Can a 3x3 magic square be made with different proper fractions so that all denominators are 9 or less? Warren Phillips and Nick Gardner found a solution to this. I sent their solution on to Will Shortz, who commented, "What a clever extension of my NPR puzzle! Very nice. Thanks for sharing it. If I hadn't already taped tomorrow's show, I might have mentioned this on the air." As a hint, all rows, columns and main diagonals add to 3/2 (ignore the typo I had earlier). Send answer.

With all the other web problems, I've been amazed at how busy my firewall has been. Zonealarm is free and effective.

Bryce Herdt's Mars puzzle was very popular. Here are the solvers, and the answer.

Rod Bogart, Joseph DeVincentis, and Darrel C Jones pointed out that the answer I posted for Scott Purdy's chess puzzle was incorrect. Darrel sent this corrected analysis.

Next week, I'll be in Japan for the International Puzzle Party.

I was very impressed by Andrew Clarke's new polyform collection.

Five X pentominoes (shaped like a +) can be fit on a toroidal 5x5 square. One property is that all 5 pieces touch each other. There are at least two hexominoes that can be placed on a toroidal 6x6 board so that all 6 pieces touch each other.

Bryce Herdt sent me a number of good suggestions and puzzles. He notes that MARS on a telephone pad is 6277. There are 14 5-digit multiples of this number. The following grid can be divided into pentominoes so that each pentomino has the digits of one of these multiples.

The Mars puzzle, by Bryce Herdt. Send Answer.
{12554, 18831, 25108, 31385, 37662, 43939, 50216, 56493, 62770, 69047, 75324, 81601, 87878, 94155}

Full results and solutions for the International Math Olympiad are available. Wei-Hwa Huang solved two problems while assisting as a proctor for the competition. He didn't have a full allotment of time. Ulrich Voigt solved two. I only solved one, as did a few others. Interesting, tough problems.

A few years ago, I gave a passing explanation about how dice work (All objects will bounce around until they run out of energy, and as such can be modeled with matrices). Dan Murray liked my model, and has built a Dice Rolling Machine to test it, with 34560 rolls per day.

Samantha's Wallpaper Page is worth a visit. She's collecting various mathematical wallpaper designs.

Wei-Hwa Huang: "The answer to the chess puzzle is this . Very nice." Matt Elder: Very cool puzzle. Different, unique, and special, too! ::) Answer. Jay Stuler sent this analysis.

Can a 21-17-10 triangle be cut up to make a 15-14-13 triangle in such a way that the sides and areas of the pieces are integers? (Both have area 84)?

Andrew Clarke noticed there are 100 dom-sliced tridominoes. Can they make a 5x100 rectangle? Yes!

Space War, a video game where two rocket ship orbited a black hole while blasting each other, turned 40 this week. The applet I link to is much larger than the original program. Many of my favorite programs were very small, a rarity these days, except perhaps for Zillions. Muppet Labs recently explored tiny programs. Something crazier is the world's smallest compiler.

I have several independent computer proofs that 11 X pentominoes cannot be placed on an 8x8 toroidal board. Here is a Matlab program by Brian Galebach that proves it.

Will Shortz used a math puzzle on NPR this week. See it here. Solve it and send in an answer! I'm intrigued by fraction magic squares. It's easy to make a magic square with proper, positive fractions so that the lowest denominator is 10 (all rows, columns, main diagonals adding to 1.5). Is a lowest denominator of 9 possible?

Many years ago, Julius Caesar communicated with his generals via encrypted messages. He decided that he would shift each letter forward three spaces, and let his generals know this. To decrypt his messages, they would shift each letter three spaces back. More here. An internet search on caesar shift will bring up hundreds of websites. The easiest encryption to code and decode is ROT13. Each letter is shifted forward (or back) 13 spaces. One encoder/decoder is here. Rot13 has been a standard on the internet since 1980.

From an official criminal complaint:
"Dmitry Sklyarov "eBooks security – theory and practice": Security aspects of electronic books and documents, and a demonstration of how weak they are: "standard" PDF encryption, Rot13 (used by New Paradigm Resources Group, Inc.)..."
"Based on the foregoing, I believe Dmitry Sklyarov, employee of Elcomsoft and the individual listed on the Elcomsoft software products as the copyright holder of the program sold and produced by Elcomsoft, known as the Advanced eBook Processor, has willfully and for financial gain imported, offered to the public, provided, and otherwise trafficked in a technology, product, service, and device that is primarily designed or produced for the purpose of circumvention a technological measure that effectively controls access to a work protected under Title 17..."

At the moment, Dmitri is being held without bail in prison. In my opinion, imprisoning codebreakers and puzzle solvers is a bad idea. I much prefer the tack taken by places like rsa.org, which offers awards to anyone that can crack their encryption services.

While looking into all this, I found the ROT13 programming page, which lists ways to program ROT13 in many different computer language. Now that I have a Linux machine (using Mandrake 8), i've started writing my own programs again. I'm 37. My first computer was a Model 3 from Radio Shack. At the time, the main point of having a computer was to write programs. The salesman -- "It has 16K memory. You can upgrade to 48K, but you probably won't ever need it." My brothers and I wrote many programs. It's been fun getting back into that. There are several "Hello World" pages (hello1, hello2, hello3). I looked for a similar page for Tetris, but didn't find one.

Anagrams are covered wonderfully at the Anagrammy page. One recent thing I saw there was an amazing two word bad review of Kiss of the Dragon, which I won't reprint here.

On my packing problem from last week, Brendan Owen conformed there is no solution for placing 11 squares. Warren Phillips, Matthew Moss, and Matt Elder sent arguments that 10 is maximal, but none sent a solid proof. The problem is equivalent to putting X-pentominoes on an 8x8 toroidal board. If we assume the problem is possible, then we can checkerboard the surface, and make the assumption that six of the pentominoes cover black squares. This would leave 3 black squares left uncovered. It's just an observation, not a proof.

Massimo M sent me a rather nice game I haven't cracked yet. Players A and B alternately remove squares of sizes 1x1, 2x2, or 3x3 (only) from an 8x8 board. The last person to remove a square loses. Does player A have a winning move? Send answer.

