WINNERS OF MATHS COMPETITION IN SPAIN

From left to right: Jesús López (1st level), Mykca Semenov (3nd level) and Juan J. Plá (3rd level)

WINNERS OF MATHS COMPETITION IN GERMANY

The three winners (in the first photo) are  Luisa Wiechert (1st), Tbias Latta (2nd) and Julian Kaping (3rd).

In the second photo  we can see Margharita Eismann, a very special student who has helped us check the results and ranking of the Maths Competition and has also given away the prizes to the three winners.

WINNERS OF MATHS COMPETITION IN HUNGARY

Photo of the winners of Maths Competition in Hungary.
From the left to the right:

Zsófia Zentai (1st), Réka Vass (2nd), Máté Fenyvesi (3rd)

They are looking forward to the international part of the challenge.

WEEK'S PUZZLE 10

PUZZLE 10 -from the 7th to the 13th of December (to be collected by the 14th of December)

THE MAGIC NUMBER

'Free me, please', Ali begged the genius who had trapped him in a cage.
‘I will free you only if you find a number which obeys certain conditions’, the genius answered.
And these were the conditions:
1.- If the number were multiple of 2, then it would be a number between 50 and 59, both included.
2.- If it were not a multiple of 3, then it would be a number between 60 and 69, both included.
3.- If it were not a multiple of 4, then it would be a number between 70 and 79, both included.
Which was the magic number?

SOLUTION
If the number is a multiple of 2, it could be:
50, 52, 54, 56, 58
If it is not a multiple of 3, the number could be:
61, 62, 64, 65, 67, 68
If it is not a multiple of 4, the possibilities are:
70, 71, 73, 74, 75, 77, 78, 79

By eliminating the only possibility is number 75.

WEEK'S PUZZLE 9

PUZZLE 9 – from the 30th November to 6th December( to be collected by the 7th of December)
Try to find all the two-digit numbers which, when divided by the sum of their digits, have a quotient equal to 7 with no remainder.


SOLUTION
The number in polynomial form is 10a+b. Because of the condition:
(10a+b)/(a+b) = 7
10a+b = 7a + 7b
3a = 6b
a = 2b
With b= 1,2,3,4 (more it's impossible), the solutions are: 21, 42, 63, 84

WEEK'S PUZZLE 8

PUZZLE 8 – from the 23rd to the 29th of November ( to be collected by the 30th of November)

A certain four-digit number obeys the following conditions:
a) The sum of the squares of the two digits on both sides is = 13.
b) The sum of the squares of the two digits in the middle is =85.
c) If we substract 1089 from that certain four-digit number, we get another figure this time with the same digits but exactly in the opposite order.
Which number is it?


SOLUTION
Because of the conditions:
The digits on both sides are: 2, 3
The digits in the middle are: 6, 7 or 2,9

There are eight possibilities, the numbers: 2673, 2763, 3672, 3762, 2293, 2923, 3292, 3922

The solution is the number: 3762 (because 3762-1089 = 2673)

WEEK'S PUZZLE 7

PUZZLE 7- From the 16th to the 22nd November (to be collected by the 23rd of November)

A FIERCE BATTLE
In a fierce battle, say that at least 70% of the soldiers have lost an eye; at least 75% an ear; at least 80% an arm; and at least 85% a leg.
What percentage, at least, must have lost all four? (Lewis Carroll)

SOLUTION:

At least 45% lost an eye and an ear (70+75 = 145)
At least 65% lost an arm and a leg ( 80 +85 = 165)
Therefore at least 10% lost an eye, an ear, an arm and a leg (45 + 65 = 110).

WEEK'S PUZZLE 6

PUZZLE # 6 -from the 9th to the 15th of November (to be collected by the 16th of November)

FOUR DIGITS
This week your task is to try and find a four-digit number (abcd) so that when writing a decimal comma* between the second and the third digit (ab,cd) we get a number which is the mean or arithmetic average of the two digits which are left on both sides of the decimal comma.
*decimal comma or decimal point (ab.cd)


SOLUTION

Mentally:
The numbers are 49 and 50, because (49+50)/2 = 49,50

Mathematically:
If the number is abcd:

((10a + b) + (10c + d))/2 = 10a + b + c/10 + d/100

50 ( 10a + b) + 50 (10c + e) = 1000a + 100b + 10c + d

490c + 49d = 500a + 50b

49(10c + d) = 50(10a + b)

The only solution is 49 and 50.

