Some numerical amusements

Goldbach's conjecture
This unproven suspicion of mathematics suggests that every even number larger than 2 can be expressed as the sum of two prime numbers (4 = 3 + 1, 6 = 5 + 1, 8 = 7 + 1 = 5 + 3 etc.).

Your mission, should you accept it (yes, the reference was intended), is to explore the pattern of even numbers which can be "Goldbached" in more than one way. Don't ask me what the answer is, because I have no idea where it might lead.

This will help you give you some extra information

Pentominoes
These are the Pentominoes (the name is a registered trade mark of Solomon Golomb). There are twelve of them, and they are each made up of five squares. You can lay them down either side up, and you can use them to make a variety of shapes.

First, get some cardboard or light board of some sort, and make up a set, using the patters you see here. Once you have done that, see if you can arrange them into a square, 8 x 8, with a small square, 2 x 2, in the middle somewhere.

Then try to see if you can arrange all of the pentominoes to make a 6 x 10 rectangle, then a 5 x 12, a 4 x 15 and a 3 x 20 rectangle. Golomb has published at least one book on the subject -- try your local library to see if you can track it down. Martin Gardner has also written on the subject.

This will help you understand

Pi via spreadsheets
There are quite a few series that converge on a value of pi, or some function of pi. Here is one of them:
(pi^4)/96 = 1/1^4 + 1/3^4 + 1/5^4 + 1/7^4 + 1/9^4 + . . . .

These instructions will help you to create a spreadsheet which will get to pi to about five decimal places in about 250 rows.

Begin with the value 1in cell A2
Now use EDIT - FILL SERIES (step 2) to get the numbers 3, 5, 7 . . 199 in the cells down to A101.
Now enter this formula in B2 : =1/(A2*A2*A2*A2)
And put this in C2: =C1 + B2 (this will give us a running total of column B, up to that row).
Then put this in D2: =SQRT(SQRT(C2*96)) to get the first estimate of the value of pi.
Now you can highlight cells B2, C2 and D2, and then highlight down to row 250, and use EDIT -- FILL DOWN to copy the formula down into those rows as well, and extend column A down to row 250 (think about this!!).
Check the answers that you get in cells D246 and D247, after you have extended the spreadsheet down to row 250. The value you should be aiming at is 3.141 592 653 589 793 238 462 . . . but that will take a while longer . . . :-)

This will help you fix any problems

The game of Eleusis
You will need an ordinary pack of cards (no jokers), and a group of four or five "players". This game is intended to help you get some understanding of how scientific theories are formed or tested. You will be either setting up a "law of nature", or you will be trying to work it out.

The game starts with the lawmaker checking the number of cards, so that each of the players gets the same number of cards, with one card left over (the lawmaker has no cards). With two other players, take out one card at random and lay it face down, with three other players, take no cards at all, with four other players, remove three cards, and so on. Do NOT look at the removed cards.

The one card left over is laid down as a starter card, and the lawmaker thinks of a "law", something like

red follows black and black follows red; or
red follows black and black follows red OR odd follows even and even follows odd;
red follows black and vice versa, unless the card is a 7 or higher, when you follow suit, and then writes the "law" down, so it can be inspected afterwards;
a court card always follows a non-court card, and vice versa; or
cards must appear in the order spades, clubs, diamonds, hearts, but you can play any card after a court card.

The players take it in turns to try to lay a card on the starter card, with the lawmaker saying whether the card can be left there or not. If you work out the "law", you can get rid of your cards sooner, so you need to watch which cards are accepted and rejected. All rejected cards are placed face up in front of the player who tried to play them, where they can be seen. All the "correct" cards are fanned out, so all of them can be seen as well.

When all the players' cards are on the pile or face-up, the lawmaker's score is made by working out how far the leading player is ahead of the other players. If there are two players, take the leading player's number of cards away from lower player's number of cards. With three players, take double the leading player's number of cards away from the other two players' number of cards, and so on.

Now you start trying to play the face-up cards, with the lawmaker checking as before. When one player has no cards, or when the lawmaker sees that no more cards can be played, the second stage finishes. The rule is now checked, and once everybody is satisfied that it has been followed, the players are scored.

