Something new: STEAM activities for the Covid19 lockdown!
Yes, the Playwiths have been converted into a book.The Playwiths began in about 1995, and a couple of years back, I was urged to make a book of them.I did, and my friends liked what they saw, but the publishers didn't. Frightening economic times, they said. Well, I went ahead and did it in three forms:
Full details of Playwiths, the book here 
Find out more about GEM 


Goldbach's conjecture
Another interesting conjecture, this time concerning odd numbers, was put forward in the middle of the 19th century by a French mathematician, A. de Polignac. De Polignac's conjecture was that "every odd number can be expressed as the sum of a power of 2 and a prime". In presenting this notion, de Polignac implied that he had tested this against all odd numbers up to three million. While most odd numbers can be constructed by adding a prime to a power of two, 127, which is also a Mersenne prime, fails the test. All of the possible constructions of this number ( 2 + 125, 4 + 123, 8 + 119, 16 + 111, 32 + 95 and 64 + 63) involve adding an odd number which is a composite to a power of two. Perhaps you can find the next number that fails de Polignac's conjecture. Back to the details  Back to the index
Pentominoes
Another interesting challenge is the "double double" where you make the same shape with two pentominoes, make a second copy with two more pentominoes, and then make a doublesized model with the remaining eight pentominoes. Why not make a set to use at home? Back to the details  Back to the index
Finding pi
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The game of Eleusis
Back to the details Back to the index Runs of composites Not too much help just yet. To test if a number x is exactly divisible by another number n, you use the form x/n=INT(x/n). I have some kludgey spreadsheet solutions, but no really good ones . . . Back to the details Back to the index
Soma cubes
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The game of Nim
Safe positions are those where all of the columns add up to even numbers: here, the solution is to take out two coins from the top row, which now reads 001, so that the total across the bottom is 202, indicating a safe position.
Well there you are: you work out how to win a game of Nim with four rows instead of three. Back to the details  Back to the index
Pascal's triangle
By the way, if you draw a slanting diagonal through the 1 at the left of the fifth row, then through the leftmost 3 in the fourth row, and the 1 on the right of the third row, the total of the numbers is 5. Now if you do the parallel diagonals above and below and add the numbers on each diagonal, you will discover a wonderful thing. Look at a few of them, and it will hit you . . . Back to the details  Back to the index
The game of Sprouts
There can be no more than 3n1 moves in a game which starts with n points, and that there must be at least 2n moves in any game. The one and twopoint games are trivial, but games with three or four points allow plenty of room for practice before you go on to larger and more complex games. There is a proof that the first player wins the three, four and fivespot games, while the second player will always win the sixspot game, provided both players always play logically . That little phrase is very popular with mathematicians who study games . . . Back to the details  Back to the index
Analysing the crossed house problem
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Analysing the prisoner's problem
And that's a hint. If you draw this figure on a torus in such a way that the hole of the torus is inside the middle cell at the bottom, it might be a bit easier  and that's a hint, too! Back to the details  Back to the index
Analysing the Königsberg bridge problem
And that's a hint. Back to the details  Back to the index
More about Penrose tiles
Last time I made some of these, I cut them from 8 mm polystyrene foam sheets, using a sharp knife, a cardboard template for each of the two tile shapes, and a piece of scrap plywood as a cutting sheet. I found that sometimes I tiled myself into a blockage which made me go back, maybe ten or fifteen tiles, and then move on again. Now where have we heard this aperiodic stuff before? The answer is in a small book by Erwin Schrödinger, called What is Life? , in which Schrödinger predicted, before DNA was identified as the genetic material, that the message would be contained in an aperiodic crystal. These days, we forget that, and only remember Schrödinger for his conundrum about the cat. Here is what Schrödinger wrote, way back in 1944: . . . the most essential part of a living cell  the chromosome fibre  may suitably be called an aperiodic crystal . In physics we have dealt hitherto only with periodic crystals . To a humble physicist's mind, these are very