0 1 0 1 - utar research portalresearch.utar.edu.my/sose2014/4.part 2 developing higher order...maths...

0 1 0 18 4 2 10 1 0 1

1 0 1 18 4 2 11 0 1 1

1 1 0 18 4 2 11 1 0 1

Malaysia Singapore

Maths magic

Puzzle

Maths craft

Game

History

Project

Please do not suppose that the only function of puzzles is to entertain. Puzzles are entertain. Puzzles are a way of teaching mathematics. Indeed, they are the best way to teach it.

- Martin Gardner -

(a) How can you divide the 7 cakes among 10 students equally (fairly) where each student gets two pieces of cake?

Hint:How can you divide the 7 cakes (shown above) among 10 students equally where each student gets two pieces of cake (one piece of each type)?

Exercise

Real problem

Enigma

(b) How can you divide the 7 cakes among 12 students equally (fairly) where each student gets two pieces of cake?

(b)

Or ------------------------------------------------------------

http://staff.imsa.edu/~vmatsko/

Can you set anotherquestion that has- only 1 answer ?- 2 answers ?- 3 answers ?- no answer ?

Prof. Vincent J. Matsko

7 cakes - 1,6- 3,4LCM for 1, 3, 4, 6 = 12

7 cakes - 2,5LCM for 2, 5= 10

= 12(b) divide the 7 cakes

among 12 students equally (fairly), each student gets two pieces of cake?

(a) divide the 7 cakes among 10 students equally (fairly), each student gets two pieces of cake?

5 cakes - 1,4- 2,3LCM for 1, 4, 2, 3 = 12LCM for 1, 4, 2, 3 = 12

(c) How can you divide the 5 cakes among 12 students equally (fairly) where each student gets two pieces of cake?

(c) How can you divide the 5 cakes among 12 students equally (fairly) where each student gets two pieces of cake?

Or ------------------------------------------------------------

t (4.5cm)

(6cm) b 8.5cm

a (4cm)(6cm) b 8.5cm

a + t > b

t (4.5cm)

(6cm) b 8.5cm

a (4cm)(6cm) b

a + t > b

PSS: Experiment with real object

tb ta

b

a + t < b

t

t

ab

a + t = b

Book

Rulers

We can use two connected rulers as the track for the cones to roll “up hill”.

Nobody can give away what he has not got.

No teacher can impart to his No teacher can impart to his students the experience of discovery if he has not got it himself.

- George Polya -

glassImage of the burning candle

Glass panel

θ

Burning candle

Observer

θ

θθ

L9a

1 23

12

3

First secret number = 3

Special cards

“Kruskal Count”

23

41

The Last Special card

This trick is attributed to the physicist Martin David Kruskal (1925-2006).

http://www.maa.org/mathland/mathtrek_12_24_01.html

1 2 3 4 5 1 2 3 4 51 A 3 A2 A B 4 B2 C 3 C4 A B D 4 A D1 C E 2 E1 C E 2 B C4 C E 5 E1 A B D 4 A B C D5 A B D 55 15 13 C E 12 4 A B C D E2 23 A B C D E 34 53 1 A B C D E4 A B C D E 1 A B C D E3 2 A B C D E5 55 3 A B C D E

7 6 7 5 6 6.2 8 8 8 7 7 7.6

L9a

1. The Magician asks a spectator to secretly pick a number (N) from 1-3. Let’s say the spectator picks N=3.

2. The Magician asks the spectator to shuffle a pack of

This trick is attributed to the physicist Martin David Kruskal (1925-2006).

2. The Magician asks the spectator to shuffle a pack of cards consist of numbers 1-6 of all suits.

3. The Magician then deals the cards faced up on the table.

L9a

3. As the cards from the deck are revealed one by one, the spectator counts the cards and considers the Nth

card (in this case the third card) revealed to be his special card. He then look at the number on the third card, let’s say it is “4”. He then consider the following 4th card as his new special card and keeps following 4 card as his new special card and keeps counting N cards from that one, and so forth...

1 23

12

34

1

First secret number = 3

Special cards

L9a

4. At some point, the spectator reaches a special card (perhaps the last card in the deck) that is not followed by enough cards for him to complete the count. This card becomes the spectator’s last “Special Card” and the Magician's mind-reading target.

This trick is attributed to the physicist Martin David Kruskal (1925-2006).

the Magician's mind-reading target.

5. When all the cards has been dealt, the Magician can confidently point out that one particular card is the spectator’s last “Special Card“. How does it work?

L9a

Magic Reveal

The Magician selects his own starting number (preferably a small number), then counts his way to the end while the spectator is counting the cards. The Magician is very likely to end up at the same “Special

This trick is attributed to the physicist Martin David Kruskal (1925-2006).

Magician is very likely to end up at the same “Special Card" as the spectator.

Remarkably, for many arrangements of the deck, every starting point leads to the same final “Special Card”. This trick doesn't work every time. For certain arrangements of the deck, one or two starting points may generate trails that never coincide with the others.