To do this, you take the number of cards you hold, multiply that by the number of other players (not counting the dealer), then take away the number of cards the other players hold. The player who laid down all of the cards gets a bonus of 6, and negative scores count as 0.

The deal then passes to the left, and continues until each player has been dealer/lawmaker twice.

This will help you play better

The Soma shapes
The Soma shapes are as shown here. There are six shapes made of four cubes, and one shape made of three cubes.

Use cardboard to make a set of templates that can be folded and glued into the shapes, or make a set from wooden blocks, or sew them from cloth and stuff them like cushions (this works if the basic cube is about 20 cm or more on a side!), and then see if you can solve the puzzle.

This will help you understand

The diabolic cube
This puzzle requires you to take the pieces you see here, and assemble them into a cube. Just make them up in the same way that you would make up the Soma pieces, and then get to work to fit them together. What could be easier?

Quite a few things, actually :-)

This may help you just a bit

The game of Nim
This is a subtraction game. Faced with several rows of matches or coins, players take it in turns to remove items in accordance with simple rules. Whoever takes the last item is the winner, although the game can also be played with the last taker to be the loser.

In either case, the restriction on players is that all the objects they take must come from the same row: in the example shown here, there are rows containing 3, 4 and 5 items.

In either form of the game, you can always win if you get to a position where there are items in only two rows, and the same number in each. Another winning position is to have one item in the first row, two in a second row, and three in a third row. The first player can always win in the game shown here by removing two items from the top row, then playing "rationally after that.

This will help you understand

The Fibonacci series
The Fibonacci series runs 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233 and so on. Each number is the sum of the two previous terms. If you draw squares like the ones here, they form up into the Fibonacci series, and quarter-circles in the squares make a beautiful spiral.

This series has some very interesting properties that you can track down with a Web search. This will deal with just a few of them. For example, the ratio of any two successive terms gets closer and closer to phi, the "Golden Mean", which has a value such that 1/phi = 1 + phi. Or the fact that the numbers of seed whorls in a sunflower counting clockwise and counter-clockwise, will always be two successive Fibonacci terms. Or if you count the leaves going up a stem, then as you go, you will slowly rotate around the stem, after you have made as many turns as one of the Fibonacci numbers, the leaves you have counted will sum to another Fibonacci number. Why? Do a Web search!

Some of the numbers in the Fibonacci sequence above are composite (divisible by other numbers), and some of them are prime. Is there any deep pattern which determines what a particular Fibonacci number is divisible by, if anything?

What is the maximum number of Fibonacci numbers you will need to add together to reach any integer? Is this always a maximum? How would you prove it? (For example, 6 = 1 + 5, 7 = 1 + 1 + 5, and so on)

The history of the Fibonacci series (like Fibonacci) is worth looking into � find out what the series has to do with rabbits!

You can look for help here

Blaise Pascal's marvellous triangle
Draw Pascal's triangle to as many rows as possible on a single A4 sheet of paper. There is an example below that will show you how to tackle it.

Or if you prefer, create it on a spreadsheet and print it out. (If you are going to do this, find out about the =IF commands, and how you can use these in a clever way. I would set up a single row to begin with, in row 2, with a single "1" in the first row to start things going.) You will probably need to ask for some hints on this one, but it is possible to solve this one completely.

Here is a hint that works in MS Works and older versions of Excel:
=IF(OR(H1>0,J1>0),(H1+J1),"").
(Note that once you have one cell done, you can use Copy, Edit - Fill right, and Edit - Fill down to do the hard work.)

And here, courtesy of Melanie Hughes of Intouch Consultancy and Peter Stewart, who each gave me half of the answer, is a value to insert into a spreadsheet at a point you need to work out:

=MOD(IF(OR(A1>0,C1>0),(A1+C1),0),$A$1). You have to work out what cell that comes from, and insert it there.

The clue that Peter Stewart gave me was a method to get rid of the excess zero symbols: go to Tools � Options � View and then untick the Zero values option box.

