tag:blogger.com,1999:blog-269091222018-08-27T23:33:24.893-07:00Puzzles, puzzles and puzzlesAs the name of this site is, 'PuzzleGuru', it is infact not run by a Puzzle Guru at all. It is run by all of us for all of us to become puzzle gurus and increase our technical know-how and improve our knowledge in problem solving. So let's join our hands and try our heads together on this.Karthik Nagarajnoreply@blogger.comBlogger26125tag:blogger.com,1999:blog-26909122.post-1167117709875470852006-12-25T23:15:00.000-08:002006-12-25T23:29:21.336-08:00How many conversations?<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://www.computerhope.com/jargon/m/mesh.gif"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;width: 200px;" src="http://www.computerhope.com/jargon/m/mesh.gif" border="0" alt="" /></a>We have a network of 6 computers that have some pre-loaded software on them that help in conversing with other computers. A conversation means exchange of information between the 2 computers and thus a single conversation helps in both computers knowing what information is present on each of their disks. If at some point of time all the computers have something new to share with the others, then at the most how many conversations should occur between the computers until all of them have all the information? By conjecture how many such conversations should occur at the most for a 'n' connected computer network? Do not assume any relation to an actual computer network problem, hence don't apply your network protocol skills here. Its a pure combinatorics problem!<br /><br />Difficulty: MediumKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com20tag:blogger.com,1999:blog-26909122.post-1151707368086023812006-06-30T15:39:00.000-07:002006-12-25T20:05:42.646-08:00Partitioning<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/Quarter.gif"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/blogger/2981/626/200/Quarter.gif" alt="" border="0" /></a>You blindfolded and let into a room. The room has an infinitely many quarters scattered around on the floor. your friend tells you that that 20 of these quarters are tails and the rest are heads. He also says that if you can split the quarters into 2 piles where the number of tails is the same in both piles, then you win all of the quarters. You are allowed to move the quarters and to flip them over, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and you cannot tell by feeling it). How do you go about partitioning the quarters so that you can win all of them?<br /><br /><span style="font-size:85%;"><span style="font-style: italic;">Source: William Wu's puzzle forum</span></span>Karthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com20tag:blogger.com,1999:blog-26909122.post-1151619037632722642006-06-29T15:05:00.000-07:002006-06-30T15:16:49.960-07:00Male-Female balance<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/symbols.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/blogger/2981/626/200/symbols.jpg" alt="" border="0" /></a>In a certain land to increase the number of females so as the females can outnumber the males, a ruler, ordered the following: "As soon as a mother gave birth to her first son, she would be forbidden to have any more chilren." the ruler argued that some families can have more girls but no family would have more than one boy thereby creating a higher ratio of girls to boys. Now do you really think the ruler's strategy would work? Why and why not?Karthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com4tag:blogger.com,1999:blog-26909122.post-1150332077033507902006-06-14T17:34:00.000-07:002006-06-29T14:57:07.953-07:00Profit and Loss<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/bike.0.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/blogger/2981/626/400/bike.0.jpg" alt="" border="0" /></a>If I sold a bike to my friend for $100 and my friend sells it back to me for $80 and I sell it back to him for $90, how much total profit or loss have I made or incurred?Karthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com9tag:blogger.com,1999:blog-26909122.post-1149632454911149902006-06-06T15:02:00.000-07:002006-06-20T18:41:11.346-07:00The Evil Shylock<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/shylock.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/blogger/2981/626/200/shylock.jpg" alt="" border="0" /></a>There lived a farmer, his daughter, and a broker. The farmer for some reason borrowed a lot of money from the broker by pawning the only farm, the farmer had. The time given to the farmer to repay the money and get his farm released was over and the broker, the evil shylock, proposed the following: he would place a black stone and a white stone in a bag, and the farmer's daughter would need to pick out one of the stones from the bag in front of the entire town. If the daughter drew the white stone, the broker would return the farmer's farm and forgive him of the money. If she drew the black stone, the broker would marry the farmer's daughter and take the farm as well. The farmer had no choice but to agree. The farmer's daughter does not trust the broker and she knows that the broker might as well place 2 black stones in the bag. How can she get out of marrying the broker and save the farm for her father?Karthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com16tag:blogger.com,1999:blog-26909122.post-1149295275843287552006-06-02T17:34:00.000-07:002006-07-04T10:15:41.146-07:00What's the deal with 0.999999999999....