QUESTIONS:
1) The Elder Twin
One day Kerry celebrated her birthday. Two days later her older twin brother, Terry, celebrated his birthday.
How could this happen?
2) Manhole Covers
Why is it better to have round manhole covers than square ones?
This is logical rather than lateral, but it is a good puzzle which can be solved by lateral thinking techniques. It is supposedly used by a very well-known software company as an interview question for prospective employees.
3) The Deadly Party
A man went to a party and drank some of the punch. He then left early. Everyone else at the party who drank the punch subsequently died of poisoning.
Why wasn't the first guy poisoned?
4) Heaven
A man died and went to Heaven. There were thousands of other people there. They were all naked and all looked as they died at the age of 21. He looked around to see if there was anyone he recognized. He saw a couple and he knew immediately that they were Adam and Eve.
How did he know?
5) Trouble with Sons
A woman had two sons who were born on the same hour of the same day of the same year. But they were not twins.
How could this be so?
6) The Man in the Bar
A man walks into a bar and asks the barman for a glass of water. The barman pulls out a gun and points it at the man. The man says `Thank you' and walks out.
Why?
This puzzle has claims to be the best of the genre. It is simple in its statement, absolutely baffling and yet with a completely satisfying solution. Most people struggle very hard to solve this one yet they like the answer when they hear it or have the satisfaction of figuring it out.
ANSWERS:
1) At the time she went into labor, the mother of the twins was traveling by boat. The older twin, Terry, was born first early on March 1st. The boat then crossed the International Date line (or anytime zone line) and Kerry, the younger twin, was born on February the 28th. In a leap year the younger twin celebrates her birthday two days before her older brother.
2) A square manhole cover can be turned and dropped down the diagonal of the manhole. A round manhole cannot be dropped down the manhole. So for safety and practicality, all manhole covers should be round.
3) The poison in the punch came from the ice cubes. When the man drank the punch the ice was fully frozen. Gradually it melted, poisoning the punch.
4) He recognized Adam and Eve as the only people without navels. Because they were not born of women, they had never had umbilical cords and therefore they never had navels.
This one seems perfectly logical but it can sometimes spark fierce theological arguments!
5) They were two of a set of triplets (or quadruplets etc.)
This simple little puzzle stumps many people. They try outlandish solutions involving test-tube babies or surrogate mothers. Why does the brain search for complex solutions when there is a much simpler one available?
6) The man had hiccups. The barman recognized this from his speech and drew the gun in order to give him a shock. It worked and cured the hiccups - so the man no longer needed the water.
The is a but a difficult one to solve. It is a perfect example of a seemingly irrational and incongruous situation having a simple and complete explanation. Amazingly this classic puzzle seems to work in different cultures and languages.
Saturday, April 2, 2011
King Bear & Poison ..Very Common & Tough Puzzle
Problem : A king has a wine cellar with 100 drums of wine. A traitor sneaked into his wine cellar to poison the drums. He could mix poison in only one of the drums when guards caught him. However, guards could not see which one drum was poisoned. Our king being a drunkard wishes to be able to drink wine as soon as possible but safely. The poison guarantees that the person drinking the wine would die in 10 days.
Being a king, he can order his servants to drink wine from the drums to find out poisonous drum. However, he wishes does not want to risk lives of too many servants.
1) King wishes to be able to drink some wine on 11th day. How many servant would be needed to take the risk of drinking wine before figuring out a harmless wine drum?
2) King wishes to throw a grand party on 11th day. How many servant would be needed for figuring out the poisonous wine drum?
Solution :
The solution to the first problem is extremely simple and obvious. Only one servant need to drink any one of the wine drum. If the servant dies, all the other drums are harmless. If he does not die, we can guarantee that at least that one wine-drum is harmless.
Well, second problem has actually two sides to it. Say for example, king wishes that number of servant to be put on stake can be arbitrarily large, but the number of servant died during the process should be less. Then, we need 99 servant for each wine drum. At most one servant would die in this case.
However, what if we want to minimize the number of servants to be put on stake? Initially I went on a very wrong direction. That was to use prime factorization of every number from 1 to 100. For example, there will be a servant drinking from wine-drums which are numbered in multiple of 2, and similarly another servant for multiple of 5. So, we would be able to detect number 10 as the poisonous drum if both of them dies. Using this multiple scheme, number of servant needed would be
2,4,8,16,32,64
3,9,27,81
5,25
7,49
all primes number between 1 to 100 which has not appeared in the above list.