This second solution has an interesting extra twist, in that it does not really generate a Pascal triangle at all, but a modular skeleton that is rather like a Sierpinski Sponge. This was a clue that I got from Melanie who does this sort of thing for a living. Any value that is divisible by the number you insert in cell A1 appears only as a zero, and all of the others are given as small integers. Usually, when you get lots of rows, the spreadsheet gets very ugly, but in this version, no cell is greater than one less than the value in A1.

Hints: Try small integers in A1 first, and to see the pattern, you need about 100 columns (I used 105: A to DA, with the value 1 inserted in cell BA1) and about 50 rows. To make the pattern cleaer, use very small type, and use the format commands to make the cells tiny � experiment with Autofit.

Another solution came to me through Adrian from Zbigniew Zurynski: start with the root formula: =A1+C1 in cell B2 and copy it all over the sheet. My excuse is that I came to this when I was looking for a way to use the =IF operator, so I missed the sensible way to do it!

Now here is the example for you to look at. (It is a table done in HTML, and a real pain to do, so feel free to snip the code!)

							1
						1		1
					1		2		1
				1		3		3		1
			1		4		6		4		1
		1		5		10		10		5		1
	1		6		15		20		15		6		1
etc.	etc.	etc.	etc.	etc.	etc.	etc.	etc.	etc.	etc.	etc.	etc.	etc.	etc.	etc.

This will help you extend your knowledge

The game of Sprouts
This gets its name from the appearance of a finished game. It begins with just a small set of points drawn on a sheet of paper. You may like to start with a three-point game, then work up through four and five-point games.

The first player draws a line between any two dots, and then adds a new dot, somewhere along the line. The next player then does the same, and the game continues until one player cannot move.

No point can have more than three lines beginning and/or ending at that point (though a line may begin and end at the same single point), and no line may cross over another.

The diagram shows a partly completed four-point game, with the blue dots representing the starting points, and the red dots showing the dots that were added later -- in a real game, it does not matter which dot is which, or the order in which the dots were made. Three of the red dots can have one more line beginning or ending there, and the blue dot can have two more lines to or from it. Remember that each extra line, while it closes down two existing nodes or attachment points, creates one new node.

This will help you understand

The prisoner and the cells
A prisoner was once told that if he could find a way to walk through all of the doors of all of the cells, once and once only, he would be allowed to go free.

The diagram on the left shows how the cell doors are arranged in this rather strange prison (with even stranger guards!).

Analyse the problem and see whether it is possible, and if it is, work out the solution. If it is not possible, prove it!

This will help you understand

The crossed house puzzle
Your task is a fairly simple-looking one: to draw this diagram by putting your pencil down on the paper, and drawing a single continuous line. You are not allowed to go over any of the lines.

You can solve this with a lot of difficult trial and error, or you can be mathematically clever, and work out a basic principle that applies to problems like this one and the Königsberg bridge problem.

This will help you understand

The Königsberg bridge problem
In the city that was once called Königsberg, there are two islands in the river, linked to each other and to the shore by bridges as you can see in the diagram. The river is blue, the bridges are white.

The problem for the citizens of Königsberg: is there any way of walking around the city and crossing each of the bridges once and once only?

This will help you understand

Penrose tiles
There are a number of regular shapes which "tile the plane", just by using large numbers of the same tile. Squares and rectangles, for example, can be laid down in a neat matching pattern that will go on forever. Hexagons, the sort that we see on bathroom floors, are another example of a tile which will cover the whole of an infinite plane, ignoring the raggedy bits around the edges (because infinite planes don't have edges!). NOTE: these diagrams are incomplete, so use the ones below, which have circle segments in them.

Other shapes do the same thing: right-angled triangles can be put together to make rectangles and squares, while equilateral triangles can be assembled into hexagons, and so on. Then there are sets of tiles that work together. Octagons cannot tile the plane by themselves, but if you add in some small squares, you have a complementary set of two shapes which will tile the plane.