<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/9.0.jpg"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://photos1.blogger.com/blogger/2981/626/200/9.0.jpg" alt="" border="0" /></a>Compare the numbers 0.99999... (infinitely many 9s) and 1. Which of the following statements is true? Why?<br /><br /> * 0.99999 ... <> 1Karthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com28tag:blogger.com,1999:blog-26909122.post-1148224021018417572006-05-21T08:04:00.000-07:002006-06-03T09:53:56.790-07:00Birthday matching<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/birthday.jpg"><img style="FLOAT: left; MARGIN: 0px 10px 10px 0px; CURSOR: hand" alt="" src="http://photos1.blogger.com/blogger/2981/626/200/birthday.jpg" border="0" /></a>At a movie theater, the manager announces that he would give a free ticket to the first person in line whose birthday is the same as someone who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday and that birthdays are distributed randomly throughout the year what position in line gives you the greatest chance of being the first duplicate birthday so you can get a free ticket?Karthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com3tag:blogger.com,1999:blog-26909122.post-1147817643625911642006-05-16T15:07:00.000-07:002006-06-30T16:55:11.276-07:00Crazy guy on the airplane<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/airplane.gif"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/blogger/2981/626/200/airplane.gif" alt="" border="0" /></a>A line of 1000 airline passengers is waiting to board a plane. They each hold a ticket to one of the 1000 seats on that flight. For convenience, let's say that the nth passenger in line has a ticket for the seat number n. Unfortunately, the first person in line is crazy, and will ignore the seat number on his ticket and picks a random seat to occupy. All of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random. What is the probability that the last (1000th) person to board the plane will sit in their proper seat (#1000)? Now what is the probability for the nth guy to sit in the nth seat number, if there an 'n' people waiting to board the plane?<br /><br />Level: Easy to Medium<br /><br /><span style="font-style: italic;">(Image credit: http://www.hasslefreeclipart.com/clipart_carttravel/page1.html)</span>Karthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com1tag:blogger.com,1999:blog-26909122.post-1147535849815270062006-05-13T08:48:00.000-07:002006-05-16T13:03:08.843-07:00000000000000000...............<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/factorial.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/factorial.jpg" border="0" alt="" /></a>How many trailing 0's does 100! have? Generalize it for any N! after you have computed for 100! <br /><br /><br />N! = Factorial N = N * N-1 * N-2 * ......2 * 1<br /><br />Level: Medium to HardKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com2tag:blogger.com,1999:blog-26909122.post-1147396157202618972006-05-11T18:06:00.000-07:002006-05-13T07:42:55.946-07:00What's so strange about me?<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/My-First-Numbers-Large.jpg"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/My-First-Numbers-Large.jpg" border="0" alt="" /></a>Pick a 3-digit number where the first and last digits differ by 2 or more...<br /><br /> * Consider the "reverse" number, obtained by reading it backwards.<br /> * Subtract the smaller of these two numbers from the larger one.<br /> * Add the result to its own reverse. <br /><br />What is the peculiarity you observe in the result? Can you explain why it is so? Is there some mathematical logic behind this?<br /><br />Level: EasyKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com2tag:blogger.com,1999:blog-26909122.post-1147136995102423302006-05-08T18:03:00.000-07:002006-05-21T07:16:23.080-07:00Playing quarters with the devil<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/devil.gif"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/blogger/2981/626/200/devil.gif" alt="" border="0" /></a>The devil has you captured and he agrees to leave you only if you win a game with him in playing quarters. The devil says "We both take turns in placing quarters on a table and the one who cannot place any more quarters looses". The devil also adds that there are no overlapping of the coins allowed. Now, what should your strategy be to win against the devil?