# Question Interview Questions

interview questions shared by candidates

## Question Interview Questions

How would you address a circumstance in which you disagree with your supervisor's decision or actions in a particular instance? 1 AnswerI responded that my responsibility is to provide my supervisor with sufficient info for her to make a good decision and ultimately I have to trust she is going to do just that, she won't ask me to do anything unlawful, and as she has the overall responsibility, I will need to accept her decision and move on with business to support her decision. |

### Loss Prevention Agent at Nordstrom was asked...

Has there ever been a time where you witnessed a fellow employee find money on the ground and kept it for themselves rather than turning it in to lost and found or human resource? If so, please explain how you would approach the employee and ask them about the money. 1 AnswerI said I have never witnessed a fellow work associate find money on the floor or ground and kept it to themselves. |

Tell me about your favorite campaign. How would you approach a new product that we would offer? What would be the first step you would take? 1 AnswerHave a solid grasp on your favorite advertising/marketing campaigns. Have a good knowledge of their product before walking into an interview. |

### Software Developer Position at RockYou was asked...

Given 5 pirates on a ship, they need to distribute a pot of gold that has 100 gold pieces inside of it. The first pirate must make a proposal of how the gold will be distributed. If he receives over 50% votes from the remaining pirates, then his proposal will be accepted and the gold will be distributed. If he receives less then 50% support, then he will be thrown off the ship and die. 42 AnswersThe answer is for the first pirate to offer the 3rd and 5th pirates 1 gold each, and then take the remaining gold for himself. Otherwise the 3rd and 5th will probably not receive any gold. this is an extremely subjective question. first the nature of the pirates must be examined. if this is truly a team, then of course each pirate should receive 20 pieces of gold. however, if there is a division of labor that affords certain pirates greater responsibility then we are looking at a vertically stratified solution. in this case those at the top, the CEO's of the pirates, must be given more of the booty. if there is no stratification, but the nature of the pirates is individualistic and greedy, i would assume divide the gold into pirates 2 and 4 get 25 + you get 20 + pirates 3 and 5 each get 15. The first answer is right. You need to think that the pirates are rational and greedy for the most they can get. Then you need to consider what would happen with two pirates and take it from there. Show More Responses i disagree with the first answer completely. his proposal needs to have over 50% votes by the remaining pirates, not including himself. this means 3 out of 4 votes. if he proposes taking 98 pieces and the 3rd and 5th take 1 each, there's probably no way any of the 4 will vote for that proposal. since he needs 3 out of 4 votes excluding himself, that one nay vote can get snubbed altogether. assuming that the pirates want to maximize as much as possible, they will not mind the fifth pirate not receiving ANY gold. i suggest that the first pirate make one of the following proposals: - first pirate receives 1 gold so that the other 3 equally receive 33 gold pieces. the nay vote pirate will get 0. this carries the least risk of him getting kicked off. - first pirate receives 25 pieces as well as three other pirates. the nay vote pirate will get 0. the 3 other pirates voting for the proposal might view this as the most equitable solution among the four with one guy losing out. i disagree with the first answer completely. his proposal needs to have over 50% votes by the remaining pirates, not including himself. this means 3 out of 4 votes. if he proposes taking 98 pieces and the 3rd and 5th take 1 each, there's probably no way any of the 4 will vote for that proposal. since he needs 3 out of 4 votes excluding himself, that one nay vote can get snubbed altogether. assuming that the pirates want to maximize as much as possible, they will not mind the fifth pirate not receiving ANY gold. i suggest that the first pirate make one of the following proposals: - first pirate receives 1 gold so that the other 3 equally receive 33 gold pieces. the nay vote pirate will get 0. this carries the least risk of him getting kicked off. - first pirate receives 25 pieces as well as three other pirates. the nay vote pirate will get 0. the 3 other pirates voting for the proposal might view this as the most equitable solution among the four with one guy losing out. i disagree with the first answer completely. his proposal needs to have over 50% votes by the remaining pirates, not including himself. this means 3 out of 4 votes. if he proposes taking 98 pieces and the 3rd and 5th take 1 each, there's probably no way any of the 4 will vote for that proposal. since he needs 3 out of 4 votes excluding himself, that one nay vote can get snubbed altogether. assuming that the pirates want to maximize as much as possible, they will not mind the fifth pirate not receiving ANY gold. i suggest that the first pirate make one of the following proposals: - first pirate receives 1 gold so that the other 3 equally receive 33 gold pieces. the nay vote pirate will get 0. this carries the least risk of him getting kicked off. - first pirate receives 25 pieces as well as three other pirates. the nay vote pirate will get 0. the 3 other pirates voting for the proposal might view this as the most equitable solution among the four with one guy losing out. He needs to shoot the ring leader(dominant) of the remaining 4 pirates and re-negotiate after reloading. There isn't an optimal solution for the first pirate. The first pirate can't come up with a solution that will give 2 other pirates a maximum of the gold. The other pirates will always vote no unless the first pirate gives one of them all 100 pieces, since it's always in their best interests to dump the guy overboard and split it amongst less people. It's the wrong question. The question correctly stated is as follows: (credit to vorg.ca) Five buccaneers have obtained 100 doubloons and have to divide up the loot. The buccaneers are all extremely intelligent, treacherous and selfish (especially the captain). The captain always proposes a distribution of the loot. All buccaneers vote on the proposal, and if half the crew or more go "Aye", the loot is divided as proposed, as no buccaneer would be willing to take on the captain without superior force on their side. If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all buccaneers will turn against him and make him walk the plank. The buccaneers start over again with the next senior buccaneer as captain. What is the maximum number of doubloons the captain can keep without risking his life? 20 pieces. The captain should divide the pie equally 5 ways (20 pieces). Then explain to the four pirates that anyone who does not agree to this fair deal of equal distribution will receive nothing, and that person's share will be retained by the captain alone. No one would want the captain to get their share. Everyone would think the deal fair. And therefore the captain's life would not be at risk at all. Be bold, be treacherous, be past tense: I took Pirate 2 and Pirate 3 aside. I told them, "We will split the gold evenly, three ways, between us. We three can overpower Pirate 4 and Pirate 5, so they do not receive any gold." Then I took Pirate 4 and Pirate 5 aside. I told them, "We will split the gold evenly, three ways, between us. We three can overpower Pirate 2 and Pirate 3, so they do not receive any gold." The vote was 100%, and I got away with all the gold, since each of the other pirates killed each other for being doublecrossed. Since I'm telling the story, it's clear I survived. Splitting the gold amongst 3 of the 5 pirates (1st pirate included) gets 50% of the remain pirate vote. In this way, the pirates get the most gold and there is a majority to vote against the first pirates death. Navy Seals kill all the pirates and return the gold to the rightful owner. Show More Responses The leader can keep 98 of the gold coins and give 1 to the 2nd ranked and 1 to the 4th ranked pirates. Let's assume 1st that they are so selfish that they care more about the money than each other's lives, and 2nd, that they follow this 50% rule without question. The lowest ranking pirate, number 5, has absolutely no reason to support any other pirates, because he knows that if all the other pirates walk the plank he gets 100 pieces. Pirate number 4 knows that if he's left with pirate 5, pirate 5 will demand all the booty or vote against the proposal to kill 4 and receive it anyway. Pirate 3 recognizes pirate 4's predicament and would only have to offer him 1 piece of gold so that 4 would come out better than being stuck with making the decision and losing everything. His support would provide 3 with a majority. If it came down to pirate 3's decision, he would have 99 pieces, 4 would have 1 and 5 would have 0. If the second ranked pirate, pirate 2, was faced with negotiating, he would need the support of 2 additional constituents. 5 will always vote against him, and 3 recognizes that he can make 99 pieces of booty if he's in control. In order to live, pirate 2 would have to offer pirate 3 99 pieces and pirate 4 1 piece, leaving him with nothing. Pirate number 1 similarly recognizes the predicament of pirate 2. Pirate 1 only needs two other supporters. That's easy. Better that pirate 2 receives his life and even one piece of gold than nothing. Pirate one offers him 1 piece of gold. Similarly, pirate 4 knows that he can get 1 piece of gold and his life, so pirate 1 offers him 1 piece of gold and has his support. After giving pirate 2 and pirate 4 one piece of gold each, he has their support and can take the remaining 98 pieces for himself. QED! Yeah baby! The rationale set forth by Marketing and Information Systems Major is well-explained and thought out, but with one fatal error (i.e. my user name). The interviewee is correct. Suppose that pirates are named by letters, with the most junior pirate in each scenario being Pirate A. I'll list the pirates in order of seniority for each case: In a two-pirate case (Captain, Pirate A), the captain always keeps the booty. Pirate A has no leverage whatsoever, since he'll never have superior force to remove the captain. In a three-pirate case (Captain, Pirate B, Pirate A), the captain only needs one vote. Pirate B wants to get rid of the Captain so he can be in the advantageous two-pirate case. Pirate A wants to AVOID that, so the Captain easily buys Pirate A's vote with one gold coin -- the status quo is maintained, and Pirate B gets shafted. In a four-pirate case (Captain, Pirate C, Pirate B, Pirate A), the captain still only needs one vote (a tie means no superior force, which means the status quo is maintained and the captain wins). We pose our standard question: "If the captain loses the vote, who gets shafted?" -- well, if the captain loses, it's a three-pirate case, which means Pirate B gets shafted. So the captain easily buys Pirate B's vote with one gold coin, and the status quo is maintained. Result: Pirates A and C got shafted. In the desired five-pirate case (captain, Pirate D, Pirate C, Pirate B, Pirate A), the captain needs TWO votes to avoid losing power. So we ask ourselves: "if the captain loses the vote, who gets shafted?" -- Pirates A and C (as determined above, in the four-pirate case). Therefore, to maintain the status quo, the captain proposes one gold coin to Pirates A and C each, and keeps 98% of the booty for himself. The downfall of these pirates is their inability to think laterally or deeply e.g. it's a machine question, not a human question. Subjective variables such as selflessness and morality would make these pirates much less predictable, forcing the captain to appease each and every one of them to the best of his ability, rather than knowing exactly whose vote to buy and at what price. by "deeply" i meant creatively, and by "e.g." i meant "i.e."!! edits would be awesome. The answer is 25. If they split it 5 ways, then its 20 each. If he needs 3 votes, offer each of them 25 and nothing to the last guy, then they will approve, since 25 is more than 20 (even split). Given that the first pirate needs the majority of the remaining vote (3/4) pirates. Then, First Pirate - 25g Crew Member 2 - 25g Crew Member 3 - 25g Crew Member 4 - 25g Crew Member 5 - 0g My logic behind this is simple. In order for their greed not to be outweighed by their survival the 4 pirates can share the booty equally with the 5th pirate as an example of the position they would find themselves in. 5th Pirate gets thrown of the ship and dies. It could stop here or they could revote, If they revote then, First Pirate - 33.3g Second Pirate - 33.3g Third Pirate - 33.3g Fourth Pirate - 0g Rinse and repeat until two pirates left sharing the loot 50g each. Wait, read the clue carefully. Let's say I'm the captain, and my pirates are 'greedy'. Option #1: I give myself and 4 pirates 20 gold each. That's good, right? Wrong. The pirates are 'greedy', and the 4 pirates will STILL vote to throw me overboard, because that would increase their 'take' to 25 Gold each, since I no longer get any. In order for me to WIN, I need to provide them with MORE than 25 Gold each, because that's what they will get for voting me off. Therefore: I keep 22 Gold, and I give pirates 1,2,3 each 26 GOLD. That means, if they vote me off, they will only get 25 Gold, and 26 is more than 25. Thus, the maximum I can keep is 22 Gold. BUT, if it only takes 2 Votes (i.e, my vote counts), then I keep 34 Gold, and give pirate 2 and 3 33 Gold each. Thus, my vote + 2 pirates >50% of the votes. Pirates 3,4 get zero. if they're going to kill you if you can't get 2 guys to back you up, then they can just as easily all four decide to kill you and then the remaining four are in the same quandry, but with only four shares. keep recursing until there is only one guy with all the gold. your only chance is to give them all the gold in the hope that they will deign to let you live, then kill them while they sleep and take it back. This problem really has no solution to it - as it clearly says the one who is proposing distribution doesn't get to vote! figure this: the lead pirate proposes a solution (98 for himself, 1/1/0/0 for others; or any other option) - but he doesn't get to vote. The others can always call this an unfair division and throw the helpless pirate out - as he doesn't have any say. Even if they decide to divide it as 20 coins each, the others can still throw the first pirate and maximise their share by 25% each - this can go on until the youngest/last pirate walks out with all the money. therefore this is an impractical situation. However, the whole situation changes if you allow the pirate proposing the distribution to vote!!! Then the solution, as proposed by some one above, of a distribution being 98/1/1/0/0 will eventually be accepted. Search for the pirate game on Wikipedia, the question is a straight lift of that puzzle - except that someone decided to change the problem by making the lead pirate non-voting.....and that will always make this problem non-solvable. PS: @ mktg and Systems major: Baby, you forgot to read the problem correctly and just googled it up!!! OK, so within the parameters as outlined, and assuming that this problem is recursive (after the first pirate doesn't get enough votes, he's killed, and they're faced with a 4-pirate situation), let's look at it thusly: They're all greedy, and they all know that the best way is to kill everyone else and grab all of the loot. However, to do this, they _must_ be the "lowest pirate" on the pole (the last one to be given the chance of dividing the loot). If this "pecking order" is fixed (everyone knows exactly how they stand), it is different from if it is random. Let's look at both. 5-Pirates, fixed order: In this case, then the proposal outlined by Gus and the others that results in Pirate 5 getting 98 coins, and 4 and 2 getting 1 coin each works out best. 5-Pirates, random order: This is different, as the problem then becomes how the order is established after each pirate gets killed. Everybody wants to be the Last Pirate Standing (tm) (Fox, call me for show marketing rights), but as the first one gets killed, there is increasing odds that they will be chosen next, and not be the LPS. Unfortunately, in the question as posted (5 nameless, rank-less pirates) we don't know how pirate 5 got chosen to choose, and we don't know how pirate 4 will be chosen after pirate 5's demise. In this case, Pirate 5's best play would be to give two pirates 34 gold (one more than 33, and him take 32) if only two votes are needed, or give three pirates 26 gold (one more than 25, and him take 22) if a majority vote is needed, and hope that they would see this as fitting, instead of being the next sap in the position of making a 4-way split or his life. Show More Responses As stated the answer is: 25 gold pieces should go to any three pirates, with the leader keeping the remaining 25 for himself. Note - Gus' solution is incorrect because he assumes that the leader can vote on his own proposal. The leader can't. Also - as some others have pointed out, this probably isn't the right question. The question with a "succession plan" has a more interesting answer. Here is the detailed reasoning: So - I start by assuming that for all pirates, its infinitely more important to be alive than to get any gold, but if they are alive, they want as much gold as possible. The pirates, other than the leader are "unorganized", and all things being equal, they would rather have as many pirates around as possible. These are risk adverse pirates :) Assume there was one pirate: the best proposal - take all the gold for himself, which would definitely pass. If there were two pirates, unless the leader proposes to give all of the gold to other pirate, his companion will vote against it, and take all of the gold. If there are three pirates: if the other two pirates have a 50% of getting nothing and a 50% chance of getting 100 gold. So the expected value for both them is 50 gold pieces. He needs two votes to stay alive, so offering either of them less than 50 will assure losing a vote, and thus his life. Therefore, the leader proposes 50 gold to each of the other pirates, and stays alive. If there are four pirates: The remaining pirates have a 33% chance of getting nothing, and a 66% chance of getting 50 gold pieces if they reject the leader's offer. Their expected value is 33 gold pieces each. Offering any two of them 33 gold pieces, and with the leader keeping 34 is a proposal that should pass. If there are five pirates: for the four remaining pirates, they have a 25% chance of getting 34 gold pieces, a 50% chance of getting 33, and a 25% chance of getting nothing if they reject the leader's proposal. Their expected value is: 25 gold pieces. Since three votes are needed, 25 gold pieces should go to any three pirates, with the leader keeping the remaining 25 for himself. Are you joking? Pirates aren't fair! One pirate cheats all the others and splits with all the booty AND your women... They're pirates for chrissake! If I were the 1st pirate, I'd consider that goal #1 is to stay alive - my life is worth more than all the gold. So I'd propose that each of the other four pirates would receive 21 pieces of gold and I'd keep 16... and my life. Odds are that the other pirates would find this acceptable, and they may even keep me around to make the first proposal again next time. It's a win-win. We all keep our lives and get more gold than we had before. I can't speak for other pirates, but that's what I'd do. The proposal has to be accepted by two of the four other pirates. The only way the first pirate can do this is interview each of the other pirates individually and ask what they think is fair. His objective is to find two that are similar enough that he can modify the two into one plan. After mofifying the two plans into one plan he then goes back and asks these two individually again if the modified plan is acceptable. If he can get these two to accept the modified plan then he presents it. It doesn't matter what the other two pirates who were left out think because he has the two votes he needs. There's a pot of gold no one's seeing. The end game is that the last two will share 50 gold pieces each (no majority to mutiny). So pick two friends, offer them the 100 gold pieces between them, and hope they give you some afterward (but don't bet on it). Keep the gold pot and the friends. You lose two votes from those disenfranchised, but you and two friends hold majority might. Here's my thought about this. 100 gold to be distributed to 5 pirates The first pirate proposes how it will be distributed. He has to get 3 out of 4 votes in order to stay alive, otherwise he will be thrown out of ship and die. Let's say pirates are in nature greedy. Second/third/fourth/fifth pirate will rather kill the first pirate to accumulate more share of gold. If the first pirate proposes anything that doesn't let him get even a gold, his proposal might be accepted and he will remain alive. Ofcourse it might not be what will happen. The other pirates kill the first pirate and they quarrel over the pot of gold. The last-man-standing gets everything. Or they all die and no-one gets anything. The answer is 97. Start with three pirates. I was once faced with this very same dillemma...I ended up splitting the gold between three of the other pirates, which got me the vote I needed. Yargh, but then me and me accomplice, the man without any gold, we shivved all the rest and threw the bodies overboard, each claiming 50 gold, yar! Attach the gold to a life preserver and throw it overboard . If they are so greedy they will all dive after it. You sail away with the ship and sell it for much more than 100 pieces of gold. First, you must realize this is not just a logic question with one right answer, but a way for an employer to probe you thinking process as you talk out loud. So if you make stupid assumptions like “I’m the captain so they will let me have more” or “I will just kill them later” or “I would strike a side deal”, then you will be looked at as a dodger or lazy or scared to address the question, so that crap is out in an interview. You might get points with this one good assumption: “It looks like the question is incomplete, I assume that the voting rounds continue until an offer is accepted or only one pirate is left.” Facts: in round 1 the offeror needs 3 of 4 votes for a majority. In round 2 he needs 2 of 3, in round 3 he needs 2 of 2 and in round 4 he needs 1 of 1. (So now you are showing the interviewer you are thinking ahead before rushing off trying to give the answer). So the offeror in the first round wants to maximize his gold and his chance of living. So he makes this offer. 33 Gold to each of 3 voters and 1 for himself. He explains why this is the best they can hope for as such. In the final round, the 1 voter either gets 100 pieces of declines the offer, kills the pirate and takes 100 anyway. He can even legally decline the offer of 100, kill the pirate and still take 100 pieces. Now you may think this is good, but the odds of one of the 4 of you being alive and being the voter is 1 in 4 or 25% (see mister interviewer, I can to simple math). So you have a 25% chance of 100 pieces, or 25 on average, so 33 and guaranteed life now is a better deal, in fact you should be happy if I offered you 26. But I want to take just one, so in the next round, the offeror will have to take 1 or less or face certain death, because he saw me die for just 1. In the next round, that means the 2 winning voters will want 50 pieces each. But they only have a 50% chance of being 2 of the 4 people being offered gold. That means they have a 50% chance of getting 50 pieces, on average, they get 25, again less than 33. If they want to try to make it to round 3, there will be 1 offeror and he will have to get both votes for a majority to live, so again he will have to be willing to give 50 to each. From round 1, the 4 have a 50% chance of being a voter now and a 50% chance of being dead or getting 0 and living. If they get 50 pieces, then 50% of that is 25. Still less than 33. Also, the offeror can argue that if they don’t accept his offer, then one of them will die and/or get 0 in next round, the other will get 100, but at this point on average would get only 50. So he might be willing to offer each much less then 50 to guarantee they live, so now they are worse off. In no average case are they better off than taking your 26-33 piece offer and the certainty of life. (Now, if I am the first offeror, I will probably max out their offer at 33 so I maximize my chance to live and fight another day). Or you just pull out your light saber and kill the other 4. Show More Responses You need to look at the question from the end first. As each pirate is thrown off the ship, there are fewer pirates to split the gold up with. At the end there will be 1 pirate who gets all 100 pieces of gold. Pirate 5 1 pirate 100 With two pirates, the situation looks like this: Pirate 4 Pirate 5 2 pirates 0 100 1 pirate 100 Pirate 5 has no reason to accept any offer other than all the gold, since if he rejects Pirate 4’s offer, he will get all the gold, anyway. With three pirates: Pirate 3 Pirate 4 Pirate 5 3 pirates 99 1 0 2 pirates 0 100 1 pirate 100 With 1 or 2 pirates, Pirate 5 has all the power, but with three pirates, 1 piece of gold for Pirate 4 is better than any of the other conditions. With four pirates: Pirate 2 Pirate 3 Pirate 4 Pirate 5 4 pirates 97 0 2 1 3 pirates 99 1 0 2 pirates 0 100 1 pirate 100 With four pirates, Pirate 4 and 5 get a better deal than with 3 pirates. With five pirates: Pirate 1 Pirate 2 Pirate Pirate 4 Pirate 5 5 pirates 97 0 1 0 2 4 pirates 97 0 2 1 3 pirates 99 1 0 2 pirates 0 100 1 pirate 100 Pirate 3 and 5 get a better deal than with 4 pirates. I have just a few things to add that seem logical in order to know how to determine the answer. The most important thing to know is whether they are distributing the gold while at sea or near land (or in port). If they are at sea there is a far greater likelihood that the captain would survive regardless of the distribution, since it takes several people to man a ship. If they are near land, the answer can only be determined by knowing the personalities of the crew and how much fear or respect the captain commands. The more fearsome the captain is, the less likelihood of a mutiny regardless of the division as long as each of the subordinates received an equitable portion and the captains share was not overly large. The idea of 'friends' among pirates is ludicrous- pirates by nature form alliances merely for numbers, and are by defination treacherous. The idea of trying to solve the problem mathmatically is interesting but unsubstantiated as the outcome of gold division vs staying alive would invariably be determined by the variable factors I outlined above. By solving the problem through manipulating the pirates desire to be the only one left alive, there is an underlying assumption that only one pirate is necessary to sail a large vessal at sea, which is not the case, and even pirates know that. you find the solution by starting from the opposite end. Start with the scenario where only pirate D and E remains alive. Then D will keep everything to himself. Knowing this (if C was alive) he would offer just one gold coin to E (since he knows what happens if they choose to kill C - then E gets nothing). Continue to the point where A is still alive and you'll find: A=98, C:1, E:1 The question should also be that: if the top ranked pirate dies, then the whole process is repeated for the 4 remaining pirates, and so on: if the 2nd ranked pirate dies, it is repeated for the 3 remaining pirates... and so forth... And the answer should be that if all pirates are strictly purely logically and think for himself only, then the top pirate will get 98 coins. This question seems to suggest 2 things: 1) Since the top ranked pirate can get the most gold coins, it justifies why the top person in a company get most of the money and the fraction of the fraction is left for the people underneath. 2) Since it assumes that people underneath will not team up, maybe that's why in some companies, it is the benefit of the top for the people underneath not to be in good relationship, so that it mimics the situation above, so that the top person can get most of the gold coins. also note that if makes a difference whether if 50% of the vote is established, whether the top pirate die or whether the proposal is accepted. > 2) Since it assumes that people underneath will not team up, maybe that's why in some companies, it is the benefit of the top for the people underneath not to be in good relationship, so that it mimics the situation above, so that the top person can get most of the gold coins. so those companies will actually want to foster a bad relationship between people underneath. And will that company want to hire "good guys" for the company? No, because the good people are good with each other and can team up. If there are many many bad guys, then they will be all greedy and hate each other and think for themselves only and not team up. Therefore, the company may want to hire bad guys instead of good guys. this is a trick question. if the point is to guarantee that he stay alive, he needs to make 3 pirates as happy as possible. the only way to do that is to split the entire share evenly between them. this means excluding one pirate AND HIMSELF. "Anyone who's with me and votes for my proposal gets an equal share of gold. Vote against me, and you get nothing while the rest of us gets an even more equal share of the booty." Assume you are the pirate who distribute the gold, you just put the gold to a heap one by one and let the rest pirate to decide when to stop picking, the first person who says 'stop' will get the whole golds in this heap and the others must distribute the other golds. |

### Trader at Morgan Stanley was asked...

If two cars are traveling in a two lap race on a track of any length, one going 60 mph and the other going 30mph, how fast will the slower car have to go to finish at the same car to finish at the same time? 30 AnswersIt's impossible, the faster car will be done the race by the time the slower car finishes the first lap. the answer to this question lies on how long the race track, we can solve its mph if we know how long the track we be. Well, this is interesting because there are no track details and makes for multiple answers through ambiguity and assumptions. i.e. One could assume that it is a circular track and that the two lanes are very wide and that one car is on the outermost furthest from the centre and the other is on the track very near the centre. The circumference of each track therefore could be such that the faster car would have to travel twice the distance that the slower car has to and therefore the two cars would arrive at exactly the same time. The is why cares on a racetrack must start at offsets to each other or have their times corrected in some other way! In real-life, this is highly unlikely however it does demonstrate my point. Show More Responses I agree with the first answer (by the Interview Candidate). When the slow car completes the first lap, the fast will complete the second lap. It does not matter how fast the slow car goes on the second lap; it cannot win... 90 mph Wouldn't the slow car just need to go 60mph? It doesn't say that the fast car is going double the slow cars speed only that the slow car is going 30 mph and the fast is going 60mph. The question is a trick. It says how fast will the slower car have to go to finish "AT THE SAME CAR" to finish at the same time? It can go any speed!! It will always finish at the same car (2nd) at the same time. The car isn't changing!!! I'm assuming that the question, as typed, was entered incorrectly and that it should be worded, "How fast will the slower car have to go to finish at the same time as the faster car?" The answer is 30mph. Because that's how fast the slower car is going. Nowhere in the question does it state that the cars are at the same point on the track. The slower car is currently halfway between the faster car and the end of the race. The two pieces of missing info are: 1. How long is the distance of the track and 2. The distance that each of the cars has already traveled on the track. If you have that info then you can figure it out. The two pieces of missing info are: 1. How long is the distance of the track and 2. The distance that each of the cars has already traveled on the track. If you have that info then you can figure it out. I totally agree with wildfire. Did you just say, "If two cars are traveling in a two lap race on a track of any length, one going 60 mph and the other going 30mph, how fast will the slower car have to go to finish at the same car to finish at the same time?" WTF? Are you having a stroke? Try to raise both hands above your head. OK, now smile for me. And would you please try to say a complete sentence? The way the question is currently worded, it does not indicate any of the following: 1. Whether the two cars started at the same place, at the same time (we can infer "same place, same time" because it is a race), 2. Whether either car has traveled any distance at all (if yes, then how far; if the slower car has traveled one lap, then the faster car has finished, and if no, then the answer is 60 mph), 3. What is the shape of the track (to Alanjai's point, a regular track requires offset starting positions, whereas a figure-8 track with fixed lanes would not), and finally 4. Why the question is worded so poorly ("to finish at the same car to finish at the same time" ... I mean, come on, that's practically not even literate). Show More Responses Speed a = Car A speed = 60 mph b = Car B speed = 30 mph t = Time Elapsed (in hours) d = Race Distance (in miles) ((t * a) = distance traveled by Car A) - d = Distance Remaining Car A = dra ((t * b) = distance traveled by Car B) - d = Distance Remaining Car B = drb x= mph that Car B has to drive for the remainder of the race (drb/dra)= y y * a = x or ((t*b)-d))/((t*a)-d)) = y y * a = x Example: t = 1 hour d = 240 miles ((1 * 60) - 240 = 180 [distance remaining Car A] ((1 * 30) - 240 = 210 [distance remaining Car B] 210/180 = 1.666666667 1.666666667 * 60mph = 70 mph, the speed that Car B has to drive for the remainder of the race. oh, yeah... in case you couldn't guess, I'm a Digg user. oh, yeah... in case you couldn't guess, I'm a Digg user. I agree with wildfire. This question is not grammatical and is unsolvable as written. The point seems to be that you should read the entire question (review the entire problem) before jumping in to solve the question that is immediately apparent. So, attention to detail is important at this company. Assuming the question was mistyped into this discussion, and they want to know how fast the second car would have to go to finish at the same time as the first car, then the answer is: infinitely fast. The question is better expressed as: A car is driving a sixty-mile path at thirty miles per hour. At the half-way point, the driver wants to speed up so his average speed at the end of the path is sixty miles an hour. How fast does he have to go? At the half-way point (30 miles) he has taken one hour for his drive. To average 60 MPH, he would have had one hour for the entire road. Therefore he has no time left, and must travel infinitely fast (for zero time) to average 60 MPH. It doesn't matter what answer you give, it is how you come to your conclusion that counts here. There is missing information on purpose because they want to see how you solve problems, not if you can solve problems quickly. The cars, the track the speed doesn't matter, it is the questions you ask and the information gathering that counts. Since there are only 2 cars in the race, the race is over and the instant one of the cars passes the finish line. One car finishes first, the other finishes second by default. The answer is that it doesn't matter how fast either of them are going, or how long the track is. They will always finish at the same time (not to be confused with "finishing with the same lap time"). I agree with SteveC. Once the either car finishes, the race is over. The question was clearly misworded. If not, most of you would have failed. The best answers here are from toolbelt_1 and dadag. Morgan Stanley needs people with exceptionally strong quantitative abilities and communication skills. The interviewer gives you a vaguely worded question to see (1) how you would gather the rest of the information and (2) how you would use it. In the course of a real workday your manager, client or other stakeholder will rarely provide a perfectly well-defined request for information. In the heat of the moment, important questions are worded quickly and vaguely, yet your performance will be judged based on how well you respond. One of your most crucial job skills is determining true requirements through timely and effective follow-up communication, intuition and experience. Show More Responses Both cars will finish at the same time if the track length was 0. This is typical of Morgan Stanley. Search a bit more and read about the lack of communication and clarity within this company--and when the result is as it should be (wasted time and effort) they blame the lower level worker as Al did above. If you ask for more information, you get more of the same -- confusion. Al might ALSO work for Morgan Stanley and makes a flimsy excuse for wasted time in having to track down pertinent information for the task. He makes no mention of the increasing frustration, lost productivity and the poor underlings that take the blame for poor managers. There are a few upper level managers who communicate and instruct their reports very well. It is a breathe of fresh air. They will tell the report the objective, quick background and the task and then you go do it. That simple. Others have more time for backstabbing, gossip and slimy character demonstrations than instructing their reports. No wonder they will never catch up to Goldman Sachs. They just don't get it. it's quite easy guys, just think: 30 mph is the current speed x is the race lenght 60 mph is the target average speed so theanswer is 30*(miles raced/ total race) + speed to achieve*(iles missing/tot. race) = 60 speed = i know that yu can dothis.....;) I'm pretty sure this is how the question is supposed to be worded which makes Mike's response correct. If two cars are traveling in a two lap race on any length track, one going 60mph for the entire race and one going 30mph to begin the race, how fast must the slower car travel for the rest of the race once the faster car finishes its first lap to finish at the same time as the faster car? If this is the case then we can do the following. distance = rate X time let d = the length of the track. After the fast car completes one lap the slow car will have completed one half lap, or .5d So the fast car has d left to go and the slow car has 1.5d left to go. since distance = rate x time, and the fast car is going at 60mph, we have d = 60t where t is time. For the slow car, if we let x be the rate it will go (so what we're ultimately trying to solve for, we have 1.5d = xt. now substituting d = 60t in we have 1.5 x (60t) = xt Since the track has some distance, t cannot be zero so we can divide t out leaving 1.5 x 60 = x = 90mph. Hence the slow car would have to travel 90mph the second lap to finish at the same time as the fast car. Make it simple, it depend on the fast car, if fast car got no problem( like break down , flat tires...), it hard to pass. The slow car just got to wait, time and opportunity is the key. Car A - 30mph Car B - 60mph One Lap - X miles Car A will have to decide whether it wants to catch up before completing the first lap. Otherwise it's over. We have two missing variables. We can't solve it. The speed will be calculated based on car's A location. Car A has to accelerate at any point prior reaching X. For instance, at 1/2X miles Car A will have to travel 90mph to finish at the same as time as Car B. But at 3/4X it will have to go faster. So the closer it gets to reaching X, the faster it will have to drive. Let's say the track is 60 miles long. Car 1 has completed a lap of 60 miles after one hour, and car 2 has traveled 30 miles. For the second lap (which is one hour for Car 1 to finish), car 2 must travel 90 mph. This works with any distance; this one is the easiest to visualize. |

### QA Automation Engineer at BitTorrent was asked...

A dwarf-killing giant lines up 10 dwarfs from shortest to tallest. Each dwarf can see all the shortest dwarfs in front of him, but cannot see the dwarfs behind himself. The giant randomly puts a white or black hat on each dwarf. No dwarf can see their own hat. The giant tells all the dwarfs that he will ask each dwarf, starting with the tallest, for the color of his hat. If the dwarf answers incorrectly, the giant will kill the dwarf. Each dwarf can hear the previous answers, but cannot hear when a dwarf is killed. What strategy should be used to kill the fewest dwarfs, and what is the minimum number of dwarfs that can be saved with this strategy? 21 AnswersThere are 2 different strategies, each dependent on whether there are an even or odd number of white and black hats in play. The minimum number of dwarfs that can be saved with the correct strategy is 9. There is only one strategy, does not matter how many white or black hats they are. They are always 9 dwarfs that could be saved. You just have to know about XOR. Well... In my opinion all can be saved! The tallest dwarf can see 9 hats in front of him ( 4 white and 5 black hats or the other way around). Then he knows the color of his hat because there has to be 5black and 5 white hats. The next dwarf by the size just has to believe to the tallest dwarf that he is right-(and he is) In addition ,he can see 8 hats in front of him and when he adds the color of the tallest dwarf he knows which color he has. Again,the 3rd one in a row knows about 2 colors before and can see 7 colors in front of him.And when he adds 2 colors he already had heard of -he knows total number of 9 hats as well.So he just has to see which color is less presented and that is the color of his hat. And so on until the last dwarf...All saved Show More Responses the question does not mention that there will be equal number of white and black hats ! Since they do not know, how many black or white hats there are, the following strategy will save min 5 dwarfs: The first dwarf asked names the color of the hat of the 2. dwarf. He has a 50% chance that that's correct. Anyway, the second dwarf then knows, the color of his hat and names it correctly. the 3. dwarf again names the color of the hat of the 4th dwarf and has a 50% chance to survive, while the 4th dwarf can name the correct color of his hat. a.s.o. => all dwarfs with an even number will survive and all the others have a 50% chance Don't over-think it. Each dwarf simply has to state the color of the hat worn by the dwarf directly in front of him. The tallest would have to sacrifice himself in order to save the other 9. Easy. The tallest is the only one that cannot be saved so instead of trying to guess his color, he yells out a sequence of letters starting from the shortest like BWBBBWWBB. Next :) Your going to definitly get 9 because when the dwarves collude the tallest tells the next tallest his hat color and so on down the line. Now you have a 50/50 chance that the tallest will get his color correct and live so you have 9 with a 50% chance of 10 Each dwarf can state the color of the hat worn by the dwarf in front of him, but the thing is, that color may not be the color of his own hat.So he may be killed by the giant.In that case, as mentioned above all odds should tell the color of next so that all even number will be alive and they have 50% chance of surviving. Think broadband communication. Exploit the capabilities of the communications medium. A minimum of nine dwarves can be saved based on the information provided in the original post I viewed. The strategy is for each dwarf to employ the expected language to communicate the color of their own hat to the giant, while simultaneously employing a vocal pitch protocol to indicate the color of the hat of the dwarf in front of him, high pitch for white and low pitch for black. The original post, indicates the dwarves may collude prior to the distribution of hats, so there is opportunity to negotiate such a simple broadband communication protocol. The tallest dwarf only has a 50/50 chance since the number of black and white hats in play is not known (rhetorical question, what are the odds the tallest dwarf's hat is black if he turns to find that all nine hats in front of him are white? I don't know, but odds are high that the giant is a sadistic bloke). The original post I viewed is here. http://www.businessinsider.com/toughest-job-interview-questions-2013-7#a-dwarf-killing-giant-lines-up-10-dwarfs-from-shortest-to-tallest-each-dwarf-can-see-all-the-shortest-dwarfs-in-front-of-him-but-cannot-see-the-dwarfs-behind-himself-the-giant-randomly-puts-a-white-or-black-hat-on-each-dwarf-no-dwarf-can-see-their-own-hat-the-giant-tells-all-the-dwarfs-that-he-will-ask-each-dwarf-starting-with-the-tallest-for-the-color-of-his-hat-if-the-dwarf-answers-incorrectly-the-giant-will-kill-the-dwarf-each-dwarf-can-hear-the-previous-answers-but-cannot-hear-when-a-dwarf-is-killed-the-dwarves-are-given-an-opportunity-to-collude-before-the-hats-are-distributed-what-strategy-should-be-used-to-kill-the-fewest-dwarfs-and-what-is-the-minimum-number-of-dwarfs-that-can-be-saved-with-this-strategy-11 So reading the answers provided they all have some assumptions e.g. that there are as much white hats as there are black hats. I think that Christian's comment on aug 13-2013 was very close but I'm thinking about communication integrity confirmation techniques. One of them is using a parity bit to confirm the message was correct. This could be applied right here to save at least 9 with a 50/50 chance of saving the 10th (and tallest dwarf). I will explain it but for ease of explanation I will use binary 1 and 0 for black and white. Number 1 being black hat and number 0 being white hat. Let's say we got a (random) hat sequence of 0001011101 with the tallest dwarf on the right and the shortest on the left. While colluding prior to the distribution of hats the dwarfs agree upon even or uneven parity. This means the total amount of 1's (black hats) must be even or uneven including the parity bit. In this case the 10th dwarf will count as parity bit. So if we'll take an even parity, the number of 1's must be an even number in total. When the questioning starts the tallest dwarf will see the hats in front of him being 000101110. The tallest dwarf counts four 1's (black hats) so to make the parity even he has to say 0 (white hat). He will get killed but the dwarfs in front of him will know the parity bit is 0 so the 2nd tallest dwarf will see the hats in front of him as 00010111. He will also count 4 and knowing that the dwarf behind him said 0 he'll know that the total amount of 1's is an even number thus concluding he has a 0 (white hat) and will state he has a white hat. Same goes for the rest of the dwarfs and so 9 will be saved. The 10th would've been saved it the dwarfs agreed on an uneven parity. That's why there's a 50/50 chance the 10th will be saved. I'm pretty confident this is the answer but if you want to understand