WEEK'S PUZZLE 5

PUZZLE 5 -from the 2nd to the 8th of November (to be collected by the 9th of November)

THE AIRCRAFT

A light aircraft has just flown 400 kilometres. The aircraft covered the first 100 kms at a speed of 150 kph, the next 100 at 300 kph, the following 100 at 450 kph, and the last 100 at 600 kph.
Can you figure out the average speed of the aircraft on its 400 kilometres journey?

SOLUTION

t = e/v (t = time, e = space, v = velocity)

t1 = 100/150 = 2/3
t2 = 100/300 = 1/3
t3 = 100/450 = 2/9
t4 = 100/600 = 1/6

T = 2/3 + 1/3 + 2/9 + 1/6 = 25/18

v = E/ T = 400 : (25/18) = 288 kph

WEEK'S PUZZLE 4

PUZZLE 4- from the 27th October to the 1st November(to be collected by the 2nd November)



A FOOTBALL
A polyhedron with the shape of a football has 32 sides: 20 of them are regular hexagons and 12 are regular pentagons, as you can clearly see in the picture above.
How many vertices and edges does this solid have?

SOLUTION:
The football has as many vertices as the pentagons have: 12x5=60. As for the number of edges, which are counted twice, there are as many as the edges of all the polygons: (20x6 + 12x5)/2 = 180/2 = 90.

WEEK'S PUZZLE 3

PUZZLE 3- from the 19th to the 25th October (to be collected by the 26th October)



A NUMEROUS FAMILY

The product of my sons’ ages is 1664. The age of the eldest is the double of my youngest son’s age. I am 50 years old myself.
How many sons have I got?
How old are they?

SOLUTION
The factorial decomposition of 1664 is the product of 13 x 2 with exponent 7. Since 13 can be neither the youngest nor the eldest son, the solution is 2 with exponent 3 = 8, 13, and 2 with exponent 4 = 16. Therefore the three sons are: 8, 13 and 16 years old.

WEEK'S PUZZLE 2

PUZZLE 2* 12th -18th October (to be collected by the 19th)


AT THE CIRCUS

The most experienced trainer of a circus needs 40 minutes to wash an elephant,
his son doing the same task in 2 hours.
How long will it take them both to wash 4 elephants?
How many elephants will each of them wash?

SOLUTION TO PUZZLE 2

In two hours the son washes one elephant while his father washes 120:40 =3 elephants in the same period of time.

Thus it takes them 2 hours to wash 4 elephants since the son washes 1 while his father washes 3.

*Solution to Puzzle 1 on Tuesday, October the 13th

WEEK' S PUZZLE 1

PUZZLE 1 5th -11th October (to be collected by the 12th)

A CHESS GAME

There were 15 players in the latest chess game in which Kazimier Kaczyinski took part following the Swiss variant (all-against-all, once).
The sum of all the points obtained by all the players except Kazimier was: 100 points. How many points did Kazimier get?

Note: 1 point for a win and ½ for a draw, each player.

SOLUTION

First of all we can argue that each person plays against the other 14 players, so the number of games should be 15x14. However, if we do it this way, each game is counted twice. This is why the total number of games is actually (15x14)/2 = 105.
Since each game means 1 point of the total amount, we would have as a result 105 points.
Consequently Kazimier got 105 – 100 = 5 points.

THE WEEK’S PUZZLE

This is a competition in which you are supposed to solve a logical or Mathematical problem, but trying to avoid those which are usually in the Maths books and the curricula.
The 5 schools taking part in ‘The School as the Integration Engine’ Comenius Project are invited to take part in it. That is secondary schools from Lüneburg (Germany), Lipník (Czech Republic), Ajka (Hungary), Geneva (Switzerland) and Valencia (Spain).
The competition will run along the months of October, November and December. Every Tuesday a problem/puzzle will be uploaded on the project blog:
http://schoolintegrationengine.blogspot.com/
The deadline to offer a possible solution will be the following Monday. The next day, a Tuesday then, the solution will be uploaded for everybody to check. The last problem is due to appear on December the 7th.
Each school will select afterwards 3 students with the most correct answers , in three different levels according to their ages or levels. For each level , there will be later in December ONE more puzzle/problem to figure out who is the champion among the chosen students of all the countries participating in each level. Before Christmas these students will get a small gift in their own schools.
The photos and names of the winners will later be shown on the blog..