In the above, I've put 10 squares on a toroidal 8x8 board, with each one centered over one of the smaller squares. It seems as if an 11th might be able to fit by shuffling the pieces around, but I haven't found it. I haven't been able to prove it impossible, either. An alternative way of stating this would be to put 11 knights on a toroidal board so that the pieces can be a knight's move apart, but no closer. If you can squeeze in another piece, or cleverly prove that it can't be done, please write me. Scott Purdy sent me the problem below. If you solve it, write me at ed@mathpuzzle.com .

For the puzzle of the week, I'll defer to the 2001 International Math Olympiad Problems. I just got my first look at them a few minutes ago, and they look more difficult than the year 2000 problems. I will accept answers until they get posted. So, if you see how to solve any of these, write me.

I was playing around with diophantine equations, and wondered how they applied to the Gaussian integers. These are complex numbers where both components are integers. Can new examples of Fermat-Catalan conjecture be found if Gaussian integers are used? Yes! Here are three examples I found: (8+5i)^2 + (5+3i)^3 = (1+2i)^7, (19+16i)^2 + (3+9i)^3 = (3+2i)^5, (20+9i)^2 + (1+8i)^3 = (1+i)^15. (Fermat-Catalan Conjecture: There are only finitely many integer solutions to a^x = b^y + c^z with coprime a, b, c and 1/x + 1/y + 1/z < 1. ). I tried, unsuccessfully, to find two Gaussian powers with a difference of one.

Jill Britton has several great sites about math. Most of the material deals with the study of patterns, something I've been fascinated in since third grade or so. As I recall, we were making paper snowflakes, and mine were the only six-sided snowflakes, initially, but soon everyone was doing it.

Karl Scherer, puzzle and game designer extraordinaire, has created and solved two chess tasks:
Part a) "Look at this curious chess position," said Holmes, "Each of White's sixteen pieces has a move which checkmates Black's king, the only black piece on the board!"
"Very interesting indeed," replied Watson, "and amongst all solutions to this problem it is the one with the least number of promoted pieces."
Can you find such a position? (Of course the position must be legal.)

Part b) "I wonder," Holmes continued after a while, "Whether there is a similar position with the additional property that each of White's sixteen pieces has exactly one single move which checkmates."
"Well, in this case one might need more than one black piece on the board." mused Watson.
Can you find such a chess position? How many promoted pieces do you need? How many black pieces do you need?
Answers are at Karl's site. Can you find them, or improve them? Also, he has construct two new retrograde analysis problems.

Michael Rios: Place the numbers 1 through 9 into a 3x3 grid so that as many perfect squares can be found in it in word-search fashion. For example:

1 2 3
4 5 6
7 8 9

Would only yield the 5 squares: 1, 4, 9, 25, and 36. (If you were lucky enough to have 169 in your grid, you'd also then have 961!) What is the maximum number of squares? Answer.

A few more computer animation links: Today's Science Tomorrow's Art, Animating Explosions & Brittle Fracture, Modeling of Fallen Snow, Escherization, The Digital Michelangelo Project, Virtual Terrain Project, Hollow, Fiat Lux.

Craig Kasper, one of the composer for the WPC has sent me another of his birthday puzzles. Send Answer.

When Stanley Kubrick was making his first film, he hustled chess at Central Park for spending money. His fascination with chess stayed with him throughout his life. You can see more about him during Stanley Kubrick: A Life in Pictures, which will show on Cinemax Tuesday the 19th. I've seen it, and I talked to Jan Harlan afterwards to ask about Evan Chan thing, but he told me he had nothing to do with that. After remembering Kubrick's intense love of chess and technology, I believe Stanley planned much of this game. www.aimovie.com has a fairly good AI chatbot.

The above scene (Shrek) has tons of cool math behind it. It got me looking at math graphics. Some of the more interesting items I encountered are Knot Energies, Evolved Virtual Creatures, Computer-generated floral ornament, Image-based Lighting, The Shape of Space, Elser's Surface, and Subdivision Surfaces in Character Animation.

The water supply of the principal city on the island of Samos in ancient Greece was inadequate for its growing population, but there was an ample supply in the mountains. To bring water from the mountains to the city, a one-kilometer tunnel was dug in the 6th century B.C. through a large hill of solid limestone. The tunnelers worked from both ends and met in the middle, more or less as planned. How did they do it? Project Mathematics has a videotape with the answer.

I tried to make a puzzle based on The Shape of Space, above. My best effort was a toroidal word square. Every row and column has a single black square. A word going over the bottom edge continues at the top; similarly, a word going over the right edge continues on the left. Can anyone top my effort here? Across 1. Speech impediment. 2. Acts of religious duty. 3. Frequently burnt earthy substances. 4. Kitchen tool. 5. It was out of this world. 6. Designating the third row of flight feathers. 7. Those who speak Hebrew, Aramaic, Arabic, or Amharic. 8. Stands up. Down. Same as across. Answer. If you like this, you'll probably like other forms at the National Puzzler's League website.

The Encyclopedia of Integer Sequences has introduced a webcam. One setting shows sequences that need extending.

This is a great week for online puzzles. It's so good, I will defer my puzzle of the week to:

1. The World Puzzle Championship Qualifier. It has some great puzzles by Craig Kasper, Sid Kravitz, Erich Friedman, Will Shortz, Michael Rios, Moshe Rubin, Steve Ryan, Robert Abbott, Pat Merrell, and Harry Nelson.
2. Peter Edgar Dant's Last Will and Testamant. A stickler for details, industrialist P.E. Dant has left a diamond bracelet for the Canadian that can solve his 12 part will.
3. On the Trail of the Murder of Evan Chan. AI has an incredible ad campaign.
5. Ken Duisenberg asks for a 6-digit number.

I like good games. One of the best I've played is Mafia. It is also called Werewolf now, and I like that better. Basically, there are only a few people left alive in a remote village -- two werewolves, one seer, and several innocent villagers. The werewolves will kill a person every night. The only defense the villagers have is to kill the werewolves, by group consensus. Sure, they may accidently kill a few innocent villagers, but what choice do they have?

