Merge branch 'master' of https://github.com/supportingami/sami-maths-…

…club

maths-club-pack/content/facilitator/apple-teaser.md

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+# Apple Teaser
+
+## Introduction  
+
+This session will require the students to read and interpret all the information provided to successfully
+work out the total number of apples in the three rooms. The students can work individually or in small
+groups. The instructions are as follows:  
+
+Have the students draw the following table on a sheet of paper:  
+
+![](../../images/apple-teaser-3.png)  
+
+Each number in the table represents the total number of apples, 9 apples in each room × 3 rooms = 27  
+
+Get the students to work out what the minimum and maximum possible numbers of apples are and cross out the numbers outside of this range. Also, the students should think about the answers to Adam and Belle’s questions and cross out any other numbers.  
+
+Which numbers are left? Circle them. For each of these numbers, can you work out how many apples there might be in each room?
+
+
+## Solution  
+
+Working out the minimum number of apples: We know that there is at least one apple in each room and each room has a different number of apples. So, the minimum number of apples is 1 + 2 + 3 = 6. Cross out numbers 1, 2, 3, 4, 5.  
+
+Working out the maximum number of apples: We also know that no room has more than nine apples, so the maximum number of apples is   7 + 8 + 9 = 24. Cross out numbers 25, 26, 27.  
+
+From the questions, we know that the total number of apples is not an even number and it is not a prime number. So, cross out the even numbers and prime numbers.  
+
+Circle the numbers that remain.  
+
+![](../../images/apple-teaser-4.png)   
+
+Case 1:  If the total number of apples is 9 and you have 5 of them in your room, then Adam and Belle have 4 apples between them. So, one has a single apple and the other has 3 apples.   
+
+![](../../images/apple-teaser-5.png)
+
+Case 2: If the total number of apples is 15 and you have 5 of them in your room, then Adam and Belle have 10 apples between them. Here, there are many possible combinations:   
+
+![](../../images/apple-teaser-6.png)  
+
+
+Case 3: If the total number of apples is 21 and you have 5 of them in your room, then Adam and
+Belle have 16 apples between them. So, one has 9 apples and the other has 7 apples.  
+
+![](../../images/apple-teaser-7.png)  
+
+**So, what is the question you should ask?**  
+
+You should ask:  *“Is the total number of apples 15?”* and reconsider each case.  
+
+Case 1: The total number of apples is 9: then the person who has the single apple knows that the
+total cannot be greater than 1 + 8 + 9 = 18 and the person who has 3 apples knows that
+the total cannot be greater than 3 + 8 + 9 = 20. Either Adam or Belle should tell Ruben
+that the answer is 9.  
+
+Case 2: The total number of apples is 15: then Ruben will answer “Yes” and any one of you can tell
+him that the answer is 15.  
+
+Case 3: The total number of apples is 21: then the person who has 9 apples knows that the total cannot be less than 9 + 1 + 2 = 12 and the person who has 7 apples knows that the total cannot be less than 7 + 1 + 2 = 10. Either Adam or Belle should tell Ruben that the answer is 21.
+
+
+
+

maths-club-pack/content/facilitator/circumference.md

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+# Circumference
+
+## Introduction  
+
+The answers to the two questions are related and can be done in any order. Question 1 uses numerical values to calculate the circumference but question 2 requires students to leave the width as a constant d.  
+
+These puzzles are interesting because the result is not immediately obvious. It will get students thinking about how their answers apply to different circles, regardless of their size. Encourage group discussions about the answers that they found. Students can leave their answers in terms of π or use the approximation π ≈ 3.  
+
+Circumference of a circle = diameter × π
+
+
+## Solution  
+
+<ins>Question 1 </ins> 
+
+Firstly, you need to work out how much fencing was needed **before** the extension. To do this, students will need to calculate the circumference of the original field = width × π = 10π  
+
+Secondly, you need to work out how much fencing will be needed  **after**  the extension. The width of the bigger field is now 1 + 10 + 1 = 12. Then the circumference of the bigger field = width × π = 12π  
+
+
+![](../../images/circumference-3.png)  
+
+The amount of fencing the farmer would need is the difference between the two circumferences worked out above. The farmer would need 12π − 10π = 2π more fencing. This is approximately ≈ 2 × 3 = 6 metres.  
+
+<ins>Question 2 </ins>
+
+The method is the exact same as for question 1 except here you leave the width as a constant d.  
+
+Firstly, you need to work out how much rope was needed when it was tied around the equator of the earth. To do this, students need to calculate the circumference of the earth = width × π = dπ  
+
+Secondly, you need to work out how much rope will be needed if we instead want the rope to be 1 metre above the earth at all points. The width of the bigger circle is now 1 + d + 1 = d + 2. Then the circumference of the bigger circle = width × π = (d + 2) × π = dπ + 2π.  
+
+![](../../images/circumference-4.png)  
+
+Like how this was worked out in question 1, the amount of extra rope needed is the difference between the circumference of the earth, dπ, and the circumference of the bigger circle, dπ + 2π. Amount of rope = (dπ + 2π) − dπ = 2π.  
+
+**What do you notice about the answers to question 1 and question 2? What would happen if you did the same with a smaller planet? Why do you think this happens?**  
+
+## Extension
+
+Extension – something to think about, not work out!
+What is even more interesting is what happens when the rope is lifted by a single point. The rope around the bottom of the earth is taut but now there is a large clearance at the top. For an additional 2 metres length of rope, the distance from the earth’s surface to the peak of the rope would be enough to fit **two Statue of Liberty’s underneath it!**  
+
+![](../../images/circumference-5.png)
+
+
+Are you surprised that such a small increase in the length of the rope will reach such a large height? Think about how this height would change when you change the size of the ball.
+
+
+
+
+
+
+

maths-club-pack/content/facilitator/coconut-trader.md

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+# Coconut Trader
+
+## Introduction  
+
+The key to this question is to realise that you should take all the coconuts from one sack until that sack is empty. Then do the same with the second sack. If each time you reach a checkpoint you take one coconut from each sack, then you would have no coconuts left by the time you reach the market! Leave students to work this out for themselves but if they get stuck then give them a hint.
+
+
+
+## Solution  
+
+<ins>Emptying the first sack<ins>
+
+You currently have three sacks so at each checkpoint you must give three coconuts each time. That means that the sack will be empty after 30 ÷ 3 = 10 checkpoints.You throw away this empty sack.  
+
+<ins>Emptying the second sack<ins>
+
+You now have two sacks left so at each checkpoint you must give two coconuts each time. That means that the sack will be empty after 30 ÷ 2 = 15 checkpoints. You throw away this empty sack.  
+
+<ins>Emptying the third sack<ins>  
+
+You now have one sack left so at each checkpoint you must give one coconut each time. You have already passed 10 + 5 = 15 checkpoints, which means you have five checkpoints left. That means you must give
+out 5 × 1 = 5 coconuts from your final bag.  
+
+Therefore, you are left with 30 − 5 = 25 coconuts to sell at the market.
+
+
+## Extension  
+
+Here are two more puzzles to solve:  
+
+1.<ins>Cars across the desert<ins>
+
+A car must carry an important person across the desert.   
+
+There is no petrol station in the desert and the car has space only for enough petrol to get it half way across the desert.  
+
+There are also other identical cars that can transfer their petrol into one another.  
+
+How can we get this important person across the desert?
+
+
+

maths-club-pack/content/facilitator/coin-game.md

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+# Coin Game 
+
+## Introduction
+
+This simple coin game serves as an introduction to probabilities in tennis. It is really worth playing the game first with the students before trying to answer the questions to ensure everyone understands the way the game works particularly when to stop playing.  
+
+Go through the answers to the first two questions before students try to fill in the table.  
+
+Watch students closely when they start filling in the table to ensure they are on the right track and help as necessary.  
+
+
+## Solution
+
+Which of the following scores are impossible in this game and why? 5-2, 7-4, 6-2,  
+
+5-2 is possible.  
+
+7-4 is impossible because that would be 11 tosses and you are only allowed 11.  
+
+6-2 is impossible because the game would have stopped before it got to this stage. Player 1 would have already won some time before.  
+
+Think of ways the game could end in a tie - Either 5-5, 6-4 or 4-6  
+
+What is the chance of winning this game?  
+
+The easiest way to fill in the table is to start in the top left corner, and each square is the sum of the number immediately to the left and immediately above. E.g. The numbers
+
+![](../../images/coin-game-2.png)    
+
+It is tempting to add up the number of ways to win (1+3+9+27) and divide by the total of all the numbers in the table. But this would be wrong. It is wrong because the game lengths are different in each case.  
+
+
+The 1 in the blue outlined box represents a game of length 3, getting three heads in a row – the chance of which is $\frac{1}{2}$× $\frac{1}{2}$ ×$\frac{1}{2}$.    
+
+
+The 3 is for a game where there was 4 heads and 1 tail. One way of this happening is THHHH. The
+probability of this is $\frac{1}{2}$× $\frac{1}{2}$ ×$\frac{1}{2}$ × $\frac{1}{2}$  . Then there are the other two ways of HTHHH or HHTHH (note that
+HHHHT is not an option because the game would have stopped at 3-0). Therefore the probability of winning in this way is 3 × ($\frac{1}{2}$ )$^ 5$
+etc.    
+
+Therefore the probability of winning is 1 × ($\frac{1}{2}$) $^3$ + 3 × ($\frac{1}{2}$) $^5$ + 9 × ($\frac{1}{2}$) $^7$ + 27 × ($\frac{1}{2}$) $^9$ ≈ 0.3418  
+
+## Extension
+
+The way the scoring works in tennis is that you have to win by two clear points, so the same strategy as used above can be used to answer the question “If you have a one third chance of winning a point in tennis, what is your chance of winning a match?”  
+
+Here is the working for just winning one game in tennis with players called Angus and
+Bathsheba:  
+
+The maths involves summing an
+infinite series because tennis
+games could technically go on
+forever!    
+
+![](../../images/coin-game-3.png) 
+

maths-club-pack/content/facilitator/collatz-conjecture.md

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+# Collatz Conjecture
+
+## Introduction
+
+The activity is interesting because it is easy to understand the rules and start generating sequences, but it
+is an unsolved problem in mathematics whether all number end up at one. Many people have tried to find
+a sequence that doesn’t go to one, or to prove that all sequences go to one and no one has succeeded.
+There is a prize of at least $2000 dollars if you succeed!  
+
+## Solution
+
+The shortest sequences are going to involve powers of 2, e.g. 1,2,4,8,16,32 etc. as these numbers
+“snowball” to 1. The longest sequence for a number smaller than 100 is for 97, which takes 118 steps but
+finally gets to 1. The number just before, 96, only takes 12 steps!
+
+
+
+## Extension
+
+A computer or tablet would be very useful to try and find very long sequences. Here are some instructions for GeoGebra, which you can find online (https://www.geogebra.org/classic/spreadsheet) or it may be installed on the tablet and you do not need the internet.  
+
+You will need to use the Mod command:  
+
+![](../../images/collatz-conjecture-2.png)
+
+to check if the number is even. Mod is short for Modulo and gives the remainder when you do a division.
+e.g. Mod[10,3] would be 1 because there is 1 leftover when you divide 10 by 3.  
+
+You will also need the If command:  
+
+![](../../images/collatz-conjecture-3.png)
+
+
+to choose what to do if it is even and what to do if it is odd.  
+
+Put together, here is the formula you should put into cell A2, once you have put a starting number in A1.  
+
+=If[Mod[A1, 2] == 0, A1 / 2, A1*3 + 1]  
+
+Then just hover your mouse pointer in the bottom right corner and click the left button and hold down and
+drag down lots of cells. You should see the sequence appear.  
+
+If you want to be super clever you could try and combine two If statements so that if the cell was equal to 1 it would stop calculating and just say "STOP".

maths-club-pack/content/facilitator/domino-tilings.md

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+# Domino Tilings
+
+## Introduction
+The idea of this exercise is to explore the possibility or otherwise of tiling (covering) certain shapes by dominos. While the question appears geometric, there the combinatorics of the situation also turns out to be important. One idea that learners should get from this exercise is that sometimes a construction is possible while other times it will not be; good students should also start to be able to make arguments (proofs!) for why a certain tiling is not possible. The proof of the most advanced example introduces a basic
+idea in combinatorial arguments, that or colouring.  
+
+ Experimentation can be helped by the availability of squared paper, or paper and scissors to make the dominos!
+
+
+## Solution    
+
+a. tilings clearly exists (this could be done at the beginning together so that students get the idea).Experimentation will suggest that in part  
+b. there is no solution, and good students will also spot the fact that the reason is that there is an odd
+number of squares. Indeed, removing one square as in part  
+c. will give a shape that can be tiled again. For part
+d. experimentation quickly gives a tiling; for part  
+e., it can be shown by trying essentially all possibilities that there is no solution again.  
+
+
+How so, since there is an even number of squares in both cases, and we removed two corners in each
+case? Well, the trick is to think of a chessboard colouring of our shape (it should be suggested to students
+to think of a chessboard, or explain to them what that looks like, but initially without further comment).
+The point is that no matter how one puts down a domino, it will always cover one black and one white
+square. So in part e. no tiling can exist since the coloured version has a different number of black and white
+squares:  
+
+![](../../images/domino-tilings-4.png)  
+
+This argument will also show that no tiling in case f. exists either; this case would be impossible to do by
+trial and error.
+
+## Extension  
+
+Students can go on to discuss which shapes can be tiled, and which cannot. For example,    
+
+
+![](../../images/domino-tilings-5.png)    
+
+clearly cannot be tiled; this is very easy to see directly, but the colouring argument also proves it. A more interesting question is whether there are any shapes that cannot be tiled, even though they contain the same number of black and white squares. This could be set as a challenge to a good class. A good student might find the following example:  
+
+
+![](../../images/domino-tilings-6.png)  
+
+(placing this “shape” on a chessboard so that one square becomes white, the other black). While in some sense this is “cheating”, this would be a good point to talk about connectedness of our shapes. A more interesting example, “doubling up” the previous example with 4 squares, is  
+
+![](../../images/domino-tilings-7.png)  
+
+While this shape has 4 black and 4 white squares, it still cannot be tiled, as a simple experimentation shows. Further references:  
+
+https://en.wikipedia.org/wiki/Mutilated_chessboard_problem  
+
+https://en.wikipedia.org/wiki/Domino_tiling  
+
+https://arxiv.org/pdf/math/0501170.pdf

maths-club-pack/content/facilitator/fence-around-a-field.md

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+# Fence Around A Field
+
+## Introduction
+
+
+
+
+## Solution
+
+![](../../images/fence-around-a-field-3.png)
+
+
+
+
+
+
+## Extension
+
+
+![](../../images/fence-around-a-field-4.png)
+
+