All of these tiles and tile sets produce a periodic tiling pattern, where if you travel long enough across the plane, you will find the same pattern repeating itself. The main thing about Penrose tiles is that they seem to be able to tile the plane in an aperiodic way. That is, you can entirely cover an infinite surface, always moving outwards, but without producing any repeating patterns.

penrose.gif - 2827 Bytes Penrose tiles are named after their inventor, Sir Roger Penrose, and they typically show a sort of five-fold symmetry, which derives from the fact that the angles are all multiples of 36 degrees. In the tile sets shown here, the angles are 216, 36, 36 and 72 degrees in the top tile, and 144, 72, 72 and 72 degrees in the lower tile.

In another tile set, the angles are 144, 144, 36 and 36 degrees for one tile, and 108, 108, 72 and 72 degrees for the other. A figure for these is to come . . . or maybe I will follow the example below and write the LOGO code for it.

Your task is to make a tile set in large numbers: at least 30 or 40 of each of the two tiles in that set, and to see if you can fit them all together.

Special credit:

Noel Wood (elwood@powerup.com.au) spotted that I had failed to put any structure inside the tiles, so that they failed to work. He also provided the bit map from which I generated the working figure shown here, as a stopgap until I can get my act together. He also gave me the LOGO code to generate these tiles. If you want the large version, it is available as a 10 kB GIF file here.

This will help you in the task

1729 and all that
The number 1729 is special for a number of reasons. One reason is that it is the first number that is the sum of two cubes in two different ways. Can you find the two distinct solutions to the equation "x-cubed plus y-cubed equals 1729"?

This will help you

Some "quickies"

The sum of all of the numbers from 1 to 100 is 5050, an answer which I just worked out in my head. Quickly, now, what is the sum of all of the numbers from 1 to 200?
When I hear the value of the cube of an integer between 100 and 200, I can give you the original number, the cube root, back immediately. It is a trick: what is the nature of the trick?
Back in the days when we still had single cents, an absent-minded bank teller switched the dollars and cents around when cashing a cheque. The customer had to accept several five-cent coins, as the amount was less than $100, and dropped one of the coins, at which time the customer had exactly twice the amount written on the cheque. What was the value of the original cheque?
There are many ways that you can get a group of numbers to add up to give different values. For this challenge, try using the numbers in 1729, in that order, to generate all of the numbers from 1 to 50.
Your task is to find solutions using the numbers 1, 7, 2 and 9 in that order, once only, combined by any mathematical symbols such as exponentiation, square roots, multiplication, addition, subtraction and division, with as many brackets as you like. You can even use factorial notation (5! = 5 x 4 x 3 x 2 x 1) or any other notation, but no other numbers.
For example,
1 = 1^729 = -17 + (2x9) = 1 -7 -2 +9 = 1^72^(sqrt(9)) etc
or 46 = 17 + 29

This series also includes some science quickies.
This will help you work out the answers (a bit!)

Runs of composites

The aim here is to explore the natural sequence of integers for runs of consecutive integers which are composite � which have factors. Series like 8, 9, 10, or 24, 25, 26, 27, 28.

I have been using a spreadsheet to help me search for runs longer than five composites, but as yet, the best I have found is seven in a row: is there any link between the central number in the first example of any run of a particular size? Note that there will always BE a central number, as the totals will always be odd. I am still playing with this one . . .

Note added in October 2002: I have now found a run of 33 consecutive composite numbers, and all of them are less than 10,000. I used a spreadsheet to do it, and I will add some helpful hints sometime fairly soon. Hint: I used the =IF function quite a bit . . .

This will help you work out the answers (a bit!)

Mediocrity
How about a level-3 game of Mediocrity? This is the game which Douglas Hofstadter called "Hruska" in his marvellous book Metamagical Themas, but it is at once very simple and diabolically complex.

Here are the rules for three players:
A level-zero game has each player secretly choose an integer between 1 and 5. The player who chooses the second-highest number wins, and adds that many points to his or her level-1 score.

A level-ngame is made up of five level-(n-1) games. The player with the second-highest total level-(n-1) score wins the level-n game, and adds that score to his/her level-n score.