Basically, the number comes out to be huge. Certainly a very very bad solution.
A good solution would be to arrange wine-drums in 10×10 maze. We would need only 20 servant. 10 servant for the row and 10 for the column. So when in 10 days i-th servant for row and j-th servant for column dies, we would know that the drum located at (i,j) is the culprit. But still we are far from the best solution.
We can visualize the maze in three dimension. 5x5x5 would need only 15 servants. Of course, there would be some spots empty in the maze, but that does not matter.6x6x3 = 15. Going ahead we can have 5x5x4=14.
It seems we can still push it further. We can visualize the maze in four dimension. 4x3x3x3 = 13. This is the least possible number I could find. I don’t know whether we can push it still further down. If you can figure out a still smaller number, please comment on this post.
Soln By –Saurabh Joshi from his blog me.maths & puzzle
i found its interesting so i pasted here
Being a king, he can order his servants to drink wine from the drums to find out poisonous drum. However, he wishes does not want to risk lives of too many servants.
1) King wishes to be able to drink some wine on 11th day. How many servant would be needed to take the risk of drinking wine before figuring out a harmless wine drum?
2) King wishes to throw a grand party on 11th day. How many servant would be needed for figuring out the poisonous wine drum?
Solution :
The solution to the first problem is extremely simple and obvious. Only one servant need to drink any one of the wine drum. If the servant dies, all the other drums are harmless. If he does not die, we can guarantee that at least that one wine-drum is harmless.
Well, second problem has actually two sides to it. Say for example, king wishes that number of servant to be put on stake can be arbitrarily large, but the number of servant died during the process should be less. Then, we need 99 servant for each wine drum. At most one servant would die in this case.
However, what if we want to minimize the number of servants to be put on stake? Initially I went on a very wrong direction. That was to use prime factorization of every number from 1 to 100. For example, there will be a servant drinking from wine-drums which are numbered in multiple of 2, and similarly another servant for multiple of 5. So, we would be able to detect number 10 as the poisonous drum if both of them dies. Using this multiple scheme, number of servant needed would be
2,4,8,16,32,64
3,9,27,81
5,25
7,49
all primes number between 1 to 100 which has not appeared in the above list.
Basically, the number comes out to be huge. Certainly a very very bad solution.
A good solution would be to arrange wine-drums in 10×10 maze. We would need only 20 servant. 10 servant for the row and 10 for the column. So when in 10 days i-th servant for row and j-th servant for column dies, we would know that the drum located at (i,j) is the culprit. But still we are far from the best solution.
We can visualize the maze in three dimension. 5x5x5 would need only 15 servants. Of course, there would be some spots empty in the maze, but that does not matter.6x6x3 = 15. Going ahead we can have 5x5x4=14.
It seems we can still push it further. We can visualize the maze in four dimension. 4x3x3x3 = 13. This is the least possible number I could find. I don’t know whether we can push it still further down. If you can figure out a still smaller number, please comment on this post.
Soln By –Saurabh Joshi from his blog me.maths & puzzle
i found its interesting so i pasted here
Friday, April 1, 2011
Another Hat Color Problem
The Island - One puzzle a day - Puzzle Buddies
100 men were captured and prisoned in an island.the island is guarded by a genie.
Everyday genie takes prisoners out.genie places them in a circle so that everyone can see each other,then all the prisoners are blindfolded. Genie then places hats on prisoners head,some of them are white and some of them are red.their folds will be opened after the genie places hats.
On genie's command the prisoners with white hat should step forward.They will be set free if only the men with white hat steps forward.
If any one without white hat steps forward,genie executes all of them
If no one steps forward,the same game will be repeated the next day,if the game is repeated,each prisoner will get the same hat that they got on the first day
Prisoners are not allowed to communicate with each prisoner.if communicated all of them will be executed.prisoners are not allowed to see the hat that they are wearing
At least one white hat is given, and all prisoners are aware of this.
how many days did it take for the prisoners to get out if genie gave "n" white hats where 1<=n<=100, and how did they do it?
Solution:
Let us say there is one white hat,the answer is simple.the prisoner wearing white hat sees no other white hat and he steps forward.
Now let us take two white hats.No body comes forward on the first day,because the prisoners wearing white hat sees one white hat.On the second day since no body has come forward they both know there must be two white hats and one he can see and other must be his own.so they come forward on second day.
Similarly if there are three white hats they will not come out on second day.it will take three days.
and so on..for "n" it will take n days
100 men were captured and prisoned in an island.the island is guarded by a genie.
Everyday genie takes prisoners out.genie places them in a circle so that everyone can see each other,then all the prisoners are blindfolded. Genie then places hats on prisoners head,some of them are white and some of them are red.their folds will be opened after the genie places hats.
On genie's command the prisoners with white hat should step forward.They will be set free if only the men with white hat steps forward.
If any one without white hat steps forward,genie executes all of them
If no one steps forward,the same game will be repeated the next day,if the game is repeated,each prisoner will get the same hat that they got on the first day
Prisoners are not allowed to communicate with each prisoner.if communicated all of them will be executed.prisoners are not allowed to see the hat that they are wearing
At least one white hat is given, and all prisoners are aware of this.
how many days did it take for the prisoners to get out if genie gave "n" white hats where 1<=n<=100, and how did they do it?
Solution:
Let us say there is one white hat,the answer is simple.the prisoner wearing white hat sees no other white hat and he steps forward.
Now let us take two white hats.No body comes forward on the first day,because the prisoners wearing white hat sees one white hat.On the second day since no body has come forward they both know there must be two white hats and one he can see and other must be his own.so they come forward on second day.
Similarly if there are three white hats they will not come out on second day.it will take three days.
and so on..for "n" it will take n days
Monty Hall Problem
You are a contestant on a game show. You have three closed doors in front of you. One of the doors has a car behind it and the other two doors have nothing. You have to choose a door to open.
You have with you a Magic Watch.
- Whenever you ask the watch a question with a yes/no answer, it will blink either red or blue.
- One of the colors represents 'yes' and the other 'no'.
- You don't know which color means what. (gotcha!)
- You are allowed to ask the watch only 2 questions before you make your decision.
Which questions would you ask and how would you choose the door to open?
The puzzle might have a number of solutions and this is one of them.
Solution:
The contestant should ask the following two questions:
1) If I asked whether the car was behind door 1, would you blink red?
2) If I asked whether the car was behind door 2, would you blink red?
There are two cases here:
a) red = no, blue = yes:
The watch can answer in 4 different ways:
1) red, blue:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2st question, car is not behind door 2.
2) blue, red:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is behind door 2.
3) blue, blue:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is not behind door 2.
4) red, red:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2nd question, car is behind door 2.
b) red = yes, blue = no:
The watch can answer in 4 different ways:
1) red, blue:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2st question, car is not behind door 2.
2) blue, red:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is behind door 2.
3) blue, blue:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is not behind door 2.
4) red, red:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2nd question, car is behind door 2.
Observe that the conclusions are exactly the same for the same answer set irrespective of the color code.
Hence, when the watch answers:
1) red, blue: Pick door 1.
2) blue, red: Pick door 2.
3) blue, blue: Pick door 3.
4) red, red: This outcome is impossible since it implies that the car is behind door 1 as well as door 2.
You have with you a Magic Watch.
- Whenever you ask the watch a question with a yes/no answer, it will blink either red or blue.
- One of the colors represents 'yes' and the other 'no'.
- You don't know which color means what. (gotcha!)
- You are allowed to ask the watch only 2 questions before you make your decision.
Which questions would you ask and how would you choose the door to open?
The puzzle might have a number of solutions and this is one of them.
Solution:
The contestant should ask the following two questions:
1) If I asked whether the car was behind door 1, would you blink red?
2) If I asked whether the car was behind door 2, would you blink red?
There are two cases here:
a) red = no, blue = yes:
The watch can answer in 4 different ways:
1) red, blue:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2st question, car is not behind door 2.
2) blue, red:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is behind door 2.
3) blue, blue:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is not behind door 2.
4) red, red:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2nd question, car is behind door 2.
b) red = yes, blue = no:
The watch can answer in 4 different ways:
1) red, blue:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2st question, car is not behind door 2.
2) blue, red:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is behind door 2.
3) blue, blue:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is not behind door 2.
