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material added 26 March 2003
 From Bryce Herdt, a riddle.  Solved by Nathan Stohler, Darwin D. Smith, Jeremy Galvagni, Al Stanger, Alastair Cuthbertson and his advanced class, Alan Lemm, Susan Hoover, and Travis Taylor.

From Serhiy Grabarchuk: Neo Matchstick Snake.  You have a 2x2 square and a pile of matchsticks. (Every matchstick has its length equal to 1.) The object is to form the longest possible snake exactly within the 2x2 square (including its border). Matchsticks must not cross each other, and a final snake must not have self-touches even at a point. It can be closed in a loop. Angles between the adjacent matchsticks can be 45, 90, 135 or 180 degrees only. It's easy to find Snake-12 or Snake-13, but there is even a longer snake. Jeremy Galvagni and Bob Wainwright both matched my own length 15 solution.  Roger Phillips found a beautiful symmetrical length-15 solution.   Jim Lewis Melby, Clinton Weaver, Susan Hoover, Michael Dufour, and Brian Trial sent length 14 solutions.  If one sticks to multiples of 30 degrees, I couldn't get past Snake-15.  I managed to put a degree-30 Snake-13 entirely inside a 1.95 sided square. Can you find it?

A lot of new small sliding block puzzles are available at Puzzlebeast.

Al Zimmermann runs one of the best programming contests.  In his latest, to win $500, you'll need to find polyhedra with optimized surface areas.

An extensive site about boardgames is the Games Journal.  I was introduced to it through a 1 2 3 part essay on game systems by Ron Hale-Evans.  I'm also quite pleased with my subscription to Abstract Games magazine.  And I'm constantly amazed by the new things invented with Zillions of Games.

Last week: "I recently learned the importance of the numbers 299792458 and 9192631770.  If you wish, you can read the official paper on units."  Several of you, including myself, were slightly nonplussed about the international standard of weight being based upon a casting made by George Matthey in 1879.  I was happy to learn of an ongoing project to replace the old standard with a perfect sphere of silicon.

I recently came across a site devoted to logical fallacies,  Stephen's Guide and at the Nizkor project.

Anyone near Hampton Court in England shouldn't miss the Puzzle Palace run by Adrian Fisher, now through April 27th.

Bathsheba Grossman now has 4 different very reasonably priced Polyhedral Sculptures available.  A different type of mathematical art is described at the Termesphere site. The art of simple programs is well represented at the Texture Garden. I'm helping to put together a gallery of program-related mathematical art for the NKS 2003 conference, please contact me if you have anything that might be appropriate.  Netlogo 1.2 has been released -- I'd love to hear an opinion of this on anyone in high school, or past high school, for that matter.  Ralph the Triangle has written many interesting papers about the mathematics of tiny mixers and perfect pyramids.  Darij Grinberg has a page on Triangle Geometry proofs.

CINQUANTE ET UN + ONZE - SIX = CENT UN + QUINZE - SOIXANTE is a true statement in French, and a two-sided anagram.

material added 17 March 2003

I made a bad mistake while drawing up solutions for Brian Trial's coin problem from 5 February.  Not only did I get the size of the card wrong, but I also calculated the radius of the dime incorrectly, so the "solution" I found was totally wrong.  (My first clue should have been the matching solutions of master solvers Clinton Weaver and Bob Wainwright.) My apologies.  A lesson for aspiring math people: always be wary of a solution with warning flags. 

material added 16 March 2003

Pack squares of size 1 to 17 into a 39×46 rectangleAnswer. Solvers. The uncovered area in only 9, which can be filled with a domino, tromino, and tetromino.  Although suggested to me by Robert Reid, Patrick Hamlyn, and William Rex Marshall, this puzzle can be traced to online sequence A038666. It's a very nice problem, I managed to solve it in about 10 minutes. Only the first 21 cases are known, the excess areas for squares up to size n are: 1-0 2-1 3-1 4-5 5-5 6-8 7-14 8-6 9-15 10-20 11-7 12-17 13-17 14-20 15-25 16-16 17-9 18-30 19-21 20-20 21-33.

