material added 26 Aug 2002
Jorge L. Mireles has written a gorgeous applet based on rotating decagons and stars. Just 12 easy pieces.
Owen Muniz has gotten help from Kadon Enterprises to make a gorgeous new puzzle-game, L-topia. It's all 12 ways you can drill a circle and a square out of an L tetromino. I'm trying to find a 7x7 square with a missing center so that every row and column has 1 or 2 circles and 1 or 2 squares.
Aad van de Wetering of Driebruggen sent me scads of solutions to various problems, all solved using his program FlatPoly2. An english version of his program can be found at the bottom of his website.
Richard K Guy solved Patrick Hamlyn's 30-60-90 problem partridge problem in glorious ascii.
Sudipta Das of Calcutta , India sent me a betting problem on the Game of Fifteen. Players A and B randomly draw balls numbered 1 to 9 from a bag, with player A starting. If A possesses three balls totalling 15 first, A wins. If B gets three balls totalling 15 first, *or* if no-one gets 3 such balls, then B wins. For example, the play might go A2, B5, A7, B3, A8, B9, A4, B1 and B wins the game, with 5+9+1=15. Note that A does not win with 8+7=15, since the specification calls for three balls. You are offered an even money bet to play A or B - which do you pick? Answer. In his analysis, Sudipta Das found 255168 possible games. Note that this is equivalent to playing "dumb" tic-tac-toe.
Alan Lemm sends an interesting observation. "I've just read your page about integer sided triangles in which one angle is an integer multiple of another. I've just constructed a list of 57 triangles (so far) for the case k=2. I've noticed that the side opposite the smaller angle is always a square number. Additionally, the smallest triangles for cases k=2 to 7 listed on your page each have a side that is the kth power of an integer (823543 = 7 ^ 7, for example). Has it been proven that for any k, one side of the triangle must be the kth power of an integer?" Answer.
The Pitch Drop experiment has been running ever since 1930. The questions of the 2002 International Math Olympiad are now available. The Open Problems Project is well worth a look.
Mathematica Teacher's Edition has just come out. It supplies specialized tools for precollege mathematics teachers, allowing the automatic creation of quizzes and answer keys. If you're a student instead of a teacher, you can get the Student Edition. If you're a hobbiest, you can get Mathematical Explorer. (And tell Customer Service that Ed Pegg sent you).
material added 18 Aug 2002
Patrick Hamlyn discovered a wonderful puzzle: arrange the 10 triangles on this page to make another 30-60-90 triangle. The solution is unique. John Conway, commenting on Hamlyn's discovery: "very nice answer ...wonderful!" (As a follow-up, make a rectangle frome the same 10 pieces.) This is a new discovery for the Partridge Problem, and was discovered by Patrick Hamlyn. You can find the answer at Erich Friedman's Math Magic site. The Partridge number of the 30-60-90 triangle is 4. Patrick found that the partridge number of a different shape is 6. If you add 5 30-60-90 triangles to the 10 above, there are 123 ways for the 15 triangles to make another drafter.
Ivars Peterson has written a wonderful article about Rush Hour and the complexity of Sliding Block Puzzles.
Nick Gardner: "I loved Brian Trial's WireWorld multiplier
so much that I had to make my own. My version is in the attached
file. It's just a little bit smaller than Brian's version. :-) There's some
brief notes about how it works in the pattern description." This is just
stunning to watch in action in MCell.
I didn't think multiplication of large numbers could be done with a simple cellular
automata in such a small size, but there it is.
I've started a page about Gaussian Integers.
material added 8 Aug 2002
Andrea Gilbert has introduced Orientation Mazes at her site clickmazes.com. In an orientation maze, you can only perform the actions that you can currently see. If you have a bit of floor space for 16 sheets of paper, you can use one of these designs to make your own walk through maze.
Mark Newbold has made a java applet for slicing 4D objects: See Hyperstar here.
From Patrick Hamlyn -- using 4 X, 8 Y, and 8 Z pentominoes, make a 10x10 square. The solution is unique. For a slightly easier puzzle, make a 6x10 rectangle with pentominoes FILLNPPPPUVY. There are 1462044 solutions. (No need to send answers).
Brendan Owen sent 7 puzzles (1, 2, 3, 4, 5, 6, 7). Each puzzle has 8 pieces. Remove a piece, and then make a 7x7 square out of the remaining pieces. In all, there are 56 unique solutions. (no need to send answers). A related puzzle is below. Remove one piece so that the remaining pieces cannot make a 6x6 square. The piece is unique. Answer.
Jorge Luis Mireles has put together a page on tiling with Lenses.