Take from the deck the King and Queen of Spades (alpha and beta werewolves), one red king (seer), and several low red cards (villagers). Deal them out to the players randomly. No-one may reveal their card to anyone else.

During the day, a discussion ensues as to who should be killed. The werewolves have been at this for awhile, and this town meeting is the first time the villagers have fought back. Who can be trusted? Conversations about who people suspect, and why. Eventually, one person will be voted to be killed. Before being killed, the person is allowed a final speech. Then there is a final vote as to where the person should be killed or not. In this first round, no-one has any information, so it tends to go quickly. Once the person is dead, they turn over their card.

At night, the person killed becomes the impartial moderator for the rest of the game. Moderator: "Everyone close your eyes and hum." An eerie droning starts. It is somewhat important to hum consistently, lest someone point out your humming mistakes the next morning. The moderator is the only one who talks during the night. "Werewolves, open your eyes. Pick someone to kill." The moderator lets the players clearly indicate someone. "Werewolves, close your eyes." The moderator waits. "Seer, open your eyes. Indicate the person you will study. I will nod yes if they are a werewolf." The seer indicates, and the moderator nods if that person is a werewolf. "Seer, close your eyes." The moderator waits. "Everyone open your eyes. During the night, there has been a murder." With that, the moderator flips over the card of the person the werewolves killed, and another discussion ensues.

The day-night cycle repeats. Either werewolves or villagers will be the last survivors, and it is considered a win for everyone on that team, even if they are dead.

There are no restrictions on speech. Any living player can say anything he wants -- truth, misdirection, nonsense, or bareface lie. Contrariwise, dead players may not speak at all. Almost anything you say can and will be used against you, but it's even more dangerous to say nothing. The seer tries to cast suspicions upon the werewolves subtly, while the werewolves try to act as innocent as possible. The innocent villagers try to sort it all out with logic.

I have a possible new character -- The Herbalist. Indicate this person with a red jack. Normally, a collection of nasty plants will kill a person, but not a hardy werewolf. A herbalist can collect nasty plants that will kill a werewolf, but not an innocent person. This herbal ordeal takes all night, and may be suggested as an alternative to killing someone during the day. Any person may claim to be a herbalist and collect greenery for the victim to eat. It is important to note who collected the plants for the herbal ordeal! There are several situations where it would be advantageous to claim to be the herbalist. A herbalist, facing death, would certainly do so. The werewolves will probably target him during the night. An innocent person might claim to be the herbalist, in the hopes that the werewolves would kill him, suspecting he is the herbalist. A werewolf might try it to save himself or his companion. Note that the villagers don't need to allow an ordeal by plant -- they can kill the suspicious person by any of the old-fashioned ways. A werewolf or a seer may still operate at night during an ordeal by plant.

The change in the night routine is minimal. The first night, the moderator says "Herbalist, raise your hand briefly." After that, the moderator indicates whether the person facing ordeal by plants survived.

Another game that looks intriguing is Star, by Ea Ea. Kadon will be producing the board later this year. A few more games have been added to the list at Variations on Go. Chess Variants is always worth a visit.

Mark Stubbings somehow solved Zucca's Puzzle.

If you register before 7 June, you can try the US and Canadian qualifying test for the 10th World Puzzle Championship. A practice test is there right now. The 1999 test is now available as World Class Puzzles, vol 2.

M. Oskar van Deventer is well known to puzzlers. He designed Metro, soon to be released by Binary Arts. (At their new products page, I recommend the 15 puzzle, That-A-Way, Flip-It, Lunar Lockout, and Leapin' Lizards). He is well known for Oskar's Cube (applet). Less known is Oskar's Disk's, available from Kadon Enterprises. At clickmazes.com, you can see his 4-bit and hysteresis mazes. He sent me a nice letter recently:

"I saw the Rolling Rhomboid by Wei-Hwa Huang on your website. It is great! It is the first time that someone breaks the "cubic-shapes-with-square-grid" principle, as seen at Robert Abbott's Things-That-Roll pages.

My recent multistate maze designs go even more steps beyond what we know

• Why should the grid be regular? My Rolling Cuboctahedron design has a cuboctahedron that rolls over an irregular grid of squares and triangles. No walls are needed in this design. The rules could read "step alternatingly on a triangle and a square."
• Why should the object that rolls be regular? My Rolling Turned Cuboctahedron features quite an irregular shape of triangles and squares, which rolls over a regular grid of triangles and squares.
• Why should the grid be closed? My Rolling Dodecahedron I design features a grid with pentagons and holes between them. The object is to roll a dodecahedron in such a way that it gets turned by 72 degrees.
• Why should there by a grid anyway? My Rolling Dodecahedron II design has a dodecahedron that has to roll from an indicated start position to an indicated finish position. No grid.
• Why should the grid be two-dimensional? My Delta Roll design has a 14-deltahedron that rolls around a 12-deltahedron. The object is to get two particular edges aligned. [Ed - I haven't made a picture yet]
• Why should there be only a single challenge? My Velcro Blocks design has two pieces, a 1x1x2 block and a 1x2x2 block that roll around each other. Shapes glued on the blocks act as barriers, preventing certain moves. There are also holes, and the barriers fit into the holes. There are four start positions, and just as many finish positions. (Velcro makes a ripping sound when you separate two parts. It is an ideal material for making things that roll)."

Neat stuff. Also at www.clickmazes.com, be sure to look at the updated Maze of Life.

The Stony Brook Algorithm Repository has over 70 of the most used algorithms. Steven S. Skiena maintains the site, and also wrote the Algorithm Design Manual. I recently found a page on Strongly Regular Graphs, but haven't found a way to make a symmetrical picture of one. John Baez's write-up of the Octonions is worth a look.

I've made some pages for the puzzles from last week. The A. Martin answer wound up being remarkably overblown -- his answer required 17 digit numbers. An answer with 2-digit numbers is possible. C Bumpkin, however, was remarkably close to the computer optimized solution. What a wonderful hobby we have, that we can be amazed by someone from 250 years ago.

A small puzzle: A B = product of first six primes. A+B = composite. A-B = composite. What are A & B? Answer.

