Week 2: Set Game Wrap Up, Starting Parity Party

Lesson plan:

1. In class review/discussion of some of the Set Game Questions: (30-35 min)

Hopefully some students will present their solutions at the board. Please note:

- Q2: How many pairs of cards can be made with the cards in the deck? Some groups got

80+79+78+…+3+2+1

It looks like a good moment to discuss with everyone why this equals 80×812.

- Q5: James is updating this question: split it into steps, add examples with pictures: if you choose the 1st card in the set, what options do you have for cards 2 and 3 under the given constraints.

- Q6 and on: Ask if anyone thought about these questions. Do as many as we can in the given time.

2. Play 1 round of Set (5-10 min)

3. Break for whom ever needs it (10 min)

4. Parity Party : Questions 1-3 as class discussion (20 min)

Questions 4-6 working in group (20 min) and homework

Resources

· Activity sheets per student: Set Game Questions and Parity Party Questions

· Set decks

· M&M boxes (skittles would work too).

For week 3

· A few sets of dominoes.

· A few chessboards with pieces. (A full set of pieces per board isn’t necessary).

Parity Party Solutions

1. Gearing up for some maths

In the picture on the right, you see eleven gears in a chain. Is it possible for all the gears to rotate simultaneously?

What if there was an even number of gears?

Solution:

Pick any gear to start off with, and let’s say it rotates clockwise. Then the gear next to it will rotate anti-clockwise, the gear next to that will rotate clockwise and so on. If we move around the chain, we see that the last gear will be rotating clockwise. But this is impossible, because the gear next to it (the first gear) is also rotating clockwise. If there were an even number of gears, they could rotate just fine.

2. Odds and Evens

In each of the following sums, find out whether the result is odd or even without actually computing it:

Even a. 1,256,827+7,571,269

Odd b.999,999-888,888

Odd c.1010+1

Even d.777×256

Odd e.131×99

Odd f.5+13+7+21+35

Even g.111+257+549+973

Odd h.1+2+3+4+5+6+7+8+9+10

Let’s gather some general rules:

odd+odd=even / odd-odd=even
even+odd=odd / even-odd=odd
even+even =even / even-even=even
If I add an odd number of odd numbers together, the result is odd / odd × odd=odd
If I add an even number of odd numbers, the result is even / even × any number=even

3. An addition problem

a) If we add up the numbers 1 to 9 we get

1+2+3+4+5+6+7+8+9=45.

However we can change the plus signs to minus signs to get new expressions like

1+2+3+4+5-6-7-8-9=-15 or

-1+2-3+4-5+6-7+8-9=-5

Is it possible to switch some of the plus signs to minus signs so that the numbers add to zero?

b) What if we have the same problem, but we take the numbers 1 to 100 instead?

Solution:

a) Since the numbers of odds and evens will always be the same no matter which signs we use, the result will always be odd so not 0.

b) One possible solution is

1-2-3+4+5-6-7+8+…+97-98-99+100=0.

They’ll might find others.

4. A weighty problem

There is a box full of 1kg, 3kg, and 5kg weights.

a. Is it possible to take exactly 10 weights that together weigh 25kg?

b. I can make 25kg by using:

So for some numbers of weights I can make 25kg. What numbers are these?

What other numbers of weights can you use to make 25 kg? List them all and explain.

Solution:

a) The sum of 10 odd numbers will always be even so not 25.

b) We cannot use even numbers of weights as in a).

We can use any odd number up to (and including) 25:

9 weights: 4×5+5×1=25

11 weights: 3×5+1×3+7×1=25

13 weights: 3×5+10×1=25

15 weights: 2×5+1×3+12×1=25

17 weights: 2×5+15×1=25 etc.

You can see the pattern.

5. A candy problem

You have lots of candies. You make two small piles, and keep the rest as reserve. You can modify the piles as many times as you like by applying these rules in any order:

(i) You can take the same number of candies from each pile.

(ii) You can double the number of candies in a pile.

a) Start with piles of 10 and 3 pieces. Try to finish the game with no candies in each pile. (Don’t eat all the candies!)

b) Can you find a way to finish the game with no candies, no matter which numbers of candies you started with in each pile? Try a few examples: 6 and 5 pieces; 9 and 4 pieces.

c) If you start with piles of 10 and 3 pieces, but modify the rules as follows:

(i) You can take the same number of candies from each pile.

(iii) You can triple the number of candies in a pile.

Can you finish the game with no candies in each pile? Explain.

Solution:

a) There are many possibilities. One is:

- Double the pile of 3 to get 10 and 6.

-Take away 5 candy from each pile: remain 5 and 1

- Double the 1 to 2 to get 5 and 2

- Subtract 1 from each pile: remain 4 and 1

- Double the 1 to 2 to get 4 and 2

- Double 2 to 4: 4 and 4

-Subtract 4.

b) One very inefficient but failproof strategy is: subtract as many candy as necessary to be left with 1 candy in one of the piles. Then double that 1, subtract 1 from both piles and repeat until all candies are gone. The students will surely find others.

c) Step (iii) doesn’t change the parity of the two numbers. Step (i) either preserves parities or swaps them, in either case we’ll keep having an even and an odd number. They can never be equal.