4) red, red:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2nd question, car is behind door 2.
Observe that the conclusions are exactly the same for the same answer set irrespective of the color code.
Hence, when the watch answers:
1) red, blue: Pick door 1.
2) blue, red: Pick door 2.
3) blue, blue: Pick door 3.
4) red, red: This outcome is impossible since it implies that the car is behind door 1 as well as door 2.
How can you get a fair coin toss
Interview question: How can you get a fair coin toss if someone hands you a coin that is weighted to come up heads more often than tails?
The story of these types of questions starts at the software giant Microsoft who pioneered the use of puzzle type question, where the aim is to test the interviewers problem solving skills and creativity. It's ironic that these type of questions are no longer asked in Microsoft job interviews, but the rest of the software industry hasn't caught on.
The solution is to toss the coin twice, which gives four possible outcomes:
H (heads) followed by H, probability of this is pH * pH
H followed by T (tails), probability of this is pH * pT
T followed by H, probability of this is pT * pH
T followed by H, probability of this is pT * pT
Looking at the above outcomes we can see that HT (H followed by T) is just as likely as TH,
So to get a fain coin toss, toss the coins and use HT or TH as fair outcomes of the toss, and discard the results of TT and HH.
The story of these types of questions starts at the software giant Microsoft who pioneered the use of puzzle type question, where the aim is to test the interviewers problem solving skills and creativity. It's ironic that these type of questions are no longer asked in Microsoft job interviews, but the rest of the software industry hasn't caught on.
The solution is to toss the coin twice, which gives four possible outcomes:
H (heads) followed by H, probability of this is pH * pH
H followed by T (tails), probability of this is pH * pT
T followed by H, probability of this is pT * pH
T followed by H, probability of this is pT * pT
Looking at the above outcomes we can see that HT (H followed by T) is just as likely as TH,
So to get a fain coin toss, toss the coins and use HT or TH as fair outcomes of the toss, and discard the results of TT and HH.
Another Hat Puzzle
I buried four fishermen up to their necks in the sand on the beach at low tide for keeping their fishing spot a secret from me. I put a hat on each of their heads and told them that one of them must shout out the correct color of their own hat or they will all be drowned by the incoming tide. I give them 10 minutes to do this. Fisherman A and B can only see the sand dune I erected. Fisherman C can see that fisherman B has a white hat on. Fisherman D can see that C has a black hat, and B has a white hat. The fisherman have been told that there are four hats, two white and two black, so they know that they must have either a white or a black hat on. who shouts out the color of their hat and how do they know?
Fisherman C shouts out.
Fishermen A and B are in the same situation - they have no information to help them determine their hat colour so they can’t answer. C and D realise this.
Fisherman D can see both B and C’s hats. If B and C had the same colour hat then this would let D know that he must have the other colour.
When the time is nearly up, or maybe before, C realises that D isn’t going to answer because he can’t. C realises that his hat must be different to B’s otherwise D would have answered. C therefore concludes that he has a black hat because he can see B’s white on
Fisherman C shouts out.
Fishermen A and B are in the same situation - they have no information to help them determine their hat colour so they can’t answer. C and D realise this.
Fisherman D can see both B and C’s hats. If B and C had the same colour hat then this would let D know that he must have the other colour.
When the time is nearly up, or maybe before, C realises that D isn’t going to answer because he can’t. C realises that his hat must be different to B’s otherwise D would have answered. C therefore concludes that he has a black hat because he can see B’s white on
Who Will Tell The Color of His Hat/halos.. First
Four angels sat on the Christmas tree amidst other ornaments. Two had blue halos and two – yellow. However, none of them could see above his head. Angel A sat on the top branch and could see the angels B and C, who sat below him. Angel B, could see angel C who sat on the lower branch. And angel D stood at the base of the tree obscured from view by a thicket of branches, so no one could see him and he could not see anyone either.
Which one of them could be the first to guess the color of his halo and speak it out loud for all other angels to hear?
Case 1: Angels B and C have same color(either both blue or yellow) Then A can be sure of his color.
Case 2: Angles B and C have different colors.
B after waiting for some time will realise that A has not spoken out the color yet ,hence B and C must have different colors. B will look at C's color and announce his color opposite to C's color.
Which one of them could be the first to guess the color of his halo and speak it out loud for all other angels to hear?