The original esteemed volume of recreational mathematics has been updated.  Winning Ways, 2nd Edition, Volume 2 (out of 4) is now available from publisher AK Peters.

I recently learned the importance of the numbers 299792458 and 9192631770.  If you wish, you can read the official paper on units.  A good website for particular numbers and big numbers is Robert Munafo's website.  Also, Erich Friedman has a page on numbers.  For vastly more data, you can peruse the files naming every location on the Earth.

Thunderball by Jonathan Welton

In Thunderball there are n balls which roll about on the surface of a squared grid. The balls move in response to a tilting of the surface, and all move at the same time in the same direction and the same distance, unless one or more of them are blocked. As the balls move they leave behind them a trail of damaged squares. Each visited square is so badly damaged that balls are subsequently blocked from revisiting it; if a tilt would bring a ball on to a previously visited square, the ball remains stationary. The balls start in a diagonal row of n, and the object is to bring them in to a horizontally or vertically adjacent row of n.

The 2-ball case can be solved by making tilting: N, W, 2S, 2E, 2S, E, 4N, 6W, 6S, 5E, S, E, 2N. This brings the balls in a line within the confines of an 8x7 rectangle. In fact it is possible to do so within a 5x7 rectangle. (How? Answers and Commentary.)

The 3-ball case is quite challenging, and can be solved with the sequence: S, 3E, 4N, W, N, 2E, 2S, 2E, 2S, E, S, 3E, 6N, 6W, E, 3S, 5E, 3W,  S, E, 9N, 11W, S. This brings the balls into an adjacent row within a 12x15 rectangle. I'm sure a more compact solution must be possible.

Can the problem be solved for all n? Well, here is a solution to the 4-ball problem.

I believe that this method of solution can be extended, with suitable changes of various aspects, to solve the problem for any larger n. This suggests that the problem is solvable for all n, but the method gives solutions which are far from minimal. By minimal, I mean a solution in which the encompassing rectangle has largest side as small as possible. (So 6x6 would be smaller than 5x7. This definition of minimal penalises long thin solutions like the 4-ball solution above more than a simple product of the sides would do.)

This type of problem, where all pieces move simultaneously in the same direction but squares may not be revisited, permits a great many variants. Start or target positions could be different, the board could have edges, or holes, off which the balls are not permitted to fall. There could be barriers or blocked squares which would help and hinder progress, or pits which trap balls. There could be anti-gravity balls, or ones which leave a trail of oil over which others may skid without stopping. Or the directions could be other than just orthogonal, for example diagonals could be permitted, or knight moves. I’ve tended to restrict myself to the pure version of the problem as above.

material added 11 March 2003

I fixed the link to the ISEE3, so I may as well point out the Shape of Space again as a separate type of manifold.  Stephen Hawking weighed in on this as well in an episode of the Simpsons: "Homer, your theory of a donut-shaped universe is intriguing." According to a recent New York Times article, Homer may have been right.

material added 9 March 2003

A conference devoted to Mathematical Art produced many beautiful objects and sculptures, which you can see at the Bridges & Isama website. I did some experiments in mathematical art myself while looking at the marvellous Nova Plexus by Geoff Wyvill.  If you don't mind rubber bands, you can make one of these with 12 pencils and 12 small rubber bands -- it's quite attractive.  If you don't like rubber bands, you can always use more pencils.

One good pencil puzzle is to arrange 6 pencils so that they all touch each other.  With more work, 7 pencils can all touch each other.  With even more work, 8 variously sharpened pencils can all touch each other.  Can you figure out how?  Send answer.

