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.
Monday, 22 March 2010
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
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