6. Domino problems

a) Is it possible to cover a 5×5 chessboard in dominoes?

What about an 8×8 chessboard? Explain your answer in both cases.

Dominoes:

b) If we take 1 corner square out of a 5×5 chessboard, is it possible to cover the chessboard with dominoes? Explain your answer.

c) If we take two opposite corner squares out of an 8×8 chessboard to make a “mutilated” chessboard, is it possible to cover it in dominoes? Explain your answer.

Solution:

a) You can’t split 25 squares into dominoes of 2 squares each.

You can cover the 8×8 chessboard by horizontal dominoes.

b) Yes:

c) No: even though an even number of squares (62) remain, 30 of these are white and 32 are black, while a domino must always cover 1 white and 1 black square, so we can only cover the same number of white and black square.

Parity Party II

1. Blackboard game

The numbers 1 to 2013 are written on a blackboard.

a) We can pick any two numbers from the blackboard, erase them and replace them with their sum. If we do this long enough, only one number will be left on the blackboard. What will that number be?

b) We start the game again, but this time we can pick any two numbers from the blackboard, erase them and replace them with their positive difference. If we do this long enough, is it possible for the only number on the blackboard to be zero?

Solution:

2. More dominoes

A domino is a rectangle made up of two squares. Each square has 0 to 6 dots on it. A set contains one of each kind of domino.

a. How many dominoes are there in a set?

b. All the dominoes in a set are lined up in a chain (so that the number of dots on the ends of side-by-side dominoes match). If the square at one end has 5 dots on it, how many dots does the square on the other end have?

c. If we take out all the dominoes that have a square with no dot from a set, can we make a chain with the left over dominoes?

Solution:

3. Chessboard problems

a) On a chessboard, a knight makes an “L-shaped” move, made up of moving 2 squares in 1 direction, and 2 squares in another direction. If a knight starts on one corner square of the board, what is the fewest number of moves it takes for the knight to get to the opposite corner of the board?

b) Is it possible for the knight to get from one corner to the other while landing on every other square on the board exactly once?

c) Is it possible for the knight to be on any square in the board (after starting from the corner) after exactly 6 moves?

d) Is there any number, such that starting from a corner square, the knight can get to any square on the board it likes in exactly that many moves?

Solution:

Zero-One Arithmetic

even+even =even
even+odd=odd
odd+odd= even
0+0=0
1+0=1
1+1=0

It’s like turning a light switch: if you turn it once, light is on. But if you turn it twice, light is off.

1. Light switch game

Four Christmas light bulbs are arranged in a 2×2 grid. The switches for this grid are so connected that whenever you turn the switch for one of the light bulbs, this also affects the light bulbs immediately up or down, right or left from it. I’d like to know all the possible patterns of light I can get.

a) Examples:

0 / 0
0 / 0

Unlit grid:

1 / 1
1 / 0

In the unlit grid, turn the left upper corner light switch:

1 / 1
0 / 1

In the unlit grid, turn the right upper corner light switch:

In the unlit grid, turn the left lower corner light switch:

In the unlit grid, turn the right lower corner light switch:

Now let’s turn the switches in succession:

1 / 1
1 / 0
1 / 1
0 / 1

Left Up + Right Up: + =

Left Up + Right Up + Right Low:

+ + =

b) How many patterns are there in total? If you start from the unlit grid, how many patterns can you get by turning switches?

1 / 0
0 / 0
0 / 0
1 / 0

c) If you start from how can you get to ?

1 / 0
0 / 0
0 / 0
0 / 1

d) If you start from how can you get to ?

Solution:

2. Coin game

Arrange 16 coins on a table, so that they all have the same face up. One player is turned away from the table, eyes closed. Another player flips any of the coins on the board any number of times, each time saying “Tap”, and at the end covers one of the coins with the hand. The first player turns around and has to guess which face of the covered coin is up.

Is there a strategy for guessing all the time?

Solution:

3. Hat game

Being an extremely cruel and vicious maths teacher, I have decided to take ten of the students and stand them in a line one behind the other. On each of their heads I will place a hat, either black or white, I don’t care how many of each colour. Each person will see the colours of all the hats ahead of them, but be unable to see those behind them or their own. The kids starting from the back must tell what colour hat they are wearing, if they pick the wrong colour unfortunately they will die. However if they pick the right colour I will let them live (this time!).

Before we start this whole process, the students are allowed to discuss and to come up with the best strategy in order to save as many as possible…and just so you know you can definitely save at least nine of the kids!!

Solution:

3. Chocolate unwrapping game

http://funschool.kaboose.com/arcade/games/game_chocolate_biz.html

a) Can you describe the game in 0-1 algebra? How would you represent each 1 move?

b) If you start with a completely wrapped chocolate and then click on each of its squares exactly once, what pattern do you get at the end?

c) If you start with a completely wrapped chocolate and then click on the corner squares and the 4 interior squares, what pattern do you get at the end?

d) How can you win the game in the least possible number of moves?

Solution: