WORLD BOOK DAY IN VALENCIA

At Valencia IES Campanar school, the celebrations of the World Book Day took place with different activities both on the 23rd and on the 30th . Last Friday, the Comenius Literary Contest Awards prepared for the 10 winning students as well as those for the Maths Puzzle Competition and EU Competition were presented during a concert which included poems recited and music played by our students.
In the photo above you can see (on the right) Pilar García, IES Campanar Head Teacher Deputy, who has just given Fernando Escuin, the Comenius Project Coordinator, the certificates previously mentioned so that they can travel to Lipník, and one of the IES Campanar students, Joan Josep Pla, receiving his Maths Puzzle Award, the book and the red rose.
During the whole of last week, all the literary contest entries from the different countries had a place in the exhibition held in the hall of the school to celebrate World Book Day.
"Per Sant Jordi, una rosa vermella i un llibre escrit entre tots" (On St George's Day -23rd April, a red rose and a book written by and among us all). Thank you everybody!
The Spanish party travelling to the Czech Republic also decided to send all the Comenius family a photo (below) looking forward to seeing them all.      

PUZZLE#5 SECOND LEG (from 11th of March to the 25th of March 2010)

With this Puzzle ends the Second Leg.

The following figure, enclosed by three semicircles tangent among them on their extremes is called "arbelos", which means “shoemaker's knife” in Greek, and this term is used because the shaded area in the figure below resembles the blade of a knife used by ancient cobblers. Archimedes himself is believed to have been the first mathematician to study the mathematical properties of this figure. Show that its area is the same as the area of the circle which has as diameter segment BD, perpendicular to diameter CA.

SOLUTION

We've received  correct solutions from:

Zsofia Zentai (Hungary)

Reka Vass (Hungary)

Maté Fenyvesi (Hungary)

Joan Josep Pla (Spain)

Luisa Wiechert (Germany)

Julian Kaping (Germany).

Look at the correct solution from Maté Fenyvesi

PUZZLE #4 SECOND LEG (from the 24th of February to the 11th of March 2010)

INHUMAN SIMPLIFICATION

Simplify up to the maximum this big fraction in which both numbers (numerator and denominator), have 2010 digits:

121212...121212

212121...212121


SOLUTION 
We've received 7 correct solutions from:
Zsofia Zentai (Hungary)
Reka Vass (Hungary)
Kristynna Mandá ková ( Czech Republic)
Joan Josep Pla (Spain) 
Luisa Wiechert (Germany)
Julian Kaping (Germany).
Tobias Latta (Germany)


The correct solution from LUISA WIECHERT is:
Simplify up to the maximum: 

121212...121212

212121...212121

Solution:             If you reduce this fraction by 3, you get 4040…0404/7070…0707    (2009 digits).                  
             Then you can reduce it by 1010…0101     (2009 digits)

                               and you get 4/7.

                               So the solution is 4/7.

PUZZLE #3 SECOND LEG

PUZZLE #3 (from the 9th to the 24th of February 2010)

A WICKED FRIEND

James asked Sharon what her telephone number was. Sharon, who is very fond of Mathematics, answered this:“It is the only 9-digit number whose digits are all different and different from zero and in which ANY two consecutive digits make a number which is divisible either by 7 or by 13.

 Well, what’s Sharon’s telephone number?

SOLUTION 
We've received 5 correct solutions from:
Zsofia Zentai (Hungary)
Reka Vass (Hungary)
Máté Fenyvesi (Hungary)
Kristynna Mandá ková ( Czech Republic)
Joan Josep Pla (Spain) 
Luisa Wiechert (Germany)
Julian Kaping (Germany).


Here, we plublish Zsofia Zentia's solution

At first, we need to find all those 2-digit numbers that are divisible by 7 or 13.

ð  Numbers that are divisible by 7:

o   14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91 and 98.

ð  Numbers that are divisible by 13:

o   13, 26, 39, 52, 65, 78 and 91.

After a little observation we find that:

ð  Neither of these numbers ends by 7à the telephone number may start with 78

ð  Only number 4 can follow number 8. à784

Two numbers can follow 4: number 2 or 9.

3rd number of the phone number	4th	5th	6th	7th	8th	9th
4	2	8
4	2	1	4
4	2	1	3	5	6	3
4	2	1	3	5	6	5
4	2	1	3	5	2
4	2	1	3	9	8
4	2	1	3	9	1
4	2	6	3	5	6
4	2	6	3	5	2
4	2	6	3	9	1	4
4	2	6	3	9	1	3
4	2	6	3	9	8
4	2	6	5	6
4	2	6	5	2
4	9	8
4	9	1	4
4	9	1	3	9
4	9	1	3	5	6	3
4	9	1	3	5	6	5
4	9	1	3	5	2	1
4	9	1	3	5	2	8
4	9	1	3	5	2	6

So Sharon’s phone number is: 784913526.

PUZZLE 2 (Second Leg)

TO BE HANDED OR E-MAILED BY THE 10th OF FEBRUARY 2010

A Ball Game

A box holds 40 juggling balls. Two friends are playing a game in which they alternate to pick up balls from the box. When it is their turn each player can take out as many balls as wanted but never more than half the number of balls in the box. The player that cannot take any more balls from the box will lose.
Who will be the winner, the first or the second friend to play? And which will be the right strategy to win the game?

SOLUTION

Let us see if there is a winning strategy. If there is one, the winner in his/ her last turn, should leave the other person only one ball as that would make it impossible for the other player to pick up any balls.

And so, going backwards, the winner in the previous movement, should have left the other player with 3 balls. This way the “loser” could only have picked up 1 ball and leave the other 2. (W(n-1) = 2·1 + 1 = 3).

In the same way, in the last-but-one movement, the winner should have left W(n-2) = 2·3 + 1 =7,and in previous turns he/she should have left: W(n-3) = 2·7 +1 = 15, W(n-4) = 2·15 +1 = 31.

However, only the person playing FIRST can get to do that by picking up 9 balls on his/her first movement and leaving in the following 15, 7, 3 y 1 balls.

The winner thus will be the first person to play and to use the strategy explained above.

PUZZLE 1 (second leg)

TO BE HANDED OR E-MAILED BY 25th OF JANUARY 2010 

Three people whose surnames are White, Red and Black happen to meet at a party. Shortly after introducing themselves, the lady says:

-‘It’s funny that our surnames are White, Red and Black and that there are here three people whose hair colours are exactly those three colours’.

The red-haired person answers:

-‘Yes, it really is but you have probably noticed that none of us has the colour corresponding to our surnames’.

-‘True!’ says the person called White.

If the lady hasn’t got black hair,  who has it red?

  

S O L U T I O N

Mr/Ms White hasn’t got red haired because he/she speaks after the one who is red-haired. He/She hasn’t got white hair because this is his family name so he/she has black hair.

Mr/Ms Red has neither red nor black hair so this is the white-haired person.

 The only possibility is for the person called Black to be red-haired. Apart from that the lady can neither be called White nor Black, who answers to her. Thus she is called Red.

SOLUTION: The person who has red hair is Mr. Black.