Case 1: Angels B and C have same color(either both blue or yellow) Then A can be sure of his color.
Case 2: Angles B and C have different colors.
B after waiting for some time will realise that A has not spoken out the color yet ,hence B and C must have different colors. B will look at C's color and announce his color opposite to C's color.
Wednesday, March 30, 2011
Weigh System With More Explaination..How Many Weight We Needs..???
Puzzle: You have a common balance with two sides. You can only keep weighs on one side. You have to weigh items from 1 to 1000 weight units with integral weights only.
What is the minimum number of weights required and what will be their weights?
Eg. With 2 weights one of 1 weight unit and other of 3 weight unit you can measure 1, 3 and 4 only.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow. You can provide your answer in comments.
Solution: This puzzle is an easy puzzle as to make any odd number you need all other even numbers and one 1.
So only one odd number is needed to make all odd numbers. Now you are left with all even numbers. All even numbers are base 2 and thus can be though of as binary representation. To make 1000 you need at most 10 bits(1111101000). You can make any number below 1000 from these bits.
So the answer is 10 weights required. 1, 2, 4, 8,16, 32, 64, 128, 256, 512.
Puzzle: This is part 2 of puzzle posted Above. The puzzle is that you have to weigh 1 to 1000 using a common balance scale with minimum number of weighs. This time you can put wights on both sides.
You can check Part 1 of this puzzle How many weights? Puzzle Part 1.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Solution: The answer is series of base 3 i.e. 1, 3, 9, 27, 81, 243 and 729.
We can start from smallest number 1. To measure 1 we definitely need 1.
To measure 2 we need to look for highest number possible. We can measure 2 by using 3(3-1 = 2). So we need 1 and 3.
Now to measure 4 we can use 1 and 3 both (1+ 3). To measure 5 again we will look for highest possible number. 9 is highest possible number (9-4 = 5). We can measure up to 13 using these weights.
Continuing the same way we can deduce that a series of base 3 is required, thus 1, 3, 9, 27, 81, 243, 729.
Puzzle: This is the 3rd part of weighing puzzles. You can check Part 2 and Part 1.
You have to weigh 125 packets of sugar with first one weighing 1 lb, second 2 lb, third 3 lb and so on.You can only use one pan of the common balance for measurement for weighing sugar, the other pan has to be used for weights.
The weights have to put on to the balance and it takes time. Time taken to move weights is equal to the number of weights moved.
Eg. So if you are weighing 5 lb using 3lb and 2 lb weight then you have to lift both these weights one by one and thus it will take 2 unit time. If you have to measure only 3lbs it will take only 1 unit time. Now if we have to measure sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 lb, with 1 unit of time, we place 3 lb along with 1 lb and measure 4kg with again 1 unit of time, and finally we move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4lb using weights 1 and 3 only.
Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Source: Classic Puzzles
Solution: Amazing explanation provided by Kasturi.
You will need the weights 1, 2, 4, 8, 16, 32 and 64 i.e. 7 weights. You can measure 125 packets in 125 movements. The strategy to be followed is this:
Put 1: measure 1, weights in pan 1
To measure further you need to use a new weight.
Put 2: measure 3, weights in pan 1,2
Remove 1: measure 2, weights in pan 2
Thus, by adding the next smallest weight to the pan i.e. 2, we can measure all weights below 4 in total 3 movements.
To measure further you need to use a new weight.
Put 4: measure 6, weights in pan 2,4
Remove 2: measure 4, weights in pan 4
Put 1: measure 5, weights in pan 1,4
Put 2: measure 7, weights in pan 1,2,4
Thus, by adding the next smallest weight to the pan i.e. 4, we can measure all weights below 8 in total 7 movements.
Following the same strategy, whenever we have measured all weights below n, add n and we can measure measure all the weights below 2n in 2n-1 movements.
Eventually, we can measure 125 packets in 125 movements.
What is the minimum number of weights required and what will be their weights?
Eg. With 2 weights one of 1 weight unit and other of 3 weight unit you can measure 1, 3 and 4 only.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow. You can provide your answer in comments.
Solution: This puzzle is an easy puzzle as to make any odd number you need all other even numbers and one 1.