David Clark has been looking at palindromic 3×3 magic squares in various number bases. Below, you can see his solution for base-4. He has found a base-3 solution. Can you? Answer. He cannot find a base-2 solution. Is it impossible?

42 51 21       222 303 111
17 38 59       101 212 323
55 25 34       313 121 202

For a much harder problem, find integers A, B, and C so that A^2 + B^2 + C^2 = X^2 and A^3 + B^3 + C^3 = Y^3. An answer was published by A Martin in 1898. Another toughie, solved by C Bumpkin in 1750: Find A, B, C so that A+B, A+C, B+C, A-B, A-C, and B-C are all square numbers. Both of these were solved without the aid of computers, but feel free to use one. Answers. On the topic of computers, I put Linux-Mandrake 8.0 on my back-up computer, and love it.

Many of my books are Dover books, so I was pleased to learn of the Dover Book Website. Meanwhile, I've added several items to my Books page. One addition, Problem Solving Strategies by Arthur Engel (\$50), is a real gem.

There is a new RSA Factoring challenge, with prizes ranging from \$10,000 to \$200,000. The Zillions-of-Games site has a new update. Claus Tondering has posted a paper on Surreal Numbers. I was asked how many intersection points were made by the diagonals of a polygon. I found the answer here.

Stewart Robert Hinsley has new fractals. Ken Duisenberg has a good write-up on the Doubly-Attacking Queens problem.

The 20 May 2001 problem, factorial splitting, is now at the Encyclopedia of Integer Sequences. Gary Mulkey noticed that 756×768×770×792×800×810 = 12!×12!, and wonders if that happens again. Nick Gardner sent an answer to one of my queries: " I think I've surprised myself by finding the definitive solution to the problem of how to divide the first 25 primes into three packages to obtain the closest products. 1321040331310 {2, 5, 7, 19, 43, 53, 59, 83, 89} 1321053487611 {3, 11, 13, 17, 23, 41, 47, 61, 67} 1321117514527 {29, 31, 37, 71, 73, 79, 97} Difference = 77183217

Pawns, Cooperation Maze 2, Factorial splitting, 12345b have all been collated with solvers' solutions.

Andrew Clarke has found a nice Dom-Sliced Pentominoes solution. Roel Huisman also has a beautiful polyform site. Patrick Hamlyn sends an octiamond arrangement -- "A nice easy puzzle: assemble these six congruent pieces into a similar-hole solution." Another fascinating thing to look for at Andrew's site is the irreptiles. I momentarily thought of giving the task of tiling the 13x102 rectangle with the L-hexomino as a puzzle, but thought better of it. Patrick has also found a beautiful solution with the one-sided heptominoes.

I was born on December 7th, so I'll probably go out and watch the movie Memento again this weekend.

Wes Hardaker and Gervais Chapuis have put together an applet for the 17 Wallpaper groups. Quite nice.

I've been allowed to tell you about The Mathematician's Secret Room. One sample solution you'll find there -- what perfect cube can you make with just the digits 0, 4, and 7? I like lists like these, so do let me know if you know of others. Here's one of my own -- a failed attempt to find a counterexample to the Beal Conjecture: {{3^3, 6^3, 3^5}, {13^2, 7^3, 2^9}, {7^3, 7^4, 14^3}, {3^6, 18^3, 3^8}, {26^3, 26^4, 78^3}, {211^3, 3165^3, 422^4}, {386^3, 4825^3, 579^4}, {307^3, 614^4, 5219^3}, {90^4, 5400^3, 630^4}, {217^3, 5642^3, 651^4}, {271^3, 813^4, 7588^3}, {602^3, 903^4, 8729^3}, {624^3, 14352^3, 312^5}, {1862^3, 57722^3, 3724^4}, {2246^3, 4492^4, 74118^3}, {1838^3, 97414^3, 5514^4}}. In all of these, the first two powers add to make the third. All of them have a sizable common factor.

Another list maintained by Abderrahmane Nitaj is the ABC Conjecture Home page. I needed to learn about rad(n): For a natural number, rad(n) is the product of all distinct prime divisors of n. E.g. if n = 2^5 × 3^7 × 11 × 17^2 then rad(n)=2 × 3 × 11 × 17=1122. rad(2^1000) = 2. If GCD[a,b] = 1, and a+b=c, then one of many forms of the the ABC conjecture is that log(c)/log(rad(abc)) will always be below some bound. The current record is 2 + 109×3^10 = 23^5, which gives an ABC value of 1.62991.

Playing around with Mathematica, I noticed 2^3 - 2 == 3! , 3^3 - 3 == 4! , 5^3 - 5 == 5! , 9^3 - 9 == 6! . This was noticed earlier by Gustavus Simmons in 1968, who conjectured it doesn't happen again. Now, it is problem D25 in R. K. Guy's Unsolved Problems in Number Theory. I tested through 1000! (= 2^994, 3^498, 5^249, 7^164, 11^98 ...) R K Guy -- "Each power can only occur in one of the three numbers (except that one could contain 2 and another a high power thereof). That such numbers don't exist probably follows from the abc-conjecture, but we may not need such a high-powered result. I suspect that this one is not beyond the present wit of person." I became curious about how close I could get three numbers whose product was a factorial. For 10!, my best effort had a difference of 22 between high and low: 140 × 160 × 162 = 10! Is this the best? My best effort for 11! had a difference of 28: 324 × 350 × 352 = 11! For my problem of the week, try to match or beat my difference of 44 for three numbers whose product is 12! Answer. I'm leaving my slew of puzzles from last week open, too, since they're really nice puzzles.

I'd love to see further results for ... how do I say it ... cutting up factorial cakes. Annealing techniques would probably work quite well for 20! A similar problem for programmers -- divide the 25 primes into three packages to obtain the closest products. Results.

JP Ikäheimonen has sent me a variant of his Cooperation Maze. JP -- "I've got another tiny co-operation maze now, as the first one was so well received. This time, the rules are slightly different, so that one pointer moves as many spaces as what the other pointer points to. This rule change was proposed by several people, and it makes backtracking much harder. Answer.