<br /><br />Level: MediumKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com8tag:blogger.com,1999:blog-26909122.post-1146874397532709092006-05-05T14:58:00.000-07:002006-05-08T17:19:01.800-07:00Name the hidden card<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/cards.gif"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/cards.gif" border="0" alt="" /></a>Let's take on the deck of cards today and see where it takes us to: You and another intelligent friend of yours offer to perform this trick together in front of an audience. There is a deck of 52 cards in front of you. Assume the deck is neat with 13 of each type. The deck is shuffled and your friend asks the audience (one of them) to pick 5 cards. Your friend then takes those 5 cards and sees the cards that the audience has picked without showing it to you. Then your friend will pick one card from that 5 cards and hand it back to the audience. Then your friend neatly piles the remaining 4 cards in an order and places it on the table downwards and you start to pick them up. After you pick all the 4 cards, you are able to tell for sure what the 5th card with the audience is. How is that possible? Can you do the same experiment with 4 cards? Rule out all silly assumptions like "ESP between you and your friend, somehow your friend signals to you what it is, etc.," It is purely logical<br /><br />Level: HardKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com6tag:blogger.com,1999:blog-26909122.post-1146666991065162482006-05-03T07:09:00.000-07:002006-05-10T07:12:02.173-07:00Am I red, or Am I blue?<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/prisoners.png"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://photos1.blogger.com/blogger/2981/626/200/prisoners.png" alt="" border="0" /></a>There are 100 criminals in a jail and the jailor gets a change of heart and wants to release them. But he does not want to release them just like that, he wants to play a game with them. He announces that the next day, he will be putting hats, either blue or red, in equal randomness, on each of the prisoner's heads and all the prisoners will be lined up facing in one direction in a single file. So the last prisoner is able to see all the others' hats except of course his, the last but one will be able to see all the others except his and the last prisoner's, and so on...Basically they all can see what is in front of them. The first prisoner is unable to see any prisoner's hat as he is the first guy. The prisoners will be blindfolded until the jailor completes the task of putting the hats and then the blindfold will be removed. Once the blindfolds are removed, the jailor starts from the back and asks the color of the hat the prisoner is wearing. If the prisoner guesses the color of his own hat correctly, he survives, if not he is shot. To make things spicy, lets say the jailor allows others only to hear prisoner say the color of his hat and not whether he was correct or wrong. The prisoners are dismissed for the night. They discuss a strategy and come up with an optimal solution of saving lots of their fellow buddies. To top it all off, the jailor has a microphone bug inside the cell and he is able to hear their plan. So what is the strategy of the prisoners and how do they make sure lots of their folks are saved in the cruel game played by the jailor?<br /><br />Once you solve the problem, lets try to get a generic methodology for a 'N' prisoner and 'K' different colors of the hats put in random. How many prisoners can be saved at a maximum and how many will be killed at a minimum?<br /><br />Level: Hard<br /><br /><span style="font-style: italic;">(Image courtesy: www.londonstimes.us/toons/index_misc.html, I have put in the red and blue hats for those prisoners)</span>Karthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com3tag:blogger.com,1999:blog-26909122.post-1146669226830599812006-05-02T08:11:00.000-07:002006-05-13T08:43:50.446-07:00Hands of a clock<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/clock.jpg"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/clock.jpg" border="0" alt="" /></a>How many times in a day do the hands of a clock overlap? Can you list out the times too? Explain your methodology and not just list the answer.<br /><br />Level: MediumKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com7tag:blogger.com,1999:blog-26909122.post-1146530906345646832006-05-01T17:40:00.000-07:002006-05-03T10:34:40.400-07:00What's our average salary?<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/dr-160.gif"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/dr-160.gif" border="0" alt="" /></a>There are 3 colleagues who want to know what their average salary is, but they do not want to disclose to each other their individual salary. How do they go about doing it? In the process of doing it what is the worst that each could learn about the salaries of the other two? Is this method fail-proof? Think about it! Would you be willing to adopt your methodology with your colleagues and find the average salary?<br /><br />Level: MediumKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com10tag:blogger.com,1999:blog-26909122.post-1146441804106907172006-04-30T16:31:00.000-07:002006-05-03T07:01:08.200-07:00Survivor of the pirates<a href="http://photos1.blogger.com/blogger/2981/626/1600/pirates.