So only one odd number is needed to make all odd numbers. Now you are left with all even numbers. All even numbers are base 2 and thus can be though of as binary representation. To make 1000 you need at most 10 bits(1111101000). You can make any number below 1000 from these bits.
So the answer is 10 weights required. 1, 2, 4, 8,16, 32, 64, 128, 256, 512.
Puzzle: This is part 2 of puzzle posted Above. The puzzle is that you have to weigh 1 to 1000 using a common balance scale with minimum number of weighs. This time you can put wights on both sides.
You can check Part 1 of this puzzle How many weights? Puzzle Part 1.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Solution: The answer is series of base 3 i.e. 1, 3, 9, 27, 81, 243 and 729.
We can start from smallest number 1. To measure 1 we definitely need 1.
To measure 2 we need to look for highest number possible. We can measure 2 by using 3(3-1 = 2). So we need 1 and 3.
Now to measure 4 we can use 1 and 3 both (1+ 3). To measure 5 again we will look for highest possible number. 9 is highest possible number (9-4 = 5). We can measure up to 13 using these weights.
Continuing the same way we can deduce that a series of base 3 is required, thus 1, 3, 9, 27, 81, 243, 729.
Puzzle: This is the 3rd part of weighing puzzles. You can check Part 2 and Part 1.
You have to weigh 125 packets of sugar with first one weighing 1 lb, second 2 lb, third 3 lb and so on.You can only use one pan of the common balance for measurement for weighing sugar, the other pan has to be used for weights.
The weights have to put on to the balance and it takes time. Time taken to move weights is equal to the number of weights moved.
Eg. So if you are weighing 5 lb using 3lb and 2 lb weight then you have to lift both these weights one by one and thus it will take 2 unit time. If you have to measure only 3lbs it will take only 1 unit time. Now if we have to measure sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 lb, with 1 unit of time, we place 3 lb along with 1 lb and measure 4kg with again 1 unit of time, and finally we move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4lb using weights 1 and 3 only.
Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Source: Classic Puzzles
Solution: Amazing explanation provided by Kasturi.
You will need the weights 1, 2, 4, 8, 16, 32 and 64 i.e. 7 weights. You can measure 125 packets in 125 movements. The strategy to be followed is this:
Put 1: measure 1, weights in pan 1
To measure further you need to use a new weight.
Put 2: measure 3, weights in pan 1,2
Remove 1: measure 2, weights in pan 2
Thus, by adding the next smallest weight to the pan i.e. 2, we can measure all weights below 4 in total 3 movements.
To measure further you need to use a new weight.
Put 4: measure 6, weights in pan 2,4
Remove 2: measure 4, weights in pan 4
Put 1: measure 5, weights in pan 1,4
Put 2: measure 7, weights in pan 1,2,4
Thus, by adding the next smallest weight to the pan i.e. 4, we can measure all weights below 8 in total 7 movements.
Following the same strategy, whenever we have measured all weights below n, add n and we can measure measure all the weights below 2n in 2n-1 movements.
Eventually, we can measure 125 packets in 125 movements.
Tiger & Sheep Puzzle..Will Sheep Survive or Not
Hundred tigers and one sheep are put on a magic island that only has grass. Tigers can live on grass, but they would rather eat sheep. Its a Magic Iceland because if a Tiger eats the Sheep then it will become a sheep itself (and hence can be eaten up by another tiger).
Tigers don’t mind being a sheep, but they would never want themselves to be eaten up. All tigers are intelligent and they want to survive. They however, don’t care of survival of others.
Will the sheep survive or will it be eaten up?
Solution:
This problem and the problem of pirates belong to the same family of Puzzles, where the puzzle is solved by simplification. Lets Consider the case when there are less Tigers
If there is 1 tiger, then he will eat the sheep because he does not need to worry about being eaten. Sheep will Not survive.
If there are 2 tigers, Both of them knows that if he eats the Sheep, the other tiger will eat him. So, The Sheep will Survive.
If there are 3 tigers, then they each of them knows that if he eats up the Sheep, then Iceland will be left with 1 sheep and 2 Tigers and as shown in the previous case, the Sheep will survive. Hence each tiger will try to eat up the sheep. The sheep will Not Survive.
If there are 4 Tigers, then the sheep will Survive.
And so on….
So, If there are even number of tigers the sheep will Survive, else it will die. Hence, if there are 100 tigers the sheep will Survive.