A nice result, from http://www.ktn.freeuk.com/9f.htm A fiveleaper is a type of generalised knight that makes moves of length 5 units, with coordinates either {0,5} or {3,4}. In Variant Chess, GP Jelliss made the following observation: “Since the fiveleaper has four moves at every square of the 8×8 board it follows that in every closed tour the unused moves are also two at every square, and therefore form either a tour (is this possible?) or a pseudotour (i.e. a set of closed circuits). The question of whether such a double tour is possible was in fact answered in the affirmative by Tom Marlow in a letter to me of 17 November 1991:

5-leaper double tour #1 . 5-leaper double tour #2
20 47 62 55 06 21 46 63 . 46 37 60 11 56 49 04 63
31 42 57 50 11 34 29 44 . 39 24 31 52 27 40 23 32
02 59 16 09 52 03 36 17 . 54 15 06 21 34 29 16 13
13 22 39 26 19 14 23 38 . 57 08 03 64 19 36 59 10
54 05 28 45 64 61 56 07 . 26 41 50 45 38 25 42 51
51 48 35 30 43 58 49 10 . 47 28 61 12 55 48 05 62
32 41 24 37 12 33 40 25 . 20 35 30 53 14 07 22 33
01 60 15 08 53 04 27 18 . 01 18 43 58 09 02 17 44

I found an article about the fantastically complicated journey of the International Sun-Earth Explorer 3 to be fascinating. I knew about the satellite years ago, back when I worked at NORAD.  Basically, the Lagrange point between the Earth and Sun is unstable, but there is a stable orbit around it.

I also very much liked a long discusion of why Boron could provide an ideal fuel source for cars, without pollution. For the less serious, you can peruse silly molecule names.

Gary J. Shannon has posted his investigations in a WireWorld like, logic-gate rich CA at his site.

material added 2 March 2003

Robert Reid sent me his efforts for packing 16, 17,20, 22, 41, 43, 45, 49, 51, 64, 65, 76, 90, and 94 consectutive squares into a square.  In each case, it is impossible to fit the squares in a smaller square.  I redrew two of them. In general, what are the smallest rectangles than can hold all the squares up to size n?  The squares up to size 42 don't quite fit into the smallest square.

DNA was discovered 50 years ago.  For most living creatures, 20 animo acids are enough: A (alanine), C (cysteine), D (asparatic acid), E (glutamic acid), F (phenylalanine), G (glycine), H (histidine), I (isoleucine), K (lysine), L (leucine), M (methionine), N (asparagine), P (proline), Q (glutamine), R (arginine), S (serine), T (threonine), V (valine), W (tryptophan), Y (tyrosine).  A few exotic organisms have 3 other amino acids.  In honor of DNA/RNA, here is a puzzle: aUcgAcgAACaUuGCUaCuGgaUaAuCgAnswer.  Once you figure out the capitalization... what is the longest word which is mostly capitalized?


Joseph DeVincentis: I finally coded up my idea for a method of searching for the longest words possible in the DNA code. I started by making a list of all the possible trigrams, by going through all 4^5 possible 5-base sequences. Then I used some grep commands to search word lists for words that consisted only of these trigrams at all positions. The use of trigrams eliminates words that would erroneously be formed, for instance, by entering 'S' at UCx and leaving via AGx, if you only used bigrams.  The longest word I found was flytrap which can be written as uUUACGCCu. Also, entrap, gently, and several less common 6-letter words showed up. Using the allwords list, I get a few other 7s: Cardiff, prerent, and Sargent, and one 8, alcargen, which appears in the list which supposedly represents the Shorter OED, but it is one of those "words" that shows up nowhere on the web except in word lists.  Michael Dufour sent me CARARARA (an obscure monkey) and Perl code.

I recently found the exact solution for the Snub Dodecahedron.  I was interested in finding a vectors {1,a,b} that could be multiplied by the Icosahedral Group to obtain all of the solids in that group.  Using Mathematica, a horrifying equation boiled down to Root[x6 + 6 x5 - 7 x4 - 9 x3 - 14 x2 - 7 x - 1 &, 4]. Withh more Mathematica, I obtained all 169 vectors that produce a convex solid with icosahedral symmetry.  Sixty of these points are for the snub dodecahedron alone.  I haven't yet plumbed all the secrets out of this set of points, but I rather like it.

(Using Mathematica again), I tried making an animation of Nick Gardner's Wireworld Binary Multiplication. The animation is fairly large though .. it's easier to download Mirek's Cellebration.

material added 19 February 2003

Robert Abbott let me know about a Washington Post article about Binary Arts, Bill Ritchie, and Andrea Gilbert's Clickmazes.com.  Serhiy Grabarchuk, who runs Puzzles.com, filled me in on how that website is joining in.  Andrea's plank maze will be a key demo at the currently ongoing Toy Fair.

Snopes.com is usually good for accuracy, but I noticed a scientific blunder on this page about diamonds.  Can you figure out what it is?  Mark Thompson matched my answer: I wouldn't call it a scientific "blunder," but I can see a way of telling whether a diamond was produce from graphite or from Uncle Fred. If it's from Uncle Fred, it will have a whole lot more Carbon 14 in it. Perhaps that could be determined nondestructively with a geiger counter, though the normal method would require destroying the gem, in order to verify it.

Brendan Owen made a very nice discovery.  Four corners of a cube can be removed to make a tetrahedron.  If a cube is divided into 4 identical pieces so that each gets an entire corner, you get pieces he calls cubecorners.  If four cubecorners -- tetracubecorners --  are connected together with full face connection,one piece that can be made is the original cube.  It turns out there are exactly 27 other shapes that can be made, and they can fit together to make a cube.  After solving all that, he put them all together in a lovely applet.

I've spent a lot of time lately trying to LURN.  Left, U-turn, Right, No turn -- these are the directions a Turmite must choose from, and I wondered what would happen if a turmite could split as an action... say Left and Right.  In addition, I added the rule that when two turmites met, they annihilated each other.  Some interesting patterns came out from my initial study.  Here is my Mathematica notebook, for those that want to study them (with some help from Eric Weisstein's MathWorld packages).  My main interest is finding turmites that will grow for a long time, then self-annihilate.

I read a paper by George Collins (Lecture Notes in Comput. Sci., 358).  If b and c are two random integers, then the probabilty that they have no common factor is six over pi squared.  Or P[GCD[b,c]=1] = (6/Pi^2).  That's a well-known nice result.  If b and c are two random Gaussian integers, then P[GCD[b,c]=1] = (6/Pi^2)/Catalan, where Catalan is the Catalan constant.

Here's one by Robert Reid.  Find a three digit number abc such that abc×bca×cab is a square number.  There are two solutions.  As a hint, abc×bca×cab is a sixth power, in one of the solutions. Answer.

Bathsheba Grossman is one of the foremost mathematical sculptors at the moment, and I've much enjoyed looking over the artist's pages.  One new project is particularly fascinating -- large sculptors from laser cut materials.  You can get the Sea Star project for just $22.  The techniques Bathsheba uses are also discussed -- most design work is done with Rhinoceros.  That program is well worth your valuable time for a look -- although it costs $900, the only limit on the demo version is that only 25 saves can be made.  

material added 5 February 2003

Nick Baxter kindly added my Two Hearts puzzle to his Sliding Block Puzzle Page, just in time for Valentines day.

Brian Trial: Take a 2 1/2 inch by 4 1/4 inch card and cover it with as many U.S. pennies, nickles, dimes, and quarters as you can to get the highest dollar amount. Coins must lie flat, must not overlap or stack onto other coins in any way, and must lie entirely on the card. For reference, a U.S. quarter has a diameter of 0.955 inches, a nickle has a diameter of 0.835 inches, a penny has a diameter of 0.750 inches, and a dime has a diameter of 0.705 inches.  Answers. (No-one, including the problem creator, got the correct answer.) Now, that's an awfully US-centric puzzle, so here's a more international version -- using the coins of your country, what is the most money in coins you can place on a 7cm by 11cm card, using the same rules?

Serhiy Grabarchuk sent me his Stars & Spirals puzzle: Link the six Stars and six Spirals with exactly 5 connected straight lines. Stars and Spirals along your route must alter (...Star...Spiral...Star...Spiral...), and each of them must be visited just once.  Joseph DeVincentis, Brian Trial, Aron Fay, Ron Zeno, Dan Tucker, Bob Kraus, Scott Purdy, Brendan Owen, Adam Fromm, Gromit, Robert Reid, Koshi Arai, Shlomi Fish, Ken Duisenberg, Mike McCraw, Earl Gose, Henry Robertson, Clinton Weaver, Jean-Charles Meyrignac ,Juha Saukkola, and Bathsheba Grossman sent in the solution.

Jean-Charles Meyrignac:  After more than 3 months of computation, I just finished the computation to the c1 solitaire problem on the french board. In the book Ins and Outs of Peg Solitaire from D.J. Beasley, it is mentioned that it is possible in 21 moves. In fact, I discovered that it can be done in 20 moves, and the solutions are very rare (only 280). More Info.

Geometry In Action Java Gallery, by Clark Kimberling, is well worth a look.  I think of all the courses I took in high school, Geometry was the one I found most useful, both in terms of proof technique and the general usefulness of geometric construction.

Bob Lukes has created the Lonely Unit Cube puzzle out of wood.  Bob says that, after all the work he went through cutting and painting the cubes, that if anyone wants a copy, the answer is "no".  It is possible, though, to get 54 8mm and 24 12mm dice, in which case you'd only need to make 1 4mm cube, 2 20mm cubes, and a 44mm box.  Any volunteers to make that?  I'd like to put together about 20 sets for general purchase.

I have updated the Neglected Gaussian page with many solutions by Fred Helenius and W. Edwin Clark.  

material added 26 January 2003

If the edges wrap, can a set of double-6 dominoes be placed in a 8x7 rectangle so that all numbers are in connected groups? The below is one of my efforts -- the threes are all connected if you consider the edges as wrapping, same with the blanks and sixes.  However, the ones and twos are both in disconnected groups.  Can everything be connected?  It turns out the answer is yes. The solution below, by Jason Woolever, is for the harder problem of connecting everything without the use of doubles. Other answers and solvers. See also my 2 September 2001 update for a related set of solutions by Roger Phillips.  The domino-connection number for the torus is thus 7. Suppose we go to 1x1x2 blocks as dominoes.  What is the domino connection number for a 3-D block?  If the faces of a block wrap, it is called a 3-torus, or a 3-manifold.  You can learn more about these at The Shape of Space.  Without doubles, what is the domino connection number of the 3-torus?  With doubles?  With 2 of each double?  In 3D space, as the number of doubles increases, the domino connection number would go to infinity (why?), but I don't know if anyone has looked at how fast. Send Answers.

James Stephens of Puzzlebeast.com has added a new variety of restricted sliding block puzzles.  Very nice idea, and wonderfully realized.

The cubicular goodness of the 2003 MIT Mystery Hunt (acme-corp.com) can be seen as an offshoot of the main MIT Mystery page.  All the puzzles are there for your perusal.

Tom Marlow notes two interesting squares: 4253907186^2 = 1809572634 7102438596, 5296031874^2 = 2804795361 0423951876.  Daniel Scher's Geometry in Motion has moved.  Bob Kraus has put Soccolot on the ZOG site (interesting game). Martin Watson has add lots of great stuff.  Ivar's Peterson has had a number of great columns lately, such as his Dearth of Primes write-up (I had no knowledge of this).

An even more interesting game is Amazons, and there is a very nice analysis of it at the More Games of No Chance page. The full book is available online, but I've seen so far is good enough to prompt me to buy it.

For g4g4.com, I wrote up an article about recent Polyform discoveriesComments are welcome.

material added 15 January 2003

Cihan Altay has started PQRST 4.  There are many clever ideas here, I especially like Puzzle #8.  Answers must be submitted by 18 January.

Robert Reid found a mistake in one of the puzzles on my old Solution page. "The first puzzle this week is by Scott Purdy.  The thick path travels from A to B, visiting every dot.  Can you remove 7 of the thin lines so that this is the only path from A to B that visits each dot?  In more mathematical terms, for K(n), what is the minimal number of edges that needs to be removed for a unique Hamiltonean path between two given points?  The case for K(8) is unsolved (as far as I know).  Partial solution to the general case by Scott Purdy and Erich Friedman."  Okay, seems okay -- but Robert Reid found a solution that removes only 6 lines ... and solved the K(8) caseby removing only 9 lines. Can you find Reid's six-line solution? Only Jim Boyce matched Reid's answer.

1 2 3 4 5 6 7 8 = 2003.  Add each of +, -, ×, ÷ exactly once to make the equation true. Answer.  This is by Yoshio Mimura. I found his page while looking to see if anyone else had noticed the octal square 177771777177771.

So ... 177771777177771. Twenty years ago,  Nob Yoshigahara noticed that 81619 × 81619 = 6661661161.  Are there larger square numbers using only two digits?  No-one knows.  The Mathematician Secret Room has more data for the 3-digit square problem.  I wondered if I could find new complex squares with that property (I couldn't), or two digit squares in other bases (easy).  Nick Baxter looked at rational squares of two digits, and found some interesting solutions.  There is enough here to figure out the significance of 34343443434344.

(40457+27469i)^2 = 882222888+2222626666i (20644+3425i)^2  = 414444111+141411400i
(13773+12464i)^2 = 34344233+343333344i   (23704+1811i)^2  = 558599895+85855888i
(9998+3334i)^2   = 88844448+66666664i    (4999+1667i)^2   = 22211112+16666666i
(3507+2369i)^2   = 6686888+16616166i     (16667+16666i)^2 = 33333+555544444i
(1754+719i)^2    = 2559555+2522252i      (2087+106i)^2    = 4344333+442444i
(1667+1666i)^2   = 3333+5554444i         (328+175i)^3     = 5152552+51122225i
(Base 9) 2534^2  = 6666677               (Base 9) 32641^2 = 1181111181
(Base 8) 13240265^2 =  177771777177771   (Base 9) 35577^2 = 1414111444

Robert Henderson has found a remarkable Latin pentacube solution.  With the following division of a 5x5x5 cube into 25 different pentacubes, find a a way to color the cube in 5 colors so that every row, column, stack, and pentacube is comprised of all 5 colors. Here is the answer.

 N N N N M    W W F N M    C W F [ M    D S F Q J    D S S S J
C T O O O C T T [ [ C W [ [ M H H F Q M D S Q Q J
B T T O O B B B \ G C W B G G D H F Q G D H H J J
V X X Z \ V V X \ \ Y V X \ G R U U I I R R U I K

Some prime numbers remain primes when reversed.  Are there an infinite number of them? John Gowland sent me the following cross-number puzzle called Reversed Squares, which is based on reversible primes.  Here are some clues, a solving strategy, and the answer.

material added 5 January 2003

Those who have Zillions of Games can try out Bob Kraus's very nice Extraction puzzles, now at the Zillions site.  Eventually, some brave soul is going to need to look at all of the Zillions files that are available, and summarize what is best.  That will be hard, because most of the files are quite good, and time-absorbing.

Erich Friedman:  Sometimes a day is the sum of the digits of the horizontally and vertically adjacent days on that month's calendar. For example, this happened on december 14, 22, 24, and 26 of 2002. There is only one day in 2003 on which this will happen but be the ONLY time that month it will happen. When is it?

Here are some small fractions that make a nice approximation to a familiar number.  22/17 + 37/47 + 88/83 != Pi.  Can anyone find a better approximation with small fractions?  I've noticed the greedy algorithm doesn't work very well at finding better solutions.

Some things, I've been keeping around.  For example -- here's a little puzzle -- I can't remember if I made it, and if I did make it, I can't remember the answer, and I haven't solve it.  So it wouldn't be all that fair to present it, would it?  Well, here it is, anyways.  And here is a whole page of other material I never quite figured out how to present, so here it is, all in one big batch. A similarly disorganized page is my g4g5 writeup.   Comments are welcome.  Answer to 4-divide (Solutions sent by Remmert Borst, Cihan Altay, Paul Cooper, Jonathan Welton, Jon k McLean, Franz Pichler, Clinton Weaver, Agaeus, Matt Elder, Joseph DeVincentis, Kirk Bresniker, Jeremy Galvagni, and Juha Hyvönen) If you like division puzzles, I don't think anyone has solved all of Mike Reid's puzzles.