 - You've got two pens. Both start at the START cell, AA. - Both pens move at the same time, but one of the pens must move horizontally and the other vertically. Which pen moves which way is up to you. - A pen can move horizontally or vertically, as many squares as the number under the other pen shows. - Just to make things clear: after the first move, your pens should be on squares marked with '2' and '3', on cells C and I. Then, on the next move, you can only move to K and O, where the the pointer on '2' moved three spaces down and the pointer on '3' moved two spaces right. Or AA, CI, KO, to start. - Goal: move both pens simultaneously to the GOAL cell, LL.

Cooperation maze by JP Ikäheimonen. 12345 Maze by Andrea Gilbert.

I'm going through my mailbox carefully, and I found this item by Andrea Gilbert -- "I also solved the 12345 maze, in 35 moves I think it was. I'm still dabbling with this concept myself, trying to see just how tightly I can pack these puzzles. 12345 is more productive than 123. Using a 6x6 grid and a small number of blockers I can achieve mazes over 30 moves. I attach (below) my best attempt so far." Answer.

Let me call 3^3 and 2^5 close powers, since they lie between two consectutive squares. 665719^5 and 5075571151^3 are close. {{5^2, 3^3, 2^5, 6^2}, {11^2, 5^3, 2^7, 12^2}, {46^2, 3^7, 13^3, 47^2}, {2536, 186^3, 23^5, +1}, {558640, 199^5, 6783^3, +1}, {572783, 201^5, 6897^3, +1}, {3362407, 22444^3, 408^5, +1}, {7928108, 39760^3, 575^5, +1}, {8928803, 43039^3, 603^5, +1}, {67460050, 1354^5, 165716^3, +1}, {106938971, 1628^5, 225298^3, +1}, {1763350849, 4995^5, 1459573^3, +1}, {2501641555, 1842822^3, 5745^5, +1}, {2756149047, 498^7, 1965781^3, +1}, {4584349318, 7320^5, 2759636^3, +1}, {5713606932, 3196001^3, 7994^5, +1}}. Can three powers be close between squares? I have no idea. Send answer.

Fermat-Catalan Conjecture: There are only finitely many integer solutions to a^x = b^y + c^z with coprime a, b, c and 1/x + 1/y + 1/z < 1. {1+2^3 = 3^2, 2^5 + 7^2 = 3^4, 7^3 + 13^2 = 2^9, 2^7 + 17^3 = 71^2, 3^5 + 11^4 = 122^2, 17^7 + 76271^3 = 21063928^2, 1414^3 + 2213459^2 = 65^7, 9262^3 + 15312283^2 = 113^7, 43^8 + 96222^3 = 30042907^2, 33^8 + 1549034^2 = 15613^3} are the ten known solutions. The Beal Conjecture, now with a \$100,000 prize, claims that every solution includes a square. I found many solutions of a^x = b^y + c^z with {x,y,z}>2, but every one had GCD[a,b,c] >1. In fact, I couldn't find an example where Min[{a,b,c}/GCD[a,b,c]]>5. What is the maximal value of this minimum?

Rodolfo Kurchan has been working on some chess problems, and has recently published a small book with Sterling Press -- Mesmerizing Math Puzzles. Most of the problems were new to me. Here's a sample puzzle: In a 5x7 grid, arrange 15 pawns in a connected shape so that no four pawns are in a line, in any direction. Pawns must be connected horizontally or vertically, diagonal connections don't count. Answer. Can anyone find the maximal number of pawns for no five in a row, or beyond?

Highland Games offers handmade wooden flexagons. They also provide the simple instructions for making your own. I've been playing with them now for a week. Flexagons are a lot of fun.

3x5 Dots and Boxes has been solved. It's a 1 point win for the second player. Details by David Wilson. A Dots and Boxes applet is here.

Other sites I've looked at recently: Mathstamps, Non-Euclidean Maze, Variations on Go by João Neto, and Play-By-eMail server.

João Neto has developed a wonderful game he calls Gonnect. It has the same rules as Go, except that passing isn't allowed, and the object is to connect two sides of the board. See http://www.abstractgamesmagazine.com/gonnect.html for more.

Are 8 and 9 the only consectutive powers? Catalan conjectured they were, and this will likely be proven within the next few years. The space in which an additional solution can exist is bounded. Ensor.org has a distributed program for finishing the problem off.

Carl Ginnow has a pattern matching problem at http://www.microprizes.com/mp43.htm that is quite nice. Carl -- "I have just started to explore different strategies for solving the puzzle. The puzzle has similarities to peg-solitaire, but the colors help the user plan ahead. For example, in a particular puzzle the user may notice that only one tile remains that has a lower panel that is red, and that only one tile remains that has an upper panel that is red. They also notice that the two tiles are separated by two rows. They conclude that the tiles will never meet (allowing for one tile to remove the other), since neither tile can change its row towards the other. Logic like this allows the user to avoid combinations of moves that will never lead to a solution, making the puzzle easier to solve. Logic can also be used to conclude that certain moves must be made. I'm sure that you and your visitors will soon develop your own strategies once you play a few puzzles."

I made a discovery with Mirek Wójtowicz' MCell. A few days ago, I decided I didn't like the Critters rule under Margolas, so I played around and came up with the Pegg rule. Three new gliders, so I sent them to Mirek, who informed me that nice gliders were unknown in Margolas! I had no trouble finding a rule with three gliders. Can anyone find a Margolas rule with 4 gliders?

Rodolfo Kurchan found a 22 square solution to the Gabriele Carelli puzzle. David Clark found a 27 square solution:

Those familiar with Clickomania will like Shatter, at www.skillgames.com. Each turn, you must remove polyominoes from a grid. once removed, surrounding squares will fall to fill the void. This game is perhaps computer solvable. Try it out, and see if you can develop a technique to win \$500 from Skillgames.

Dave Clark developed a puzzle 18 years ago. "Place integers evenly around the circumference of a circle so that each number is either the sum or the product of its two nearest neighbours, and no two numbers have the same magnitude." I have one solution where all integers have a magnitude under 16. Can you find it? Michael Dufour, Juan Montalvo Bressi, Gary Mulkey, Matt Elder, Michael A. Rios, Chris Lusby Taylor, Frances Watkins, Don Woods, Aron Fay, Bob Kraus, and Aram Hakobyan found the solution to this puzzle.

