon Enterprises, and Cleverwood. Kate Jones also had some extra Partridge Puzzles that Erich Friedman had commissioned (my favorite is the long rectangle). Oskar van Deventer, with some advice from Prometal and Bathsheba Grossman, made some knotted gears in printed brass, you can buy one from his first production run for $85 (to email him, M.O.vanDeventer with planet.nl). It's a limited edition object -- the order deadline is 31 August 2003.

Mosaic expert Ken Knowlton made a variety of to his Geometric Pieces, including Solomon Golomb in pentominoes, Bob Wainwright as a Partridge puzzle, and Will Shortz as a crossword. If you'd like to try out the crossword, you can see it at crosswordtournament.com.

I picked up Derrick Schneider's winning design from last year -- put three 2x2 squares together so that they are joined together by half-sides. There are four ways to do this. Now, pack the 4 pieces into an 8x8 square, and into a 7x9 rectangle. Each solution is unique. The tray pictured below is two-sided. If you would like one of Derrick's puzzles for $17, you can email him at (derrick with misfit.com). As a puzzle of the week, try making it yourself and solving it both parts. Answer. Answer2. Solvers.

There are 140 different magic tours of the chessboard by a knight, but none is diagonally magic. You can read all about it under Eric Weisstein's Mathworld News. In a slightly related note, at the following site you can see 20 TwixT puzzles. Jean-Charles Meyrignac solved Serhiy's non-crossing leaper tour problem from last week, and found the optimal solution of 17 moves (I missed it, too). He also looked at the octagonal board, which has a non-crossing tour of length 18. Can you find it? Answer1. Answer2. Solvers. Jean-Charles also noted a unique non-crossing knight tour of length 19 in the English Solitaire board, and length 20 on French Solitaire.

At a site devoted to brilliant numbers, I learned that 10¹⁰⁵ + 4293 is the product of the following two numbers:

Over 400 years ago, Kepler hypothesized that the densest way to pack spheres was the face-centered cubic packing. In 1998, Thomas Hales proved this conjecture, but the referees couldn't quite give a 100% endorsement that the proof was correct. As a result, the Flyspeck Project was started, to provide a computer verification of the proof. Nature wrote an article about it. In another packing result, some principles developed by John Conway and NJA Sloane turned up in real-life materials. For a different aspect of sphere packing, consider the Waterman Polyhedra.

For anyone doing research in Combinatorial Games, a new resource is the Combinatorial Game Suite. In the book On Numbers and Games, John Conway describes a way of assigning values to many different types of games, which winds up having greater depth than the normal number system. This is explored further in Winning Way I, II, III, and IV.

Serhiy Grabarchuk noticed similarities between the Matchstick Snakes and non-crossing leaper tours. He came up with the board below, and found a length-16 path with a knight that doesn't cross itself. Can you find it? (Jean-Charles Meyrignac found a length 17 answer). For an inappropropriate use of knights, consider DARPA's self-healing minefield. (Laying netting down and exploding just one would allow for easy removal, once all of the mines had leapt and self-entangled.) The length 16 path is given by Joseph DeVincentis.

A €1000 prize is offered to the winner of the upcoming Amazon tournament. Jenazon Cup 2003 will be held in September, and will pit the best human and computer teams against each other. You can read about previous Jenazon competitons, the Amazong applet, or a page about the game. Anyone may apply (deadline 31 August, 2003), but amongst all applicants at most 16 teams will be permitted.

Several people have asked me about my favorite screen saver ... it's currently ant.scr by Peter Balch. It makes Turmites. I found it at KidsFreeware.

Pascal's Triangle is closely related to the Binomial coefficient. I was doing a search center column related primes ( Do[k = -1;While[k = k + 1; Not[PrimeQ[1 + k Binomial[2 n, n]]]]; {n, k} >>> "binom.txt", {n, 1, 10000}]; ) and got some interesting data. In that {n,k}data, k Binomial[2 n,n] + 1 is prime. I noticed a lot of cases where k=n-1, so I looked at numbers where (n-1)Binomial[2n,n]+1 was prime: 2 3 4 5 6 7 9 13 17 18 22 23 28 31 48 49 52 80 99 167 201 295 372 381 391 638 653 720 779 887 1047 1454 1647 1719 2405 3234 3257 3542 3623 3765 3796 4337 4490 5228 6507 8544 9990 10000. With n=10000, one gets a 6023 digit prime.

Now, in the last paragraph I say that I found a 6023 digit prime, but that is based on Mathematica's PrimeQ. From the Help Browser: "PrimeQ first tests for divisibility using small primes, then uses the Miller-Rabin strong pseudoprime test base 2 and base 3, and then uses a Lucas test. As of 1997, this procedure is known to be correct only for numbers less than 10^16, and it is conceivable that for larger a it could claim a composite number to be prime." So, I don't have a proof my number is prime. For certain numbers, a proof of primality is relatively easy. For more general numbers, primality proving is much more difficult. In Mathematica, there is ProvablePrimeQ, but it is much slower than PrimeQ. For the Binomial primes above, ProvablePrimeQ[294Binomial[590,295] - 1] takes 45 seconds. PrimeQ[] for the same number takes .06 seconds. There is also the Primo program by Marcel Martin, which was recently used to prove that (32*10^6959 - 23)/99 is prime.