Some people use rather elaborate rules for avoiding ties, involving adding or subtracting 1/3 from certain player's scores depending on which round it was, but try it out for yourself.

This will help you understand

And now for some help

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
You can also make a 5 x 13 rectangle which leaves a hole in the shape of one of the pentominoes, while other challenges include making scaled-up versions of some of the pentominoes, using nine of the other pieces to make a model three times as high and three times as wide.

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 double-sized 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
If anything does not work, check the values in column A, which has to contain only the consecutive odd numbers. Then work your way across, checking each of the formulas in the instructions, until you spot the mistake you made. The formulas given here work in MS Works and MS Excel, but have not been tested on other spreadsheets.

Back to the details| Back to the index

The game of Eleusis
The scoring may seem complicated, but it means that the lawmaker is under pressure to come up with a rule which not everybody will guess quickly, but easy enough for one person to guess quickly.
As a group, decide how this is similar to the way in which scientists investigate natural laws. How is it different?
The game of Eleusis was invented by a Robert Abbott of New York, and popularised many years ago by Martin Gardner in the pages of Scientific American. The article is to be found in Gardner's "More Mathematical Puzzles and Diversions".

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

Later, maybe -- now do it yourself!

Back to the details | Back to the index

The game of Nim
Every position is either safe or unsafe. You can evaluate a position by using binary arithmetic, where 1 is represented by 1, 2 is represented by 10, 3 is 11, 4 is 100, and 5 is 101. Set up a table to show the position like this:

	4	2	1
	-----	-----	-----
3	0	1	1
4	1	0	0
5	1	0	1
	-----	-----	-----
	2	1	2

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
Pascal's triangle has blocks of numbers which are all divisible by the same number. It is possible to write a computer program to plot these into a diagram: try the numbers divisible by three first, and then move up to larger numbers later.

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 is a certain amount of information available on the Web about this game. Use a suitable search engine to find out more, or better still, play it yourself, and prove what the winning strategies are (if any!!). Names to look for: John Horton Conway, Michael Paterson. They invented the game on February 21, 1967.

There can be no more than 3n-1 moves in a game which starts with n points, and that there must be at least 2n moves in any game. The one and two-point 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 five-spot games, while the second player will always win the six-spot 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
The question you have to ask yourself is: how many times do I enter or leave from one of the key points. There is something very special about the points with odd numbers of starting and finishing points. The rest is up to you, but the problem does have a solution.

Back to the details | Back to the index

Analysing the prisoner's problem
Well, if you had any sense, you would have done the problem above this one first!

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
Well, if you had any sense, you would have done the problem above first!

And that's a hint.

Back to the details | Back to the index

More about Penrose tiles
The figures you see here are PENROSE1.GIF and PENROSE2.GIF, and each is only shown here at one third of its true size. Save those two files, which are copyright-free, and print them out if you wish, or draw them up with mathematical instruments, using 5 cm for the long edges. Make the tiles from cardboard, flooring vinyl, or some other object with enough thickness to stop the shapes slipping under one another.

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 five-fold pseudo-symmetry, 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 CD-ROMs or on the Web.

LOGO code for darts (by Noel Wood)