Gabriele Carelli developed a new puzzle: Find biggest shape in which each tetromino can block the placement of a second tetromino of the same shape. For an example, see below for a figure with 20 squares. This is not the maximum! What is? The answer is above. Many thanks to Livio Zucca for forwarding this on to me.

Puzzle World has been updated. Charles Ashbacher let me know that issue 30(2) of the Journal of Recreational Mathematics went to press last week, so those that have subscriptions will see it soon. Digital Sundials is a clever invention. David Parlett is a professional card game designer, and I liked his site.

The 4.5-ominoes can be placed in 9 identical shapes. The 66 octiamonds can be arranged into 11 simultaneous shapes, each three colored. Solutions by Roel Huisman and Patrick Hamlyn.

I learned of the work of Kees van der Laan, an expert in postscript, and really liked it. You can look at his papers on Tiling, Turtle Graphics, Bridge Strategy, and METAFONT.

Ambigram by Carlos Ernesto Penedo.
The restored 2001 is spectacular. You can see my report here.

My puzzle of the week: Fill a 3x4 grid so that triangular numbers read across and down. The answer is unique. Answer. If you'd like a tougher challenge, Fermat made the claim that any number could be represented as the sum of at most 3 triangular numbers. Euler tried, unsuccessfully, to prove this (I did say it was tougher). Gauss proved it on July 10, 1796. The entry in his diary was "EUREKA! num = D+D+D." With a bit of Mathematica code, I identified all the ways 3 triangular numbers could sum to 2001. {{2001,2,9,62}, {2001,6,20,59}, {2001,6,44,44}, {2001,7,34,52}, {2001,9,35,51}, {2001,9,39,48}, {2001,10,10,61}, {2001,10,28,55}, {2001,11,14,60}, {2001,12,17,59}, {2001,14,24,56}, {2001,14,30,53}, {2001,14,41,45}, {2001,16,25,55}, {2001,20,30,51}, {2001,21,39,44}, {2001,24,36,45}, {2001,25,34,46}, {2001,34,37,37}, {2001,35,35,38}}. Do any large numbers have unique representations?

I've been looking at the Game of Life recently, catching up on all of the recent discoveries. ********.*****...***......*******.***** is one of many neat things listed in Life Lexicon. I've been using Mirek Wójtowicz' MCell to do some explorations of some of the things there, and things at Hickerson's Life Page, Eppstein's Interesting Rules and Weisstein's Treasure Troves. One of my favorite things to watch has been Mirek's own Star Wars rule. Once you have Mirek's program installed, you can click on anything at Eppstein's page and watch it run, or you can paste in patterns from Life Lexicon and watch them run.

The latest Enigma had a nice article about the latest MIT Mystery Hunt. Alexamder Kruppa & Tony Forbes have found a factor of 2^(2^31)+1 -- 46931635677864055013377.

For a grand Easter Egg puzzle hunt, nothing can beat Spielberg's A.I. work. Just start with Jeanine Salla, who is listed as a consulant on the A.I. trailer (which has at least two easter eggs). A whole virtual world has been created. The words in the trailer are cirrus, Socrates, particle, decibel, hurricane, dolphin, and tulip. I'm not sure what those mean yet, but they're beautiful words.

From Erich Friedman. Tour the 48 states, without repeating a state. Answer

Joseph Becker sent a related question. What is the longest English word that can be formed from the initial letters of the names of states visited sequentially in such a tour? For example: NONUNION = Nevada Oregon Nevada Utah Nevada Idaho Oregon Nevada. Answer by Joseph DeVincentis.

The site logicalpuzzle.com has some very nice work by Aram Hakobyan of Yerevan State University.

Koshi Arai has found more wonderful patterns with the New Kite figures. How many solutions are there for the 14-piece figure?

Andrew Ofiesh, Bob Kraus, Dan Hennessy all sent comments on the Triangle Paradox. They're quadrilaterals. One is slightly concave, and the other is convex.

An incredible heptomino solution by Patrick Hamlyn

Group Games is a new page by Alessandro Logar and Alessandro Budai.

I see Joseph DeVincentis on all the puzzle sites I visit, so I asked him to list his top sites he likes. Here's his list:

1. Erich Friedman's site at is excellent, but you already know that.
2. Ken Duisenberg's site is good. The puzzles vary quite a bit in difficulty, though, with a few on the too-easy side.
3. Tim Edmonds's site is also good.
4. The Ponder This site features a monthly puzzle which is generally good and ranges from moderately hard to fiendishly difficult.
5. Math Chat on the MAA site at is good, and the other columns are often interesting to read.
6. I often try the Quiddler daily puzzle at the Set game site but rarely if ever do I get the high score needed to put my name in the displayed top scores.
7. I usually at least attempt the weekly NPR "Weekend Edition" puzzle at though I don't often catch the radio show.
8. Paint by Numbers site has hundreds of these puzzles. They are user-submitted, and as a result vary in difficulty and quality a bit. If you stick to puzzles with a 2.7 difficulty or better, you should enjoy the site.
9. The Stamford tournament's site at has a links section with links to many good (and some not-so-good) online crosswords.

I've put together solution pages for the Cooperation Maze and for the April Fools puzzle.

Larry Rosenheim asked if the order tetrasticks could cover a 5x5 grid if the square was left out. It turns out that they can, if the is left out (it is a square). Brian Barwell's solution is given below. Joseph DeVincentis solved the problem with a different piece missing. Matt Elder and Joseph both sent nice parity arguements about why removing the other square piece would not provide a solution.

I've added oneacross.com to my links -- it has many nice solving aids for anagrams, cryptograms, and more.