Oyler has sent me a zip file with 20 number logic puzzles by Rhombus, from The Listener. Here is list of known errors. "They all appeared between 1960 and 1980. I've typed them in rather than scanning them as some of the photocopies are not very good. I was out with a magnifying glass for some of the puzzles in the Rhombus3 folder to try and discriminate between c and e and also 3 and 8 in some puzzles so be warned. I think they are all correct. I haven't myself got round to trying them as yet!! The numbers after the title refers to the number of the puzzle in The Listener Crossword series. Rhombus's last puzzle was All Square ( 2547 ) and appeared on 8th May 1980." If you have trouble with DOC files, openoffice.org has a good substitute program.
Cleverwood has added many new wooden puzzles at attractive prices (my commendations to them for offering this selection). No-one solved my trivia question on Hafnium -- it makes the tips of plasma cutters, which you can buy inexpensively at any welding supply store. Jeff Barnes pointed out that Hafnium is an impurity (often 5%) in cubic zirconia. In a slightly related note, a nice set of spectographs for the elements is here.
While I'm on elemental trivia, I just found out that Cadmium is slightly radioactive. An average sample of cadmium, such as in a NiCad rechargable battery, is about 12.22% 48-Cd-113, which has a half-life 7.7 x 10^15 years. That's older than the estimated age of the universe. Hmmm... so ... how many clicks per hour does a cell phone give off? Answer.
I play with my Fill-Agree set a lot. There are 20 different ways to puncture a tetromino with two holes. I show 18 of them below. Note how the two figures have identical hole placements. I asked readers to select 18 of the 20 punctured tetrominoes, and make 3 identical figures. Helmut Postl solved the problem, and I'm presenting it as a new problem. Divide the three figures into 18 different twice-punched tetrominoes, so that each line is between two pieces. Answer.
material added 6 Aug 2002
Brian Trial took up my challenge to make a working Multiplier in WireWorld. It can be run in the free program Mirek's Cellebration. Brian: "Without Mirek's excellent application I couldn't have begun to put this together. Note that multiplication is done using a SINGLE full bit adder, with the carry bit and output fed back in as feedback loops. The entire operation takes about 18,000 generations to complete, so go get a cup of coffee. Don't forget to light the FUSE or the result won't get captured properly. The file as shown will multiply $A3 (value in b7 - b0) by $95 (value in a7 - a0) and put the result in c15 - c0." The MCL version of this construction is right here. It uses lots of OR, XOR gates and AND gates.
WireWorld is a cellular automata, with a simple rule. Black (background) goes to Black. Red (electron head) goes to Blue (electron tail). Blue goes to Yellow (wire). Yellow goes to Red if it touches exactly 1 or 2 Reds. That's all. Here is the .mcl version of the various WireWorld gates pictured below.
A bigger update comes later tonight.
material added 30 July 2002
From David Terr -- For this week's puzzle, you need to know what
Mod (or Modulus) is. In Mod 10, 6+5=1, instead of 11. For Mod 10, you keep subtracting
10 until you have a a number less than 10. With a negative number, you'd keep
adding 10 until the number was positive. So 5 + 12 is equal to 7, Mod 10. 5
- 12 is equal to 3, Mod 10. That's how Mod works.
Puzzle: What is remarkable about the first nine digits of Pi (3 1 4, 1 5 9, 2 6 5), if you use Mod 10? Answer.
Martin Gardner mailed me another Pi curiousity : 123456789101112131415.
Astronomy: On August 18, an 800 meter diameter asteroid will pass 1.3 times the distance from earth to the moon from earth. At this distance, the asteroid will move across the sky pretty quickly, moving across two constellations in 4 hours. It should be visible in binoculars. For more on this story, visit NASA. (courtesy Jeff Bryant)
Erich Friedman has some interesting circle placements by Clinton Weaver and Dan Hoey here. They need to be analyzed further. Some of these might make better rigid packings. I particularly like 16b. Erich's fairly certain that all the packings at his main rigid packing page are rigid.
If you have an interesting math item, you can always send it to me. Alternately, you can send it to the new Online Journal of School Mathematics, an effort sponsored by the National Council of Teachers of Mathematics. Unlike me, they will pay up to $150 for each article. John Mahoney is one of the editors. If you have questions, he'd be glad to answer them. He's at (@) sidwell.edu, and his user name is mahoneyj.
material added 24 July 2002
Clint Weaver managed to improve several results on the Rigid Packing page. In particular, he found a much better result for 13 circles.
Ken Duisenberg has created a new puzzle type he'd like people to try at his weekly puzzle site (link).
Erich Friedman: Take an 18x18 grid of squares, and remove two of them so that the resulting figure contains exactly 2002 squares of all different sizes made from the remaining squares. Clinton Weaver sent the answer below. Clinton: "I have what I believe is the full list of solutions for Erich's 18x18 puzzle. I wrote a program to compute them all (doing it by hand would take ages). Unless I made a major error in my program (which, I'm willing to admit, is entirely possible) these should be correct. These solutions represent the basic pairings. Reflection and rotation of these basic pairings will produce solutions as well. Again, I'm pretty sure these are all of them. My thanks to Erich for a very tough puzzle." If correct, this is an improvement on Erich's answer .. he only found the first two.
