How to do it
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.
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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 noncourt 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 faceup, 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 faceup 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 quartercircles 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 counterclockwise, 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 threepoint game, then work up through four and fivepoint 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 fourpoint 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!
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The crossed house puzzle
Your task is a fairly simplelooking 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: rightangled 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 tiles are named after their inventor, Sir Roger Penrose, and they typically show a sort of fivefold 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 "xcubed plus ycubed 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 absentminded bank teller switched the dollars and cents around when cashing a cheque. The customer had to accept several fivecent 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 level3 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 levelzero game has each player secretly choose an integer between 1 and 5. The player who chooses the secondhighest number wins, and adds that many points to his or her level1 score.
A levelngame is made up of five level(n1) games. The player with the secondhighest total level(n1) score wins the leveln game, and adds that score to his/her leveln 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.
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