For my math puzzle of the week, I'll mention a long unsolved math problem that involves representing integers with only 1, +, ×, and parentheses. What is the smallest number of ones needed to represent a number? 7 is the smallest number needing 6 ones: (1+1)×(1+1+1)+1. 41746319 is the smallest number needing 58 ones. What is the smallest number requiring 100 ones? No-one knows. The data that Piotr Fabian added to the Encyclopedia of Integer Sequences shows a tendency towards -1 (mod 120) with larger entries (everything after 540539). Here's the data: 1, 2, 3, 4, 5, 7, 10, 11, 17, 22, 23, 41, 47, 59, 89, 107, 167, 179, 263, 347, 467, 683, 719, 1223, 1438, 1439, 2879, 3767, 4283, 6299, 10079, 11807, 15287, 21599, 33599, 45197, 56039, 81647, 98999, 163259, 203999, 241883, 371447, 540539, 590399, 907199, 1081079, 1851119, 2041199, 3243239, 3840479, 6562079, 8206559, 11696759, 14648759, 22312799, 27494879, 41746319. Below is a Logplot I made with Mathematica. This data is just days old, perhaps a reader can spot a pattern in it that I missed.

Logplot of the number of 1's needed to represent a number.

I've received the following paradox several times in the past few weeks. Rearranging the pieces of a triangle, you obtain the same triangle, but with an extra square. Where's the flaw? I'll post the solution next week.

Taus B-N -- "Erik Hermansen has made his excellent puzzle game DROD public domain. This is without a doubt one of the best puzzlegames I have ever seen (and I consider myself a true puzzle connoisseur). Erik is currently making a true PD version, but you can still download the commercial version. Also, Hobbit's Room is a site where Hirofumi Fujiwara has ported some of his puzzles (and made new ones) for the Palm."

I've recently looked at Polysticks (or polyedges)-- as have other people, like Alfred Wasserman and Livio Zucca. For a moment, I thought that an equivalent set might be possible with octagons and squares, but that was wrong. Guenter Stertenbrink took a look at the related pieces, with the square missing. You can see more of these at Miroslav Vicher's site. Polyedges are different, though. Brian Barwell wrote about them in the the Journal of Recreational Mathematics (v22, n3, 1990). One of his discoveries was that sets of order 1-4 could cover a 6x6 grid. Larry Rosenheim recently wrote to me and asked if the order 4 polysticks could cover a 5x5 grid if the square was left out. Can they? Yes, if you count the stick 4 as a square.

Erwin Eischner and Joseph DeVincentis solved Erich Friedman's pentagon problems. Erwin Eischner, Micheal Malak, Joseph Becker, Darren Rigby, and Joseph DeVincentis solved the States problem . JP's Cooperation maze proved to be the most amazingly difficult tiny maze I've ever seen. Many wrote to me to say it was impossible. It was solved by only by Koshi Arai, Yuval Gabay, Matt Elder, Carl Ginnow, Mike Fee, Erwin Eichner, Juha Saukkola, and Igor Krivokon. AA,DM,LP,IL,AI, DA,BM,JP,BO,JN, LF,IN,AP,MO,AN, MP,AO,MN,PF,ON, NF,FH,NG,FF,HN, GF,CH,OG,NC,PO, LN,IF,AH,MG,AF, DN,BF,JH,BG,JF, BH,JG,LC,IO,KK

Patrick Hamlyn has found an amazing octiamond solution:

Peter Esser has used his solver to find a very nice solution for Roel Huisman's dom-sliced pentominoes.

I've added a link to the journal Word Ways, which is well worth a look. Many sample articles from its pages are there.

I got another great puzzle from Erich Friedman -- Take a tour of the lower 48 states, always moving from a state to an adjacent state, and never visiting a state twice. If you start in MAINE, visit MARYLAND 16th, WASHINGTON 31st, KANSAS 39th, and end in CALIFORNIA, in what order did you visit the states? The answer is unique. What is it? Send answer. A state meta-puzzle: what is the fewest number of states you have to name in such a puzzle?

Another good Erich puzzle: What's the fewest number of convex pentagons that a triangle can be dissected into? how about the fewest number of convex pentagons in a square?

Karl Scherer's Squaring the Square problem has led to many beautiful solutions. You can see them at his page. He is now being helped by DeVincentis. Can a large enough grid square always be divided into smaller grid squares so that no two squares of the same size touch? Karl has proved this result, and is working to refine it.

I have a new book recommendation -- The Particle Physics Booklet. 272 pages of distilled science. Even better -- It's totally free! Just see ordering information at http://pdg.lbl.gov.

JP Ikäheimonen has created the Cooperation Maze. Despite the small size, it's surprisingly difficult.

The rules are as follows:
1. You've got two pointers, two pens maybe.
2. Both pens start at cell marked 'start', in this maze top left (number 3).
3. Goal is to have both pens arrive simultaneously to cell marked 'goal'
4. A pen can move horizontally or vertically, as many squares as the current number shows.
5. Both pens move at the same time, but one of the pens must move horizontally and the other vertically. Which pen moves which way is up to you.
6. Just to make things clear: after the first move, your pens should be on squares marked with '2' and '3', where one pen moved three squares to the right and the other moved three squares down. For the next move you will have twochoices. One possible move is to move the upper pen two squares downwards while the other pen moves three squares right, and the other choice is to move the upper pen two squares left while the other moves back up again. And so on.
7. This particular maze has no number under 'goal', so getting just one pointer in 'goal' is considered to be an illegal position,where you cannot continue further. A dead end, in maze terms.

Karl Scherer has been investigating square and rectangle dissections where no two squares share a full edge. For starters, try to divide an 11x11 square with this rule. He's made some discoveries about these using Zillions, you can find more than 100 programs by Karl here, including Square the Square.

by Peter F. Esser's wonderful solver . Half-dom+2 set, Half-Trom+2 set, and domino. Do the same pieces allow a more irregular solution?

Koshi Arai has assisted my with my investigation into new shapes.

In the Monty Python sketch, Buying a Bed, the following two lines occur. DWIII calls this the Monty Python Paradox, since the two people are refering to each other, which would mean that there numbers aren't accurate. What are the real numbers? DWIII solved it, as did Impulse on alt.math.recreational.
Lambert: Excuse me, sir, but before I go, I ought to have told you that Mr Verity does tend to exaggerate. Every figure he gives you will be ten times too high.
Verity: Yes, remembering of course that you have to multiply everything Mr Lambert says by three. It's nothing he can help, you understand. Otherwise he's perfectly all right.