Each square has 2002 squares not containing the red squares.
Another good puzzle by Erich, from his Math Magic pages: Divide a 20x20 (or 23x23) square into 9 squares or dominoes with sides 1-9. Divide a 25x25 square into 10 squares or dominoes with sides 1-10. Clinton Weaver provides the answers.
Jaap's Puzzle Page has many path/rubic type puzzles, histories, and solutions, including many unfamiliar to me. Cubism For Fun has a new website.
Oyler -- I found your site whilst surfing the net and was intrigued by 34567 by John Gowland. This sort of puzzle is very similar to the mathematical puzzles that appear about 4 times a year in The Listener Crossword series published every Saturday in The Times ( London ) newspaper. I have been lucky enough to have set a number of these puzzles over the past few years - the last one, Euler's Spoilers, appearing a couple of months ago.
Seven A Side by Oyler
Cihan Altay: Three talented mathematicians called A, B and C participate
in a competition. In a box, there are 99 cards with a different integer from
2 to 100 written on them. Each mathematician will pick a card and try to find
the numbers of the other two before them and as soon as possible, by asking
some questions. A referee will decide the order of the questions by looking
the numbers of the cards; the mathematician with the biggest number will ask
first, the one with the smallest number will ask last. Each question should
be answered correctly by the mathematicians except the one who asks. So, the
questions, in order, are:
A: Is your number a perfect square?
B: Is your number greater than 50?
C: Is your number a perfect cube?
Just a second after the last answers, they all say "I've found", simultaneously. What were the numbers of the three mathematicians? Answer.
After being highly intrigued by our wonderful Titanium and Zirconium samples (theodoregray.com), I decided to track down some Hafnium. It turns out you can buy Hafnium at type of store found in almost any town. What kind of store, and what are you looking for? Well ... no-one figure this out. Plasma cutters have hafnium tips. You can pick up a plasma cutting tip at any welding-supply store. (This might be a good "show and tell" item for school.)
For Mac Users, LifeLab 4 is available. CA FAQ by Tim Tyler is a good source for many such items, and has an absolutely gorgeous realization of flame with a cellular automata. If you'd like to experiment with large constructions, I'll award $10 to the first person that can make a WireWorld construction capable of multiplying two 8 digit binary numbers.
Nathan Stohler sent me this nice diagram showing the solution of my Plus maze. Other solvers.
material added 16 July 2002
With the help of David Cantrell, Erich Friedman has vastly expanded the Rigid Packing Page. Erich suspects many of these can be improved.
48 49 7^2
120 121 11^2
360 361 19^2
840 841 29^2
1680 1681 41^2
5040 5041 71^2
The number 24 has 8 divisors. No smaller number has that many divisors, so we call 24 a highly composite number. All of the numbers in the first column above are highly composite. By adding 1, we squares of prime numbers. I checked to see if this happens again, and I was surprised to find another example. Can you find it? Answer.
Preda Mihailescu wrote to me, and let me know that he proved Catalan's Conjecture with a pure proof. No computer searches were involved (as I had mistakenly reported). His proof is still being checked, but I can mention that the core argument is just 12 pages. Whether Catalan's conjecture holds for powers of Gaussian Integers is still unknown.
For those that like computer searches, 59,974,310,000 zeros are know known for the Riemann Hypothesis.
With the work on the Periodic Table that I've done, I can't help but comment on the recent debate in the US about the Yucca Mountain nuclear storage area. So far as I can tell, the original sensational news story about radiation concerned a very wealthy industrialist by the name of Eben Byers. Back in the late 1920's, there was a "health tonic" by the name of Radithor, and it's makers attached all sorts of amazing health benefits to their radioactive thorium/radium drink. (Beware of hyped-up health benefits from dubious sources). Eben drank two or more bottles daily for more than two years. Eventually, all that caught up with him. The press loved the ongoing story - there was a full page spread in Time magazine when he died. Because of his case, the Food and Drug administration was given control over radiation treatments. And ever since, radiation stories have had a special place in press sensationalism.
Today, thorium is still avilable in stores, in lantern mantles. Americium is in smoke detectors. Radium paint is in many antiques. All of this is far more radioactive than the police escorted trucks that will be on US roads in 2010. I am ashamed of National Public Radio for playing the rant of one activist who claimed that "thousands of bodies will be on the roads in 2010 [because of radioactive material shipments]." If it weren't for the activists planning to hurl themselves in front of the speeding trucks, it would all be perfectly safe.
Theo and I decided to see how flammable Magnesium really is. After 10 minutes with a blowtorch, it lit right up. Zirconium is much easier to ignite. Randall Fullman sent us the sample ... the photograph is of zirconium shavings.