The various integer-sided blocks with sides 0<a<b<c<7 will fit into a 7x7x15 block, as shown by Erich Friedman.

Warning: Big File!  Edward Brisse compiled all of the Triangle Centers into one big text file.  I won't give a direct link, but you can find it at EdwardBrisse.txt, or EdwardBrisse.zip, on this site.

material added 31 December 2002

While playing around with Pick's Theorem, I came up with a deceptively tricky little problem. Divide a 5x5 square into 5 regions which have identical perimeters but differing areas.  All lines must be straight, and must connect grid vertices. Answers. Another by Livio Zucca. I used a trick in my solution (1) -- I was surprised when Joseph Devincentis sent me different solution (2): "This was a very nice puzzle, and a wonderful demonstration of the usefulness of Pick's Theorem. With the theorem I could quickly see that I needed to somehow split up the regions using lines that crossed only 6 of the internal points, with the five regions containing 0,1,2,3,4 of the other internal points. It still took me a while to find the correct perimeter and arrangement of lines to make it work."   After that, Daniel Scher and Martin Bernstein sent solutions (3) and (4). Taus Brock-Nannestad sent a 5th answer, and William sent 20 more (using the 3-4-5 triangle trick in a normal grid).  

Theo Gray and I managed to collect all 90 stable elements.  We have started testing some of the samples.  One of the most bizarre -- a weird rock I found when I was six years old has turned out to be 38% titanium. On the other hand, a "Titanium" tennis racket wound up having no titanium whatsoever.  

Stephen Wolfram has made some of the Historical Notes from A New Kind of Science available. I like what he does with WireWorld.

The Retrograde Anaylsis Corner has made some nice new updates.

material added 25 December 2002 (Merry Christmas)

I've updated my Prize Puzzle page.

NetLogo 1.2 has been freely released at the Center for Connected Learning and Computer-Based Modeling, at Northwestern University.  The Logo language is frequently known as the language for "programmable turtles."  A turtle with the instruction set {move 1, turn 90 degrees} would make a square.  There is much, much more, and this releases is filled with lots of excellently documented programs in art, biology, chemisty, physics, computer science, earth science, mathematics, social science, and more. No programming knowledge is necessary, you can just start up the models and what how they work. With slight programming, the models are easily modified.  It's a wonderful package for learning.

Jean-Charles Meyrignac found an old puzzle that involved the 18 ways to three-color a tromino. How many can be placed in a 7x7 square so that only like color touch?  It seems 13 is the answer. In an 8x8 square, he can place all but 1 piece, and isn't sure if all 18 is possible. He wrote a program to find how many of the 40 4-colored trominoes could be placed in an 11x11 square, and he found two solutions with 33 pieces placed.  He does not know if these can be improved.  It's a nice task to split the grids into the different trominoes.

1 1 1 1 1 2 2 2 2         1 1 2 2 2 1 1 1 3   1
1 1 1 2 2 2 2 3 1 1 2 2 2 1 1 3 1 2
1 2 2 1 2 2 2 2 1 3 3 2 2 1 1 1 1 1
3 3 3 1 3 3 3 3 3 2 2 2 4 4 4 4 1
3 3 3 4 4 4 3 3 3 2 1 2 4 4 1 1 1 1
4 4 4 4 4 4 4 3 3 3 1 2 4 4 4 3 1 2 3
3 2 2 2 2 2 2 1 3 1 2 2 3 3 3 3
3 1 2 1 1 4 4 4 4 4 3 2 3 2 2 2 3 3 3
1 2 1 1 1 1 4 4 3 3 2 3 4 4 4 4 4 4
4 1 2 4 4 4 1 4 3 3 2 2 3 4 1 4 4 3
2 2 4 1 1 4 3 1 2 4 2 4 1 2 4 4 1 3
After picking up some domino cards, I designed a little game I call Double-9 Trumpery. Seems to work fairly well. I still like my Ups&Downs game -- it works better with domino cards.

