Dear pupils!
Not as soon as expected... but, finally, here you have these brilliant pictures from our trip to London!! We had a really good time there, didn't we? Ha ha ha!! :)
As Mercedes told you, I passed my exam in June and now I am working in my city: Elda!! Yippeeeeeeeee!! But, anyway, Valencia will always be special for me!
It was nice to work with all of you at CAMPANAR!! Hopefully, we will see each other again one day!! I wish you all the best!!
Kind regards
Susana!
PS: Make the most of your classes with Mercedes, she is a fantastic teacher and a very special person for me!!
Sunday 10 October 2010
Monday 24 May 2010
FORTNIGHT'S PUZZLE 16
PUZZLE 16: 24th-30th May (to be e-mailed by the 31st May)
POSTED BY IES CAMPANAR
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.
POSTED BY IES CAMPANAR
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.
Monday 10 May 2010
FORTNIGHT'S PUZZLE 15
PUZZLE 15: 10th-16th May (to be e-mailed by the 17th May)
POSTED BY IES CAMPANAR
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.
POSTED BY IES CAMPANAR
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.
Monday 26 April 2010
FORTNIGHT'S PUZZLE 14
PUZZLE 14: 26th April - 2nd May (to be e-mailed by the 3rd May)
POSTED BY IES CAMPANAR
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.
POSTED BY IES CAMPANAR
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.
Monday 12 April 2010
FORTNIGHT'S PUZZLE 13
PUZZLE 13: 12th-18th April (to be e-mailed by the 19th April)
POSTED BY IES CAMPANAR
ANIMALS
You have a fox, a chicken and a sack of grain. You must cross a river with only one of them at a time. If you leave the fox with the chicken he will eat it; if you leave the chicken with the grain he will eat it. How can you get all three across safely?
SOLUTION
Take the chicken over first. Go back and bring the grain next, but instead of leaving the chicken with the grain, come back with the chicken. Leave the chicken on the first side and take the fox with you. Leave it on the other side with the grain. Finally, go back over and get the chicken and bring it over.
POSTED BY IES CAMPANAR
ANIMALS
You have a fox, a chicken and a sack of grain. You must cross a river with only one of them at a time. If you leave the fox with the chicken he will eat it; if you leave the chicken with the grain he will eat it. How can you get all three across safely?
SOLUTION
Take the chicken over first. Go back and bring the grain next, but instead of leaving the chicken with the grain, come back with the chicken. Leave the chicken on the first side and take the fox with you. Leave it on the other side with the grain. Finally, go back over and get the chicken and bring it over.
Monday 22 March 2010
FORTNIGHT'S PUZZLE 12
PUZZLE 12: 22nd-28th March (to be e-mailed by the 29th March)
POSTED BY IES CAMPANAR
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.
POSTED BY IES CAMPANAR
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.
Wednesday 17 March 2010
Monday 8 March 2010
FORTNIGHT'S PUZZLE 11
PUZZLE 11: 8th-14th March (to be e-mailed by the 15th March)
POSTED BY IES CAMPANAR
NAMES
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?
SOLUTION
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.
POSTED BY IES CAMPANAR
NAMES
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?
SOLUTION
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.
Monday 22 February 2010
FORTNIGHT'S PUZZLE 10
PUZZLE 10: 22nd-28th February (to be e-mailed by the 1st March)
POSTED BY IES CAMPANAR
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, 58If it is not a multiple of 3, the number could be:61, 62, 64, 65, 67, 68If it is not a multiple of 4, the possibilities are:70, 71, 73, 74, 75, 77, 78, 79By eliminating the only possibility is number 75.
POSTED BY IES CAMPANAR
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, 58If it is not a multiple of 3, the number could be:61, 62, 64, 65, 67, 68If it is not a multiple of 4, the possibilities are:70, 71, 73, 74, 75, 77, 78, 79By eliminating the only possibility is number 75.
Monday 8 February 2010
FORTNIGHT'S PUZZLE 9
PUZZLE 9: 8th-14th February (to be e-mailed by the 15th February)
POSTED BY IES CAMPANAR
DIGIT NUMBERS
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) = 710a+b = 7a + 7b3a = 6ba = 2bWith b= 1,2,3,4 (more it's impossible), the solutions are: 21, 42, 63, 84
POSTED BY IES CAMPANAR
DIGIT NUMBERS
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) = 710a+b = 7a + 7b3a = 6ba = 2bWith b= 1,2,3,4 (more it's impossible), the solutions are: 21, 42, 63, 84
Monday 1 February 2010
Monday 25 January 2010
FORTNIGHT'S PUZZLE 8
PUZZLE 8: 25th-31st January (to be e-mailed by the 1st February)
POSTED BY IES CAMPANAR
DIGIT NUMER
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, 3The digits in the middle are: 6, 7 or 2,9There are eight possibilities, the numbers: 2673, 2763, 3672, 3762, 2293, 2923, 3292, 3922The solution is the number: 3762 (because 3762-1089 = 2673)
POSTED BY IES CAMPANAR
DIGIT NUMER
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, 3The digits in the middle are: 6, 7 or 2,9There are eight possibilities, the numbers: 2673, 2763, 3672, 3762, 2293, 2923, 3292, 3922The solution is the number: 3762 (because 3762-1089 = 2673)
Monday 11 January 2010
FORTNIGHT'S PUZZLE 7
PUZZLE 7: 11th-17th January (to be e-mailed by the 18th January)
POSTED BY IES CAMPANAR
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).
POSTED BY IES CAMPANAR
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).
Wednesday 6 January 2010
Tuesday 5 January 2010
Monday 4 January 2010
Sunday 3 January 2010
Saturday 2 January 2010
Friday 1 January 2010
FORTNIGHT'S PUZZLE 7
PUZZLE 7: 7th-13th May (to be e-mailed by the 14th May)
POSTED BY IES CAMPANAR
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
The correct solution from LUISA GARCÍA is:Simplify up to the maximum:
121212...121212
212121...212121
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.
POSTED BY IES CAMPANAR
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
The correct solution from LUISA GARCÍA is:Simplify up to the maximum:
121212...121212
212121...212121
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.
Subscribe to:
Posts (Atom)