This should paste directly into your LOGO engine

to dart
setpc 9
fd 50 * 1.618
rt 180 - 36
fd 50
rt 180 - 216
fd 50
rt 180 - 36
fd 50 * 1.618
rt 180 - 72
fd 50 * 1.618 * 1 / 2.618
rt 90
setfc 5
pu
fd 1
fill
bk 1
pd
setpc 4
repeat 72 [fd 50 * 1.618 * 1 / 2.618 * 2 * 3.14 / 360 rt 1]
repeat 72 [bk 50 * 1.618 * 1 / 2.618 * 2 * 3.14 / 360 lt 1]
setpc 9
lt 90
bk 50 * 1.618 * 1 / 2.618
fd 50 * 1.618
rt 180 - 36
fd 50 * 1 / (1 + 1 / 1.618)
rt 90
setpc 3
repeat 216 [fd 50 * 1 / 1.618 / (1 + 1 / 1.618 ) * 2 * 3.14 / 360 lt 1]
repeat 216 [bk 50 * 1 / 1.618 / (1 + 1 / 1.618 ) * 2 * 3.14 / 360 rt 1]
setpc 9
lt 90
bk 50 * 1 / (1 + 1 / 1.618)
lt 180 - 36
bk 50 * 1.618
end
to kite
setpc 9
fd 50 * 1.618
rt 180 - 72
fd 50
rt 180 - 144
fd 50
rt 180 - 72
fd 50 * 1.618
rt 180 - 72
fd 50 * 1.618 * 1.618 / 2.618
rt 90
setfc 4
pu
fd 1
fill
bk 1
pd
setpc 5
repeat 72 [fd 50 * 1.618 * 1.618 / 2.618 * 2 * 3.14 / 360 rt 1]
repeat 72 [bk 50 * 1.618 * 1.618 / 2.618 * 2 * 3.14 / 360 lt 1]
setpc 9
lt 90
bk 50 * 1.618 * 1.618 / 2.618
fd 50 * 1.618
rt 180 - 72
fd 50 * (1 / 1.618) / (1 + 1 / 1.618)
rt 90
setpc 6
repeat 144 [fd 50 * 1 / (1 + 1 / 1.618 ) * 2 * 3.14 / 360 lt 1]
repeat 144 [bk 50 * 1 / (1 + 1 / 1.618 ) * 2 * 3.14 / 360 rt 1]
setpc 9
lt 90
bk 50 * (1 / 1.618) / (1 + 1 / 1.618)
lt 180 - 72
bk 50 * 1.618
end

Ignorant about LOGO? Peter Ware on the Oz-teachers list gave me the address for a freeware version of this language which I know and use, but had not downloaded for a couple of years. Try http://www.softronix.com/logo.html and note that it may NOT be sold for profit -- it is free.
Back to the details | Back to the index

1729 and all that
You want help, you lazy beast? Do it yourself! Write down the cubes of all of the numbers where the cube is less than 1729, and look at them -- you will soon see the answer.

Probe

Fourth powers
There 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
But you may not find the deep pattern! I have no idea what the answers are to these questions -- yet! Maybe you can email me if you work out an answer.

Probes

Lucas numbers
The Lucas sequence runs 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199 . . . Apart from the first couple of numbers, can you prove that there is no number common to both series?

Q numbers
Q numbers run 1, 1, 2, 3, 3, 4, 5, 5 , 6 , 6, 6, 8, 8, 8, 10, 9, 10, x . . .

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 :-)
Still here? What about trying to work out if the series always gets bigger, or whether it gets back to zero at some point? What about trying a Q series with two other starting numbers?

