Material added 17 August 2003

NJA Sloane has a great
article at ams.org about the OEIS.

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

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

Material
added 15 August 2003

Theo Gray's
latest Popular Science
column goes into the fun you can have with a microwave. For
example, you can make Grape
Plasma Fireballs (picture near bottom), or incandesce compact
discs. The main part of the article talks about melting metals in a
microwave. Speaking of heat, the Wired
article on artificial diamonds is fascinating.

Ivars Peterson's
current project is Science
News for Kids. It's part of a large project to add depth to
education. Other involved parties include Thinkfun, Scott Kim, and the Slocum
Puzzle Foundation. Scott's latest Discover
column talks about some familiar mazemakers.

R. William Gosper: Last month I had the
pleasure to visit the Microsoft Theory Group. Mike Sinclair and Gary
Starkweather in a neighboring group generously helped me laser-cut a
puzzle consisting of a 8" diameter circular tray and twelve unequal
circular
disks with diameters {1.14662, 1.24041, 1.33278, 1.52265, 1.65921,
1.74812,

1.83679, 2.36627, 2.54162, 2.63057, 2.73347, 3.03309 ©2003
Bill Gosper}. To my amazement, *no one* has yet succeeded in
fitting all twelve disks back into the tray, although Christian Borgs,
a Microsoft physicist, solved two slightly easier versions. This puzzle
is fairly resistant to computer solution--I'm not even sure how to
quantify the combinatorics of the search.

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

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

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

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

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

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

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

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

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

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

Material
added 7 August 2003

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

A working brass knotgear, designed by Oskar. See animation.

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

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

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

Find a non-crossing path of length 18 in the above graph of knight
moves.

At a site devoted to brilliant numbers,
I learned that 10^{105} + 4293 is the product of the following
two numbers:

12842269977506686435684442645814821681335988332440829

77867853716788858218366878517755270539950026013843817

77867853716788858218366878517755270539950026013843817

Material
added 29 July 2003

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

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

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

For picture puzzles, one of the best is
Blehm's Penmen series (Penmen,
Space,
Music)
where one must find the two identical figures within a large poster.

Material
added 17 July 2003

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

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

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

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

For my Yahoo Groups
picture, I use the
sun, as seen by the SOHO
satellite. Another interesting satellite is ISEE3. A
good source of astronomy pictures is the Astronomy Picture of the Day.

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

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

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

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

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

Jorge Mireles
made a nice graphic of how pentomino sets can mimic each other.

Material
added 14 July 2003

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

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

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

SIAM News has
many interesting articles, for example a discussion on
origami. An interesting origami program is Tess.

Patrick Hamlyn has a spiffy polyform
solution in the enneiamonds.

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

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

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

Material
added 6 July 2003

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

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

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

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

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

Here is a new puzzle by John
Gowland. Answer and solvers, SolutionPart1, SolutionPart2.

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

Older material below. One
ongoing note ... you can always subscribe to my moderated mailing list
at http://groups.yahoo.com/group/mathpuzzle/

**Older Material - 1 Apr 2003 to 29 Jun 2003**

**Older Material - 25
Dec 02 to 26 Mar 03**

Older Material - 19 Aug 01 to 4 Dec 01

Older Material - 17 Feb 01 to 5 Aug 01

Previous Puzzles of the week are here.

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