Grace - Added more problems

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

@@ -38,7 +38,8 @@ Like how this was worked out in question 1, the amount of extra rope needed is t
 
 ## Extension
 
-Extension – something to think about, not work out!
+**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)

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

@@ -10,15 +10,15 @@ The key to this question is to realise that you should take all the coconuts fro
 
 <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.  
+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.  
+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
+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.

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

@@ -11,17 +11,17 @@ Watch students closely when they start filling in the table to ensure they are o
 
 ## Solution
 
-Which of the following scores are impossible in this game and why? 5-2, 7-4, 6-2,  
+*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  
+*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?  
+*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
 

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

@@ -19,16 +19,17 @@ finally gets to 1. The number just before, 96, only takes 12 steps!
 
 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:  
+You will need to use the Mod command:   
+
+Mod[ <Dividend Number> , <Divisor>Number ]
 
-![](../../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)
+If[ <Condition> , <Then> , <Else> ]
 
 
 to choose what to do if it is even and what to do if it is odd.  

maths-club-pack/content/facilitator/counting-chickens.md

@@ -0,0 +1,27 @@
+# Counting Chickens
+
+## Introduction
+
+This activity looks at the interesting concept of having a circle with n points on its perimeter and what happens when you joint those points up.   
+
+Three examples are given on the question sheet so students can see how to visualize the problems. The activity questions rely on students being able to develop on these primary examples to solve more complicated cases. The extension looks at spotting a pattern and developing a rule.   
+
+It is advisable that students try drawing out each case. It should be noted that the use of colours can be a
+way for students to picture the different regions more easily.
+
+## Solution
+
+![](../../images/counting-chickens-2.png)
+
+## Extension
+
+Is there a pattern between the number of fence posts and the maximum number of chickens that can be kept?   
+
+Hint: Fill out this table (some of it has been filled out for you)   
+
+![](../../images/counting-chickens-3.png)   
+
+Using Pascal’s triangle, what will happen if you add the numbers to the left of the line in each row? Can
+you spot the pattern now? Can you explain this strange pattern?    
+
+![](../../images/counting-chickens-4.png) 

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

@@ -1,20 +1,35 @@
 # Fence Around A Field
 
 ## Introduction
+This problem probably appears more difficult than it is at first. Learners may wish to first try with an example using numbers for the length and width of the field, however this is not necessary to solve the problem.
 
 
 
+## Solution   
+This is a question about perimeter. The key to a quick and easy
+solution here is to understand that the edges of the fence parallel
+to the edges of the field will always be the same WHATEVER THE
+SIZE OF THE FIELD.   
 
-## Solution
+The extra fence is only needed at the corners (shown in the
+diagram).   
 
-![](../../images/fence-around-a-field-3.png)
+In total there will be 8 extra pieces, all measuring the same
+length, perhaps called *w*.
 
+![](../../images/fence-around-a-field-3.png)   
 
+We know that the extra length of fence is 1 metre which is the total length of the 8 pieces. Therefore if 8w = 1m, **w = 12.5 centimetres**. The path is too narrow for a tractor but a mouse could easily run along it.   
 
+## Extension   
 
+This problem sets up nicely for another called the *belt around the earth*,
+which goes as follows:   
 
+Imagine a piece of rope is tied the entire way around the earth, along the
+equator, so that it fits perfectly. If we instead wanted the rope to sit 1 metre above the earth at all points, how much longer does the rope have to be?   
 
-## Extension
+Hint: you do not need to know the size of the earth!   
 
 
 ![](../../images/fence-around-a-field-4.png)

maths-club-pack/content/facilitator/frog-party.md

@@ -0,0 +1,32 @@
+# Frog Party
+
+## Introduction
+
+This is a really fun problem, and can be used to build many much harder problems. It is important when starting to problem to make sure the rules are clear. You might want to do this by showing examples of moves which are and are not allowed, using coins on top of pieces of paper or similar.
+
+## Solution
+
+There are many ways to solve this, and it is possible to have a party on every single lily pad. Two examples are given below, however you should ask students to also share their solutions
+
+![](../../images/frog-party-3.png)
+
+## Extension
+
+There are lots of ways you can make this problem more challenging and interesting,
+which include:   
+
+  - Add more lily pads, is it still possible with 6, 10 or N?   
+
+    ![](../../images/frog-party-4.png)
+
+- Choose one of the frogs to be the Queen frog, who must be at the top of the
+party   
+
+    ![](../../images/frog-party-5.png)   
+
+- Swap one of the frogs for a lazy toad, who refuses to move. Is it still possible?   
+
+    ![](../../images/frog-party-6.png)   
+
+Note, the strategy used in the second solution of working from the inside out also works
+for any number of lily pads!