Tigers don’t mind being a sheep, but they would never want themselves to be eaten up. All tigers are intelligent and they want to survive. They however, don’t care of survival of others.
Will the sheep survive or will it be eaten up?
Solution:
This problem and the problem of pirates belong to the same family of Puzzles, where the puzzle is solved by simplification. Lets Consider the case when there are less Tigers
If there is 1 tiger, then he will eat the sheep because he does not need to worry about being eaten. Sheep will Not survive.
If there are 2 tigers, Both of them knows that if he eats the Sheep, the other tiger will eat him. So, The Sheep will Survive.
If there are 3 tigers, then they each of them knows that if he eats up the Sheep, then Iceland will be left with 1 sheep and 2 Tigers and as shown in the previous case, the Sheep will survive. Hence each tiger will try to eat up the sheep. The sheep will Not Survive.
If there are 4 Tigers, then the sheep will Survive.
And so on….
So, If there are even number of tigers the sheep will Survive, else it will die. Hence, if there are 100 tigers the sheep will Survive.
A Fox, A Sheep, and A Sack of Hay
A farmer is travelling with a fox, a sheep and a small sack of hay. He comes to a river with a small boat in it. The boat can only support the farmer and one other animal/item. If the farmer leaves the fox alone with the sheep, the fox will eat the sheep. And if the farmer leaves the sheep alone with the hay, the sheep will eat the hay.
How can the farmer get all three as well as himself safely across the river?
Answer
1. The farmer takes the sheep across the river, then returns back.
2. The farmer takes the fox across the river.
3. The farmer takes the sheep back to the first side of the river.
4. The farmer leaves the sheep back on the first side of the river, and takes the hay to the other side.
5. The farmer returns to the first side of the river.
6. The farmer brings the sheep back to the second side.
How can the farmer get all three as well as himself safely across the river?
Answer
1. The farmer takes the sheep across the river, then returns back.
2. The farmer takes the fox across the river.
3. The farmer takes the sheep back to the first side of the river.
4. The farmer leaves the sheep back on the first side of the river, and takes the hay to the other side.
5. The farmer returns to the first side of the river.
6. The farmer brings the sheep back to the second side.
Pills Puzzle
A person was prescribed to take two pills (tablets), one each, from the two bottles viz. Bottle A and Bottle B, daily. The tablets are exactly look alike.
The medicines have to be taken exactly one tablet from each bottle, neither less nor more, else the medicines will not be effective.
One fine day, the patient popped out one tablet from Bottle A, but while taking the tablet from bottle B, by mistake, two tablets spilled over. Now he has three tablets in his hand, and he can't put back the extra tablet to Bottle B as all the tablets are identical in looks.
He has to ensure that he takes exactly one tablet from each of the bottle and at the same time he must avoid any wastage of the medicine.
Constraints:
1. Both bottles have equal number of tablets, say 30.
2. Tablets from both the bottles look exactly identical.
3. Medicine is very costly, so any kind of wastage is not affordable.
Problem Statement: How would you ensure that, in the above situation, you take exactly one tablet from each bottle, at the same time ensuring no wastage of the medicine.
Answer
- Take (1) Pill A from the bottle and add it to the 3 unknown pills. You now have (2) Pill A and (2) Pill B in your pile.
- Take each of the 4 pills and cut them in half.
- For each pill, put one of the halves in a pile on the right and one of the halves in a pile on the left.
- Each pile now contains 2 halves of Pill A and 2 halves of Pill B, which is the same as (1) Pill A and (1) Pill B in each pile.
Tic Tac Toi Who Will Wind the Game
Who will win in this game of Tic-Tac-Toe (Refer picture), and what was the last move played? No additional information is available apart from the picture.
Answer
see the game fron the begining.firsst chance will be of 0,which would be placed at [1,2].Now x would be placed at [1,1].Now the 0 would be placed at [3,1].At this point x has two choice ,either at position [2,1] or [3,3].consider x is placed at [3,3].but if it is placed at [3,3] then 0 would be placed at middle ie [2,2].but it is not shown in the fig.then we have 2nd option i.e [2,1].now 0 would be placed at [3,2],which is blocked by x at[3,3].now the last turn will be of 0 which iss placed at [2,2].hence 0 will win the match.
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