interesting and complicated objects; they constitute one of the most fascinating and complex material structures by which inanimate nature puzzles his wits. Yet, compared with the aperiodic crystal, they are rather plain and dull. The difference in structure is of the same kind as that between an ordinary wallpaper in which the same pattern is repeated again and again in regular periodicity and a masterpiece of embroidery, say a Raphael tapestry, which shows no dull repetition, but an elaborate, coherent, meaningful design traced by the great master.Erwin Schrödinger: What Is Life?, Canto edition, (with a foreword by Roger Penrose!), 1992, page 5. If you want to explore this sort of thing further, and see how aperiodic crystals form in three dimensions, search the Web for references to quasicrystals . Interestingly, these quasicrystals also show a fivefold pseudosymmetry, never before seen in nature  except in the echinoderms! A further reference: if you do not already own a copy of Fractint 19.3 or better, go and get one  it is out there on the Web, and four of the lsystem fractals are Penrose tilings in this form. Fractint is freeware, and easy to find on magazine CDROMs or on the Web. LOGO code for darts (by Noel Wood)This should paste directly into your LOGO engine
to dart
1729 and all that
ProbeFourth powersThere is a solution to the problem of finding a number which is the sum of two fourth powers in two different ways. The numbers 133^4 and 134^4 are the numbers for one solution: the rest is up to you. It should not be difficult to write a small computer program, given this information, to extract the rest. Back to the details  Back to the index
Fibonacci numbers
Probes
Lucas numbers
Q numbers
To find the next number (shown here by "x"), count back the number of places from the x position shown by the two numbers before the "x". This will bring you back ten places to the bold italic 5 in the series, and nine places to the bold italic 6. Adding these together will give you 11, the next term in the series.
That's all you get: make up your own problem :) Back to the details  Back to the index
A few days later, Frank was back again, asking Nevertheless my question is if I might use double factorial function as well? For example, 8!! = 2 × 4 × 6 × 8 = 384. A definition for n!! can be found here http://en.wikipedia.org/wiki/Factorial. This would give us more flexibility: 32 = (1+7)!!/2/(Sqrt(9)! So now I guess Klaus will come back and search for more! I hope he does! But for the record, like the knife fight scene in Butch Cassidy and the Sundance Kid, there are no rules! Just mathematics. Post script 3: Frank Schnell came back— and brought the missing answers with him! On January 24, 2011, I had this email: Hi Peter, seems the missing numbers are 32, 34, 38, 49 and 50, right?
(1+7)!!/2/(Sqrt(9)!)=32 Please check that. Meeting is over now... Mit freundlichen Grüßen / Best regards Frank Schnell A small admission from me: in the long time this page has been running, my life has changed in many ways. The key changes for here: I have turned myself into a hardworking writer, and I have become a grandfather. Within a few days of getting this email, I was due to complete and print out the first draft of a new book, so I can take it to New Zealand while we visit the grandchildren. So, with time tight, I have pasted the solutions in, unchecked, but I trust Frank! Back to the details  Back to the index
Hofstadter is less than helpful at first about ties: "We found a makeshift way of handling that case". Later he says: "Let each player have a slightly shifted range, relative to the other two. More concretely, let player A pick integers from 1 to 5, B picks numbers of the form n + 1/3, and C picks numbers of the form n + 2/3, where n runs from 1 to 5." In this version, the winner (the middle player) gets his/her number, while the other players get the nearest score they can to zero. So if the numbers are 1^{1}/_{3}, 2 ^{2}/_{3}, and 4, the scores will be ^{1}/_{3}, 2 ^{2}/_{3}, and zero. From there, it's over to you  or you can look in Hofstadter's book at page 711. Back to the details  Back to the index
The Diabolical cube
So there you have it: the puzzle can be solved: all you have to do is work out where to start. 
I don't plan to reinvent the wheel. For an excellent set of links to interesting mathematical sites, go to Nina Heal's Cool Math Sites and rummage around. It is well worth an hour or five of your time.
One homebrewed mathematical site with some remarkable effects is Adrian Bruce's Symmetry page. Go look, both for clever Webbery and also for some excellent mathematics.
Note well: That email exists, but only as a spam trap. To get in touch without delay, add my first name on the front of the email address. Robots can't do that.
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