maths-club-pack/content/facilitator/locks-and-keys.md

@@ -0,0 +1,37 @@
+# Locks and Keys
+
+## Introduction
+
+This problem is not specified very precisely, and in this case it is intentional. The aim of this problem is to work with assumptions. The problem has different solutions under different assumptions, and it might be
+easy or even impossible depending on the assumptions. For example, if we assume that the messenger can break the box or the lock, then the problem is impossible (and somewhat pointless).   
+
+There is no right or wrong solution because there is no right or wrong set of assumptions. However, if we specify our assumptions clearly, then for the given set of assumptions any solution is either right or wrong.
+The facilitator should encourage students to decide on what should be allowed or not so that the problem is not solved too easily or made impossible.   
+
+This activity involves lots of discussion between groups. It could help to have paper cutouts of locks, keys and boxes so students can demonstrate rather than only explain. You should encourage students to demonstrate their solutions, and to challenge each other to see if they can find a way to intercept.   
+
+## Solution
+
+There’s are lots of different strategies that could be tested and argued, here we will only outline a few possibilities, it is up to the students and facilitator to decide if their own solutions work or not.   
+
+We will start with a few assumptions:   
+
+         1. The messenger cannot break the box or the lock.   
+
+         2. The messenger cannot make copies of keys.   
+
+**Case 1:** Alice locks the box and sends it to Bob. Bob sends the messenger back to Alice. Alice now gives the messenger the key to send it to Bob. The messenger gives the key to Bob and Bob unlocks the box.   
+
+**Intercept:** The messenger only pretends to deliver the box to Bob, but instead keeps it. He then returns it back to Alice and obtains the key to unlock the box and read the message.   
+
+**Case 2:** Alice sends the key using a different messenger than for the box.   
+
+**Intercept:** The messengers collaborate. You would need to decide in the rules whether this would be possible.   
+
+**Case 3:** Alice locks the box with her lock and sends it to Bob. Bob attaches his own lock and sends it back.Alice takes her lock off the box and sends it back again. Finally, Bob takes his own lock off to read the message.   
+
+**Intercept:** In this case there would be no way to intercept the message without breaking the box or copying keys!
+
+## Extension
+
+Bob opens the box and see it is instructions to play a game. To start the game Alice and Bob will flip a coin to decide who goes first, however they will still need the messenger to communicate the outcome. Who should flip the coin? Who should choose heads or tails? How can they communicate in a way so that neither person can cheat?

maths-club-pack/content/facilitator/magic-cards.md

@@ -1,11 +1,12 @@
 # Magic Cards
 
-## Introduction
+## Introduction   
+
 Place the cards randomly on the desk and ask a student to find all the cards with their birth date on (e.g. they might be born on the 12th of November 2005 so they would be looking for all the cards with a 12 on it).   
 
 Ask them to tell you the colour of all the cards that their number is on.  
 
-Then all you need to do is add together the top left number of each of those cards and it will instantly tell you what date they were born on.  
+Then all you need to do is add together the **top left number** of each of those cards and it will instantly tell you what date they were born on.  
 
 Do this a few times with the students. They can pick any new number for you to guess.
 Then ask them to work out how you are doing it so quickly.   

maths-club-pack/content/facilitator/making-squares.md

@@ -1,6 +1,6 @@
 # Making squares
 
-## Introduction.
+## Introduction
 
 This game is a great way to practice using coordinates and develop geometrical reasoning (understanding of shapes). It might be a good idea to play the game once altogether, splitting the class into two teams. Each team takes it in turns to mark a point on the game grid until one team can join four of their own points together and make a square. A typical game on a smaller grid might look like:
 

maths-club-pack/content/facilitator/monkey-business.md

@@ -0,0 +1,50 @@
+# Monkey Business
+
+## Introduction
+
+This is another problem where the solution seems impossible because there are too many things to consider (1000 monkeys!), but again becomes achievable once simplified and able to see patterns. To start the problem you should ensure students understand the rules, perhaps imitating with 4 students and the same number of coins or cards that can be flipped to face up or down.   
+
+The first student flip all cards up (u):                         [u, u, u, u]   
+
+The second student flip the evens down (d):            [u, d, u, d]   
+
+The third student would flip card 3:                           [u, d, d, d]   
+
+The fourth student would flip card 4:                         [u, d, d, u]
+
+
+
+
+## Solution
+
+1.For the first part of the problem you only need to consider 10 monkeys (as the 11<sup>th</sup> monkey and later will not press any of the first 10 switches). So which monkeys will press switch number 10? Well, as 10 is divisible by 1, 2, 5 and 10 there are exactly 4 monkeys which will press the switch   
+
+* Monkey 1 turns it on   
+
+* Monkey 2 turns it off   
+
+* Monkey 5 turns it back on   
+
+* Monkey 10 turns it back off   
+
+So switch 10 will be **off**.   
+
+2.The next part is more challenging, but again relies of thinking about how many monkeys will press each switch.   
+
+Switch 10 was off because every time one monkey turned it on another turned it off. We were able to split the number 10 into factor pairs (1x10, 2x5), which will always result in the light being turned off.   
+
+Most numbers will be pressed by an even number of monkeys, e.g. 24 has factors (1x24, 2x12, 3x8, 4x6), so monkeys 1, 2, 3, 4, 6, 8, 12, 24 will press the switch, and as this is an even number the light will be off.   
+
+Square numbers are the only ones which do not, e.g. 16 has factors (1x16, 2x8, **4x4**). Monkeys 1, 2, **4**, 8, 16 will press the switch, and as this is an odd number the light will stay on (monkey 4 won’t press it twice!).   
+
+**So we just need to work out how many square numbers there are between 1 and 1000?**   
+
+If we think about the larges square number less than 1000, **<span style="color:brown">30x30=900, 31x31=961,</span> <span style="color:red">32x32=1024</span>**  
+
+So there will be **31 switches left on**! (switch numbers 1, 4, 9, 16, 25, 36, ... , 900, 961)   
+
+## Extension   
+
+How many lights would stay on if only the even numbered monkeys decided to take their turn pressing the switches?
+
+