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/pirates.jpg" border="0" alt="" /></a>There are 5 pirates on an island. They have a big box of treasure stashed (1000 diamonds) on the island and now they must distribute the treasure amongst themselves. They are extremely greedy pirates and they do not agree for an even split as each one wants to have more treasure. So they play a game - SURVIVOR! The senior most pirate proposes a methodology to split the treasure and everyone votes. If the senior most pirate can get at least 1/2 votes for his porposal, then they split the coins that way. If the senior most pirate cannot get majority for his proposal then, they kill the senior most pirate and start over with the next senior most pirate. The process continues until one proposal by one surviving senior pirate is accepted. So how does this game amongst the pirates proceed? Who gets to keep the stash and how much? What would be an intelligent proposal by "a senior" pirate? Can you then tell how this will work for an 'n' pirate scheme? Assume that the pirates are all very mathematical and logical for the sake of this question.<br /><br />Level: MediumKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com5tag:blogger.com,1999:blog-26909122.post-1146412549964550682006-04-30T08:46:00.000-07:002006-04-30T17:10:00.636-07:00Village 'Adultery'<a href="http://photos1.blogger.com/blogger/2981/626/1600/village.gif"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/village.gif" border="0" alt="" /></a>There is a village of 50 couples. Every man in the village has been unfaithful to his wife. Every woman will know when a man other than her husband has committed adultery. The rule of the village - if a woman knows her husband has been unfaithful, then she must kill him that very night itself. Life goes on in Village Adultery and one day, the queen, who always tells the truth, announces that at least one husband in the village has committed adultery. Does this change anything in the village? You would need to explain clearly what happens if anything at all changes in Village Adultery.<br /><br />Level: MediumKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com3tag:blogger.com,1999:blog-26909122.post-1146334439542868642006-04-29T11:06:00.000-07:002006-04-30T07:52:12.063-07:00The drunken jail guard<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/Jail.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/Jail.jpg" border="0" alt="" /></a>There is a circular jail with 100 cells numbered from 1 to 100 consectively. Each cell has an inmate and all the doors are locked to begin with. One night the jailor gets drunk and starts running around the jail in circles. In his first round he opens every door. In his second round he visits every 2nd door (2,4,6,...) and shuts the door. In the 3rd round he visits every 3rd door (3,6,9,...) and if the door is shut he opens it, if it is open he shuts it. This process of crazy drunken-ness continues for 100 rounds and finally the jailor gets exhausted and falls down. Let us assume the prisoners are all kind enough to wait until the crazy jailor stops and finally they decide to escape. Now how many prisoners find their doors open and can you also tell which cells' these prisoners belong to? People who post the answer should also post the logic behind their answers!<br /><br />Level: MediumKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com2tag:blogger.com,1999:blog-26909122.post-1146248631564692312006-04-28T11:15:00.000-07:002006-04-29T10:53:18.823-07:00The crazy juror<a href="http://photos1.blogger.com/blogger/2981/626/1600/juror.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/320/juror.jpg" border="0" alt="" /></a>We have a 3-person jury. I know jury is not a 3-person team, but lets just stick to our problem for now. In this 3-person jury, there are 2 jurors who are very serious. However there is a crazy juror, the third juror, who likes to flip a coin and based on the outcome agree with either juror's decision. Now, each of the serious jurors can independently give the correct decision with a probability of 'p'. You are called to inspect this jury and make suggestions. You wonder if a single-person-jury with a correct-decision making probability of 'p' is better off than this 3-person jury with the crazy flippant juror. What is your observation and what do you say to the judge about the 3-person jury?<br /><br />Level: EasyKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com4tag:blogger.com,1999:blog-26909122.post-1146161679956408012006-04-27T11:06:00.000-07:002006-04-28T08:08:44.983-07:00Second best as runner up<a href="http://photos1.blogger.com/blogger/2981/626/1600/tennis.1.gif"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/tennis.1.png" border="0" alt="" /></a>Let us take a little turn today - Probability! Lets try a question on chance in a tennis match. There are 8 players for a tennis tournament. They are to play in pairs with one other player in the first round and the winners play in the next round and so on until we have a winner and a runner up. There is a player amongst these 8 who is the best and the odds of him winning the tournament is very high. There is another player who is considered the second best. The players draw numbers to take on their counter players for the first round. What is the chance (probability) that the second best player can be the runner up when the tournament is over? Once you get the probability, work a generic mechanism to find the chance for a 'm' player tournament. Substitute m = 8 in that equation and see if you get the same answer as you worked out.<br /><br />Level: Easy to MediumKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com10tag:blogger.com,1999:blog-26909122.post-1146089884882906912006-04-26T15:09:00.000-07:002006-05-06T06:35:12.746-07:00Bulb dropping<a href="http://photos1.blogger.com/blogger/2981/626/1600/bulb_1.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/bulb_1.jpg" border="0" alt="" /></a>You need to figure out how high a bulb can fall from a 1000 story building before it breaks. What is the minimum # of bulbs you will ever need for this and what is the largest number of drops you would ever have to do to find the right floor where the bulb would break? Lets say you are given an unlimited supply of the bulbs. But figure out how much you will actually ever need and how many drops you will ever have to make to figure out the right floor. Lets take this a little further - For a 's' story building and 'b' # of bulbs, find the optimal strategy that will minimize (worst case scenario) the number of drops required to determine the floor on which the bulb will break.<br /><br />Level: Medium to HardKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com25tag:blogger.com,1999:blog-26909122.post-1146084974971963722006-04-26T13:14:00.000-07:002006-04-27T08:14:33.896-07:009 balls yet again...<a href="http://photos1.blogger.com/blogger/2981/626/1600/9ball.gif"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/9ball.gif" border="0" alt="" /></a><a href="http://puzzleguru.blogspot.com/2006/04/9-balls-and-only-2-weighings.html">Ok lets revisit the 9 ball question again</a> - only this time lets make it a tricky - we do not know if the faulty ball is heavier or lighter. So how many weighings are required to find out the faulty one (minimum # of weighings) and, and can you find out if the faulty one is lighter or heavier?<br /><br />Level: Medium to HardKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com6tag:blogger.com,1999:blog-26909122.post-1146013634967943622006-04-26T00:00:00.000-07:002006-04-26T13:13:26.163-07:00Water and Milk<a href="http://photos1.blogger.com/blogger/2981/626/1600/mwater.0.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/mwater.0.jpg" border="0" alt="" /></a>You have two containers. One contains 1 gallon of milk and the other contains a gallon of water. If you were to take one cup of milk from the first one and pour it to the second one and after mixing, you were to pour one cup of the mixture from the second to the first one, which containers now have a pure concentration of their original contents?<br /><br />Level: MediumKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com4tag:blogger.com,1999:blog-26909122.post-1145991940531726342006-04-25T12:04:00.000-07:002006-05-02T16:56:08.426-07:00The bad king's wine cellar<a href="http://photos1.blogger.com/blogger/2981/626/1600/king.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/king.jpg" border="0" alt="" /></a>A bad king has 1000 bottles of wine. A neighboring king plots to kill the bad king, and sends a servant to poison the wine. the servant is able to poison only one of them before he is caught. The guards don't know which bottle was poisoned, but they do know that the poison is so potent that even if it was diluted 1,000,000 times, it would still be fatal. Furthermore, the effects of the poison take one month to surface. The king decides he will get some of his prisoners in his vast dungeons to drink the wine. At a minimum how many servants does he need to kill to find out the poisioned bottle of wine and will still be able to drink the rest of the wine in 5 weeks time? As an added note, he does not want to bring in 1000 of his servants as he knows there are a lot lesser number of prisoners he can bring in to begin with.<br /><br />Level: HardKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com31tag:blogger.com,1999:blog-26909122.post-1145984003723060442006-04-25T09:49:00.000-07:002006-04-25T18:02:55.903-07:009 balls and only 2 weighings<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://photos1.blogger.com/blogger/2981/626/1600/balls.jpg"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;" src="http://photos1.blogger.com/blogger/2981/626/200/balls.jpg" border="0" alt="" /></a>You are given 9 balls and they look exactly the same in color and physical appearance. You are also given a scale. To make things easy you are given that one of the balls amongst these 9 balls is heavy. With ONLY 2 weighings how can you find out which ball it is.<br /><br />Level: EasyKarthik Nagarajhttps://plus.google.com/104972713938717882861noreply@blogger.com3