Circles can be arranged to touch each other in a square matrix in an obvious way. Mike Christie asked me for the best way to pack the shapes between the circles, and I suggested something like the following. He later sent me an analysis of the best packing density this method could get. Is there a better method?

George Jelliss has made a page about non-intersecting leaper tours. Finding the best non-intersecting tour for a (2,3)-leaper on a 10x10 board seems like it would be very difficult to solve by hand.

A good program for looking at polyhedra is Stella. Robert Webb decided to extend that to 3D Minesweeper, for those who prefer playing platonically. Slightly different is the Slidy Rubik's Cube.

Serhiy Grabarchuk extended some of results on Erich Friedman's Snakes page. In particular, he looked for the longest snake on a unit cube.

Eric Solomon's Hexagrams is similar to a set of pieces Oskar van Deventer developed in 1981. Kate Jones (gamepuzzles.com) calls them Fjords. Oskar: "I am still looking for solutions for some interesting shapes, like the 5x5 diamond shape, the 7x7 triangle (omitting one piece with six-fold symmetry), the 3x3 hexagon and other shapes. As Kate says, a puzzle can only be brought to the market if there exist interesting challenges that have solutions. The attached drawing presents the full set of pieces, ready to cut out and play with. Perhaps some mathpuzzle readers would like to give it a try?" You can contact M.O.vanDeventer by appending planet.nl, if you get some nice solutions

Solve for integers x and y: 57|x³ - 3y³| < x. (This isn't recommended for hand solving.) Answer. Vaguely related: (35^(1/3)+1/11)³=38, almost.

Cihan Altay: The PQRST 06 puzzle competition will start on July 19th Saturday at 20:00 (GMT+02). There are 10 puzzles to be solved and rated in 7 days.

Taprats is an applet for Islamic star patterns. I read up on XI HyPErONs, for some reason.

Here's a toughie ... identify the number sequences I have pictured below. The Online Encyclopedia of Integer Sequences might help. Your choices are Recaman Sequence, Gray code, the Primes, Binary, Ternary, the Primes, Zeckendorf's Fibonacci representation, and 5n.

In 1838, Dirichlet showed that the average number of divisors for all numbers 1 to n approaches Log[n] + 2 Euler - 1. If you look at the average number of divisors for all complex integers 1+i to n+i, the function is far more chaotic.

Pascal's triangle (at Eric Rowland's page, for example), has a lot of interesting properties. If you look at all the entries Mod 2 (that is, the evens go to zero, and the odds go to 1), and then look atthe result as a series of Binary numbers, the first 31 rows read as follows: 1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295. Why is that remarkable? Send Answer (if you didn't know).

In the March 1978 issue of Games&Puzzles, J O P Sweeney wrote a letter about a dice trick he'd purchased many years earlier. The five dice are labeled 186-285-384-483-681-780, 179-278-377-773-872-971, 168-366-564-663-762-960, 459-558-657-756-855-954, 147-345-543-642-741-840. One rolls the five dice, and immediately announces the sum of the five topmost numbers. Of course, there's a trick to it. What is it? Answer and solvers.

In the Summer 1979 issue of Games&Puzzles, Eric Solomon (the inventor of Black Box) described a set of tiles he called Solomon's Hexagrams. They are pictured below. It's a fairly nice set of pieces. Two puzzles, both based on matching edges: 1. Make a figure with inverse symmetry, 2. Make a puppy. An applet for playing with these is at Eric Solomon's site. Andrea Gilbert sent these solutions. Clinton Weaver sent (1 2 3 4 5 6)

Eric Solomon is still an active game designer, and has recently implemented a version of Hexagonal Black Box. It's a very nice applet, well worth a look.

Oscar van Deventer: Thank you for bringing the 3D work by Bathsheba Grossman to my attention (and to all the MathPuzzle readers). Inspired by George Miller's talk on 3D printing, I started designing puzzles/objects to be 3D printed. My first design was a kinetic object (doodle): Knotted Gear. One attachment shows an animation of the object: two threefoil knots moving though each other like gears. It was just takes a matter of minutes to draw the basic design in Rhino. I used Rhino to optimize the sizes of the parts, such that they touch, but do not intersect. The other attachment shows the model that I ordered from a local 3D print shop for more than $100. The first thing I did with it was dropping it on the floor and breaking it. Those 3D plaster prints are rather fragile. Fortunately, I could glue the pieces. The object gears almost as smoothly as the animation. [Ed -- I'm very eager to see some of Oscar's works in printed bronze.] Also, Robert Abbott has updated his "Things that Roll" page with my proposal for for a mechanism to implement these puzzles.

Intrigued by the tune Tic-Toc-Choc, I came across the oddity Mxnxstrxndxsx, another work by the same composer.

$20 Challenge. I like the WireWorld automaton a lot, mainly because I can see what is happening. I met Brian Silverman last week at NKS2003, and heard from him that he invented it after getting frustrated with the Game of Life. He only had a 40×40 screen at the time, and designed WireWorld so that he could actually see something. One big item in Cellular Automata research is proof of universality, which basically boils down to proving that a given system can emulate a Turing machine. Various systems have been proven to be equivalent, but they usually have a size that makes them difficult to watch. For example, Paul Rendell's system with Game of Life. I'm certain that WireWorld can be used to make tape-like cells that mimic Turing machines. If you can make one, please let me know. MCell might be useful.