I finally have the solvers for Erich's string problem listed. I have more results and corrections to do, I hope to finish those soon.

Ellen Ripstein won the American Crossword Puzzle Tournament. Hurray, En!

The solution above is based around the 1-3 triangle. Tossing in the 1-2 triangle produces the Kites&Bricks puzzle. I've tried to find related puzzles, but haven't gotten anything polished yet. However, I have found an interesting class of quadrilaterals.

I restricted myself to edge lengths of 1, 2, 3 (green) or sqrt(2), 2 sqrt(2) (blue). All the angles are from 1-1, 3-4, or 1-7 triangles. I'm not sure what to call this set of angles, but they work nicely which each other. At the moment, I'm not sure if the above shapes can make a rectangle, or anything. There are 6 more quadrilaterals with only 90 and 45 degree angles, and 25 rectangles and squares. If a length of sqrt(5) is added, and angles from the 1-2 and 1-3 triangles, 23 more quadrilaterals are possible. Add sqrt(10), and 36 more quadrilaterals appear.

I asked Erich Friedman what the optimal method for putting seven circles on a flattened torus.
He didn't have an answer, but he did send me his best answers for 1-6 circles

Plank Maze 15 ("Fiendish") was solved by many people, and was very popular. I haven't solved it myself yet. It's likely the best puzzle of the year, though, so I'll keep at it.

Paul A Valiant has solved one of my longstanding problems -- It is impossible to make the Petersen Graph with a single length of strut.

Andrea Gilbert is another master mazemaker, and she has added a new Plank maze to her site. Her Plank Maze 15 is fiendishly difficult, and I'm making it my Math Puzzle of the Week. Write me at ed@mathpuzzle.com if you find the secret code.

Robert Abbott's maze from last week was extremely popular. Here are comments from all the solvers. Yoah Bar-David has put a solver and creation tool at http://www.bardavid.com/.

For my Math Puzzle of the Week, Master maze creator Robert Abbott has added eight new mazes to logicmazes.com, the Alice Mazes. The seventh one is the hardest, and if you manage to solve it I'll get a secret code. At his site, the mazes are fully interactive. The below picture is what the first maze looks like.

Portal to logicmazes.com, where eight new mazes await you.

Aron Fay found some very nice results for my Observer question from 17 February. In the image below, there are 20 observers, and each one can see exactly 15 other observers. He also found a way to arrange 12 observers so that each one sees exactly 9 others. Can you match him? Send Answer. I'd love to see further non-trivial examples of n observers that each see exactly k others.

An observer solution by Aron Fay. Each observer cannot see exactly 4 other observers.

Patrick Hamlyn packed the solid octominoes in an impressive way, with a three-coloring!

In related news, Peter Esser has made available a polyform solver (link to zip file) which can create and pack dom-sliced polyominoes. More information can be found at the websites of Roel Huisman and Andrew Clarke.

Roger Phillips found a solution for an 8x8 grid of dots with 9 arcs. He just found a solution for 8 arcs through a 7x7 grid of points.

Eight arcs through a 7x7 grid of points, by Roger Phillips.

I've added a few new books to my Books area. Proofs from THE BOOK by Aigner and Zeigler is a wonderful glimpse at a metaphysical artifact. Quoting from the preface: "Paul Erdös liked to talk about The Book, in which God maintains the perfect proofs for mathematical theorems, following the dictum of G. H. Hardy that there is no permanent place for ugly mathematics. Erdös also said that one need not believe in God but, as a mathematician, you should believe in The Book." The beautiful Aigner & Zeigler glimpse is now in a second edition.

Problem Solving Strategies by Arthur Engel is aimed at people involved at math competitions. It has 1300 problems and solutions, many of them new to me (which is saying something). Often a problem becomes easier if you know of a certain trick or method. The Herglotz trick, for example (see Proofs from The Book, chapter 19). Problem Solving Strategies concentrates on these methods.

Donald Knuth has calculated the number of 47-ominoes as exactly 272,680,844,424,943,840,614,538,634. More exciting are his plans for Art of Computer Programming, Volume 4. He's done extensive research on puzzle kings Loyd and Dudeney, and much of that research is indexed at Knuth's site, such as the Dudeney-Loyd columns for Tit Bits.

Erich Friedman noticed that 941 is the only 3 digit number that is the reverse of the sum of its proper substrings. 94+41+9+4+1 = 149. There is a four digit number with this property. What is it? Joseph DeVincentis, Ben Langton, Gerami F. Havewala, Gerard Schildberger, Cedric Lo, Andrej Jakobcic, and others added 891, 1392, 51070, 147970, 1,330,550 and 1,523,870.

More mazes can be found at joker-games.com. The Maze of Rah and Just Skidding are both good logic challenges.

I've moved the Logistic Map to my Prizes page as a \$50 problem.

I've located the Retrograde Analysis Page again.

Observer puzzlers and solutions were moved to the Solutions page.

Arcs through Dots were moved to the Connect the Dots page.

Domino graphs were solved by many people. My intended answer for the 9-circle problem (arrange the dominoes 1-3 1-4 1-5 2-3 2-4 2-5 3-4 3-5 4-5 in a circle so that no domino shares a number with its neighbor) involves the Petersen Graph. Just look at the 9-sided representation. Add the 1-2 domino at the center, and the rest of the problem is easy. The Petersen graph corresponds with the order-5 dominoes. The below two graphs correspond to the order 6 dominoes. I'm not sure if they have a name, but I probably won't get into too much trouble by calling them "domino graphs."

Order six domino graphs. Each vertex can be labeled uniquely with a pair
of numbers from {1,2,3,4,5,6} so that no adjacent vertices share numbers.

Patrick Hamlyn investigated augmented pentominoes.

Archimedes and Aristotle might soon have some new papers published. Details.

From rec.puzzles: "Take the cards from a standard deck and turn them face up one at a time. You can stop at any time, at which point you are paid in dollars the number of red cards so far minus the number of black cards so far. Assuming optimal play, what is the expected payoff from this game?" Answer by Mark Thornquist.

"Lucy and Lily" is a game by Rich Schwartz that sneakily demonstrates the usefulness of four dimensional lattices.