Updating my list of weblinks, here are a few free programs everyone with a PC should have: Mozilla (best browser), Irfanview (opens anything), Zone Alarm (excellent firewall).  

The Game of Clobber now has some award winning problems.

I rather like Pascal's Triangle at The Sound of Mathematics. More algorithmic music is at Tune Toy. Somewhere on my site I need to update details of Don Wood's 20 questions -- a fixed version is at his site.  Listening to various carols recently, I noticed a six letter word with 18 syllables. Kevin Wald noticed the same word is sung with 21 syllables in "Ding Dong Merrily Along." There is a 36-syllable five-letter word in "Poor Wandering One" from Gilbert and Sullivan's Pirates of Penzance. Beating that, in Handel's "For Unto Us a Child Is Born," the word "born" is sung with 57 syllables.  Gordon Bower: "The longest one of these that comes to mind is from Mozart's Magic Flute, about two-thirds of the way through the Queen of the Night's aria "Der Hoelle Rache kocht in meinem Herzen": "Bande" gets 86 notes in the queen's part (85 for Ban- and 1 for -de). Admittedly some of these are slurred together, so I suppose it is only 70-ish syllables."

The CRC Concise Encyclopedia of Mathematics, Second Edition by Eric Weisstein is now available. If you use the link I provide, Eric will get a percentage of every sale.  It gives a good glimpse at the scope of his MathWorld site. Hopefully, in a future edition, CRC will give Eric standard royalties for his work, and full control over how it is printed.  That would allow for a much better book, more up-to-date, with better pictures. It's very sad that CRC has done so much to give Eric a hard time about the material he created.

In contrast, over the past few weeks I've been wending my way through Mummy Maze Deluxe. The game has an interesting history. Originally, Robert Abbott designed the Theseus and Minotaur maze for Mad Mazes, which was later turned into a Java applet by Toby Nelson. Popcap inadvertantly copied the idea for their game Mummy Maze.  Soon after realizing their error, the Popcap company apologized to Robert, paid him, and now are giving him credit. Legally, they didn't have to do any of that.  Some of the mazes in MMD are quite tricky, if you get stuck, you can use the bardavid solver. Some reviews.

Mummy Maze

I took a look at minimizing h in a2+b2 = c2+d2 = e2+f2 = 2*g2  ,  a2 + c2 + e2 = 3*g2 + h  ,   b2 + d2 + f2 = 3*g2 - h.  With all variables as different ppositive integers, Claudio Baiocchi, Jordi Domènech, Juha Saukkola, Abdullah Gencay and Denis Borris found: {a=2465, b=1015, c=1363, d=2291, e=1651, f=2093, g=1885, h=120} and {a=3767, b=119, c=635, d=3715, e=2591, f=2737, g=2665, h=120}  Can h be lower than 120? (With g under 60000, there is no better solution.)  Send Answer.

Fred W. Helenius noticed that 2542645806624 is 100101000000000001100000000000011000100000 in binary, and 100000002000100200000002000 in ternary.  At most 4/21st of the digits in the ternary/binary representations are non-zero. Is there a positive integer with a lower percentage of non-zero digits in both the binary and ternary forms? My best finds were 1208614932, 2453670144, 17448310278, and 22083026976.  Send Answer.  Update:  Robert Harley search up to 2^64, and found 451521135633235968, which gives 11001000100001000000001010010100010000000000000000000000000 and 10000020000020200000020000000001000000.

Martin Watson rediscovered a nice puzzle -- pack 11 F pentominoes into an 8x8 square, or a 4x4x4 cube. Both solutions are unique. The puzzle is actually sold commerially, at Polzeath, Cornwall.  Patrick Hamlyn found the puzzle to be too easy for computer solving, so he offers a counter challenge: Fit 20 solid Z-pentominos plus five other pentominoes into a 5x5x5 cube. Send Answer.

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