Back to the details | Back to the index

Gauss was the first person to work out the answer, as a small boy. The first thing he did was to add 1 and 100, then 2 and 99 -- now work it out from there :-)
The first thing is to know that the last digit of the cube tells you the last digit of the number that was cubed -- and the pattern is very neat! The next thing is to know that there is a range, starting at 1 million, for the cubes of numbers between 100 and 110: with that knowledge, I know which "decade" to look in, and I know the last digit, so I have the answer.
This can be solved with a Diophantine equation: the cheque is for x dollars and y cents, which gives us
100y + x - 5 = 2 (100x + y) which reduces to 98y - 199x = 5, for which there is only one solution with integer values.
As an alternative, the customer has 2x dollars and 2y cents after dropping the coin, and the total number of cents will be x-5. If y is 50 cents or more, we have one set of simultaneous equations, while if y is less than 50 cents, we have another equation. One set gives x a minus value, so the other set is the one to go with. Good luck :-)
I have solved every integer to 31, but I am missing 32, 33, 34, 38, 39, 40, 44 and have not tried for 48 and above so far. Some integers have multiple solutions: which integer has the most solutions (1 has large numbers of trivial answers, 6 has three good ones), and what are the solutions for my missing integers? Please email me if you find answers to any of these.
Post script: Dr Klaus Marx at Robert Bosch GmbH in Germany quickly solved four of the numbers that I could not reach:
33 = (-1+7)^2 - sqrt(9)
39 = -1 + 7^2 - 9
40 = (-1+7)!/(2*9)
44 = 1 + 7^2 - sqrt(9)!
This leaves 32, 34, 38, and numbers above 48, and as nobody has sent solutions to these in over two years, I conjecture that there are no valid solutions.
You can get at least four fairly pleasing solutions for 1. Are there any numbers with more solutions? Are there any more solutions to 1?
This is a great way for putting yourself to sleep, though having solved 14, 20 and 21 after putting the light out one night, I awoke about four times during that night, worrying that I might forget the solutions. I didn't.
Klaus added these solutions the next day, which he correctly described as "rather amazing":
23 = sqrt(17 + 2^9)
33 = sqrt(1 + (7-2)!) * sqrt(9)
42 = sqrt(1 + 7!) - 29
36 = 1/7!/2*9!
This raises the interesting point: 4! and 5! and 7! are all just one off a perfect square. Is there a pattern here? I have a proof for this, one way or the other, but the margins of this Web page are too narrow to contain it . . . :-)
OK, that's a bit mean. I am looking at quite a few factorial numbers which seem to be less than the next perfect square by a number that is itself a perfect square, and a power of 3, at that, though as you test this out, you will find that this is only half the story. I have to say that 12 is a bit of a worry . . .

Back to the details | Back to the index

Mediocrity
Why does Hofstadter call it Hruska? It was the surname of Senator Roman Hruska of Nebraska in the early 1970s, who argued that Nixon's proposed appointee to the US Supreme Court should not be ruled out just because he was mediocre. Mediocre people are also entitled to representation, Hruska said.

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 1¹/₃, 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
I saw this in a particularly cute version in Edinburgh at a Festival of Science some years ago. The pieces were made from cushion material and stuffed, and as I watched, two 8 or 9-year-old children battled the pieces into place.

So there you have it: the puzzle can be solved: all you have to do is work out where to start.

Back to the details | Back to the index

The extensions in this series include most especially (if you are starting form here), my investigations into Triangular and other polygonal numbers, but refer also to the home page for the Science Playwiths collection, or see my continuing update accountor the general menu

Useful mathematical sites

Here is an amazing Indian site. Total Gadha tipped me off about it, and is the main contributor, but there is a lot of lovely stuff there. If you like the puzzles here, start with Some Mathenatical Curios: you won't be sorry!

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 home-brewed mathematical site with some remarkable effects is Adrian Bruce's Symmetry page. Go look, both for clever Webbery and also for some excellent mathematics.

This file is http://members.ozemail.com.au/~macinnis/scifun/number.htm, first created on August 20, 1997. Last recordedrevision (well I get lazy and forget sometimes!) was on August 17, 2008.

Worried about copyright? You need to go look at my fine print. Well, maybe you don't after you read the next paragraph, but do it anyhow . . . and to see some more ideas, look at the start of that same page

©The author of this work is Peter Macinnis -- macinnis@ozemail.com.au, who asserts his sole right to the product as it is packaged here, recognising that many of the ideas are common. Any non-profit educational or home use is completely acceptable without let or hindrance. Copies of this whole file or site may be made and stored or printed for personal or educational use. The work used here derives from on-going research and development which will one day lead to a book on brain food ideas.

GEM, The Gateway to Educational Materials

This is one of a number of pages on this site which contain metadata to make access easier for people using the GEM, the Gateway to Educational materials. The whole idea of metadata is to make searching easier and more effective. GEM is a brilliant one-stop, any-stop access to educational resources on the Internet, and sponsored by the US Department of Education, while welcoming outsiders like me. They provide lesson plans for teachers, ideas, brain food and more, and they make searching a breeze.

To get an idea of how it works, view the source for this page, and see the metadata at the top. To get an even better idea, click on the link above, or the link at the top of the page.

This site had 219,000 hits on the index page from 1999 to January 2007 and an unknown number on other pages. In January 2007, a combined counter was placed on all of the pages, counting page hits which now total