Discussion: The reason this works is that as you’re both going through the deck, if you ever happen to hitthe same card as the victim, from then on you will both be locked into the same “path” through the restof the deck and will certainly arrive at the same final card. With a double deck, you just have about twice4as many chances of hitting the same card the victim does during the counting.If you do the trick as presented in class, where jack, queen and king represented the numbers 11, 12, and13, then you’ll only win about 2/3 of the time.You can make a mathematical estimate as follows. The “average” advance will be by:(1 + 2 + 3 + · · · + 9 + 10 + 5 + 5 + 5)/13 = 5.38cards, so you will take, on average, about 52/5.38, or about 8 steps. The density of the victim’s cards iscards, so you will take, on average, about 52/5.38, or about 8 steps. The density of the victim’s cards isabout 1/5.38 = .186, so there’s a 1 − .186 = .814 chance of a miss. But you have to miss all 8 times,and this will occur about (.814)8 = .174 so you’ll win about .836 of the time.I wrote a computer program to simulate this, and after a million trials, the winning percentage was .84169.If jack is 11, queen 12 and king 13, the winning percentage goes down to .68284 (again with a milliontrials). Finally, if we just spell out the names of the cards, so “ace” advances us by three cards, et cetera, Isimulate the probability to be .95347. With two decks, the percentages are, respectively, .97708, .90219and .99823.To make it much more likely to win, have the victim spell out the name of the card, as “ACE”, for threesteps, et cetera. The final section contains the simulator code that I used to obtain these numbers.This trick is apparently due to the physicist Martin Kruskal. Look up “Kruskal count” or “Kruskal cardtrick” for more information.

1-6Average advance = (1+2+3+4+5+6)/6 = 21/6 = 3.5Average Steps taken = 24/3.5 = 6.857, about 7 Average Steps taken = 24/3.5 = 6.857, about 7 stepsThe density of the victim’s cards is 1/3.5 = 0.286, so there’s a 1 − 0.286 = 0.714 chance of a miss. But you have to miss all 7 times,and this will occur about (0.714)7 = 0.0945 so you’ll win about 0.905 of the time.

1-5Average advance = (1+2+3+4+5)/5 = 15/5 = 3Average Steps taken = 20/3 = 6.667, about 7 stepsAverage Steps taken = 20/3 = 6.667, about 7 stepsThe density of the victim’s cards is 1/3 = 0.333, so there’s a 1 − 0.333 = 0.667 chance of a miss. But you have to miss all 7 times,and this will occur about (0.667)7 = 0.061 so you’ll win about 0.939 of the time.

1-3Average advance = (1+2+3)/3 = 6/3 = 2Average Steps taken = 12/2 = 6, 6 stepsAverage Steps taken = 12/2 = 6, 6 stepsThe density of the victim’s cards is 1/2 = 0.5, so there’s a 1 − 0.5 = 0.5 chance of a miss. But you have to miss all 6 times,and this will occur about (0.5)6 = 0.016 so you’ll win about 0.984 of the time.

1 Twinkle, twinkle, little star,2 How I wonder what you are.3 Up above the world so high,4 Like a diamond in the sky.

5 When the blazing sun is gone,5 When the blazing sun is gone,6 When he nothing shines upon,7 Then you show your little light,8 Twinkle, twinkle, all the night.

9 Then the traveller in the dark,10 Thanks you for your tiny spark,11 He could not see which way to go,12 If you did not twinkle so.

Lyrics by Jane Taylor (1783–1824)

13 In the dark blue sky you keep,14 And often through my curtains peep,15 For you never shut your eye,16 Till the sun is in the sky.16 Till the sun is in the sky.

17 As your bright and tiny spark,18 Lights the traveller in the dark.19 Though I know not what you are,20 Twinkle, twinkle, little star.

21 Twinkle, twinkle, little star.22 How I wonder what you are.23 Up above the world so high,24 Like a diamond in the sky.24 Like a diamond in the sky.

25 Twinkle, twinkle, little star.26 How I wonder what you are.27 How I wonder what you are.

2+2 = 2x21+2+3 = 1x2x3

Here are two examples where the sum and the product of the same set of numbers are the same. Can you think of three more sets of numbers of this nature?

1+1+2+4 = 1x1x2x4

1+1+1+2+5 = 1x1x1x2x5

1+1+1+3+3 = 1x1x1x3x3

1+1+2+2+2 = 1x1x2x2x2

1+1+1+1+2+6 = 1x1x1x1x2x6

1+1+1+1+1+2+2+3 = 1x1x1x1x1x2x2x3

http://www.youtube.com/watch?v=F6hTrzw3EJM

x x

41x

3=

1

Area = 4

1X 4

= 1

x¼

41

X 3x =

= 43

= 1

x

8cm

L8a

8cm

L8a

L8a

4.2cm

4.2cm

L8a

4.2cm

4.2cm

L8a

?

8cm √64.16

Area of the new frame= 64.16

= 8.01cm

Area of the small square in the new frame= 64.16-64

0.4cm

= 8.01cm= 64.16-64= 0.16

Width of the small square in the new frame

0.16 = 0.4cm=√

1+1+2+4 = 1x1x2x4

1+1+1+2+5 = 1x1x1x2x5

1+1+1+3+3 = 1x1x1x3x3

1+1+2+2+2 = 1x1x2x2x2

1+1+1+1+2+6 = 1x1x1x1x2x6

1+1+1+1+1+2+2+3 = 1x1x1x1x1x2x2x3

glassImage of the burning candle

Glass panel

θ

Burning candle

Observer

θ

θθ