maths-club-pack/content/facilitator/number-challenge-1.md

@@ -12,13 +12,11 @@ The solutions are **not important**, it is about the thinking process learners g
 
 1. The first two numbers add up to 11 so we can try different pairs. If they were 5 and 6 then the other number would have to be 12 or 13 but then none of the pairs add up to 22, so it's wrong. If we keep trying values we see that the first numbers can only be 3 and 8 and the other number will be 14 which works with all the information. However, if the numbers were bigger this would take much longer.  
 
-    A better way is to use **algebra** and **simultaneous equations.** Let the three numbers be x, y and z , and . If the first pair is x and y , then we know that x + y =13. If the second pair is x and z  then,  x + z = 17  and the last pair must be y  and z so y + z = 22. This gives us three simultaneous equations:
-  x + y = 11 ;  x + z = 17    ;   y + z = 22  
+    A better way is to use **algebra** and **simultaneous equations.** Let the three numbers be x, y and z , and . If the first pair is x and y , then we know that x + y =13. If the second pair is x and z  then,  x + z = 17  and the last pair must be y  and z so y + z = 22. This gives us three simultaneous equations:   
+
+        x + y = 11; x + z = 17; y + z = 22  
 
-   Equation 2 - Equation 1 gives: z - y = 6 . Now add this to Equation 3 to get 2z = 28. 
-Which means z = 14 . Using the original equations we can then work out that  x = 3 
-
-      and y = 8.
+   Equation 2 - Equation 1 gives: z - y = 6 . Now add this to Equation 3 to get 2z = 28.Which means z = 14 . Using the original equations we can then work out that  x = 3 and y = 8.
 
    Now try the question if the totals were 73, 140 and 109. Which method is faster now?
 
@@ -34,8 +32,7 @@ Which means z = 14 . Using the original equations we can then work out that  x =
 
       Do the same and you will see it's not possible to have a 2 or a 4 between the 1s. You can solve it from there.
 
-3.  99 + $\frac{9}{9}$ = 99 +1 = 100 or even $\frac{99}{99}$ = 100.  
-   This question requires some creativity!
+3.  99 + $\frac{9}{9}$ = 99 +1 = 100 or even $\frac{99}{99}$ = 100. This question requires some creativity!
 
 
 

maths-club-pack/content/facilitator/picture-puzzles-1.md

@@ -0,0 +1,13 @@
+# Picture Puzzles 1
+
+## Introduction
+
+These problems can all be solved in different ways, but it is recommended you encourage students to use logic and reasoning skills instead of guessing numbers. Usually each line gives you a piece of information that you can use in the next, so you should think carefully about each piece of information before moving onto the next. The problems also get harder as you go, so encourage students to start with the top-left puzzle first.   
+
+## Solution
+
+![](../../images/picture-puzzles-1-2.png)
+
+## Extension
+
+Ask students to try and make their own picture puzzles!

maths-club-pack/content/facilitator/russian-multiplication.md

@@ -19,7 +19,7 @@ It will work for any two numbers. An insight into why it always works can be giv
 
 (taken from http://mathforum.org/dr.math/faq/faq.peasant.html)   
 
-Binary numbers are numbers written in base two instead of base ten. This means that place value depends on powers of two instead of powers of ten: instead of ones, tens, and hundreds places, base two has a ones place, a twos place, a fours place, and so on. For example, fourteen in base two is 1110:  
+Binary numbers are numbers written in [base two](https://mathforum.org/dr.math/faq/faq.bases.html) instead of base ten. This means that place value depends on powers of two instead of powers of ten: instead of ones, tens, and hundreds places, base two has a ones place, a twos place, a fours place, and so on. For example, fourteen in base two is 1110:  
 
         1110 (base 2)
 
@@ -77,7 +77,7 @@ We are trying to multiply 12 by 13. One way to do this would be to use long mult
 
 
 
-Notice that we are adding 2*13 and 10*13 to get our final answer. This works because of the distributive property:  
+Notice that we are adding 2*13 and 10*13 to get our final answer. This works because of the [distributive property]():  
 
         12 * 13