it better (maybe my explanation is a bit vague) go search for "parity bit" on the internet. @Kristen - you have so underthought it! What if the dwarf behind you says "black" and the dwarf in front of you has a white hat???? @JustJanek- Your solution is close, but not correct. Every dwarf needs to consider the parity of a number composed of all the following dwarfs and all the dwarfs behind. The first dwarf says the parity of the 9-bits number in front of him. Assuming that all the other dwarfs know the trick and they stay alive, each dwarf needs to compare the initial parity with the parity of an 8-bit number composed of the hats in front of him and the hats behind (assuming that the dwarfs behind gave the right answer), if the parity is the same, he knows that he has a white hat, otherwise he has a black hat. Show More Responses You can save 9 dwarves at least. Dwarves agree that the tallest one says he is wearing black hat if he sees odd number of black hats in front of him and he says white hat if he sees even number of black hats in front of him. So, the tallest one has 50-50 chance of survival and other dwarves survive 100%. What is the minimum number of dwarfs that can be saved with this strategy? 9 First of all, let's numerate the dwarfs as N1, N2, N3, etc. with N10 being the tallest. Now, N10 will state the color of N9 as his own answer, "My hat is WHITE". Based on this answer, N9 will state his color with a positive statement if the color of N8 is the same as his, "My hat is WHITE". Based on N8's answer, N8 knows that his color is WHITE, now, he will state his color depending on N7. Let's say N7 is black, so N8 will state, "My hat is NOT BLACK". N7 knows that his color is BLACK, but N6 is white, so he will use a negative statement, "My hat is NOT WHITE" and so on. Full example: N10 = BLACK N9 = WHITE N8 = WHITE N7 = BLACK N6 = WHITE N5 = WHITE N4 = WHITE N3 = BLACK N2 = BLACK N1 = WHITE N10: My hat is WHITE (Dies) N9 = My hat is WHITE N8 = My hat is NOT BLACK N7 = My hat is NOT WHITE N6 = My hat is WHITE N5 = My hat is WHITE N4 = My hat is NOT BLACK N3 = My hat is BLACK N2 = My hat is NOT WHITE N1 = My hat is WHITE N10 will have a 50/50 chances of survival... I'm sorry N10, I couldn't save you :'( What is the minimum number of dwarfs that can be saved with my strategy? - 10 My strategy doesn't make any unmentioned assumptions (such as equal number of white and black hats, etc). At the same time, it doesn't add any unmentioned constraints either. The strategy is simple to the point of appearing simplistic. But it meets all the requires. The strategy is: When asked, every dwarf answers "Not RED". This is not an incorrect answer and the Giant, if he were an honourable giant that is :), would have to let the Dwarf live. On a different page altogether, I went through all of the above answers. Not to sound patronizing, but some of the solutions were quite brilliant. I was thinking if I could even come up with that even after years of pondering. But I must admit, all of the above strategies are made by an outsider (ie. us) who is not impacted by this fate. Whereas in the casestudy, the strategy has to be devised by the 10 dwarfs, who face the impact of this strategy. Not to bring in factors such as emotion, the bell curve distribution of intelligence, and other such anal considerations. But I thought it was important to bring in Game Theory, and that all Dwarfs are rational, and that all rational people do not want to harm themselves in any way. In other words, when a strategy's success depends on the conformance to that strategy by ALL the participants, the strategy should also benefit ALL the participants if it is to succeed (or in the least, should not harm even ONE participant). In all the above strategies, since even in the best case scenario the poor Tallest Dwarf has only a 50% chance of living, can we assume that he would conform to this strategy? Each dwarf should pronounce color of hat of dwarf before him. This way they can save atleast 9 dwarf out 10. Well, Once the dwarfs are lined up in descending order. Without any kind of assumption, 9 people can be saved with a 50% probability of the 10th (tallest). Here is how it can be achieved, The strategy is to call the color of odd number of hat. Say for example, the tallest dwarf sees 3 Black and 6 whites, it will call out Black(it may or may not die with 50% chances). Now, the 9th tallest dwarf knows what the 10th dwarf could see and if it (9th) dwarf sees the same odd number of black hats, it will know it has white hat. Next, 8th dwarf knows number of odd (black) caps 9th dwarf could see and if it(8th) finds 1 less black cap, it would know it is wearing a black cap.. and so on. 10- White (calls out Black because it sees odd number of black hats) -dies(assume) 9- White (calls out White because it could still see 3 black hars) 8- Black (sees that there is one less black hat as mentioned by 9th, hence identifies that is wearing black) 7-White (calculates that 8th is wearing black and he could see 2 black, hence identifies itself wearing whiteO( 6-White 5-White 4- Black 3-Black 2-White 1-White Sorry if there is confusion in the way I have answered. all can b saved... They will exchange their hats in Circular form...person10 can see the person1 color of hat and after exchange every dwarf will tell color to their previous ones and person 10 knows the color before changing it from dwarf 1. d10->d9->d8->d7->d6->d5->d4->d3->d2->d1->d10 CIRCULAR EXCHANGE OF HATS First lets look at number of back and white hats...To satisfy the condition "black and white" hata there is minimum one black or white hat present. So its minimum (9 black & 1 white) or (1 white & 9 black) hat being distributed randomly amoung dwarfs. The story says dwarfs are alowed to speak before execution. Lets make a strategy of saying only one colour before execution eg) black or white. Minimum probability of (9 black & 1 white) hats and all the dwarfs say "white"...In this case one is saved but all nine dead. If all dwarfs say "black"..Nine are saved but one is dead. The same applies for the minimum probability of (1 black and 9 white)hats. Thus minimum one dwarf is saved and maximum nine dwarfs can be saved. If each dwarf say the colour of hat in front of him..(dwarf can hear previous answer) then at least five people are saved. |

### Store Manager at Kohl's was asked...

why are you looking for a positon with Kohl's? 4 AnswersI need a job? I am looking for a challenging career in which I can use all of my 16 years of merrchandising, competitive, selling skills to use. I The previous posting, while good, focuses on the candidate. and can be applied to any position. I would start with that phrase, but then follow it by answering the question about why do you want to work HERE? I suggest: "I want to work here because this is a place that I know and respect." This shows that you are already familiar with the company and its products. It also provides you the opportunity to demonstrate any research that you may have done. Show More Responses I shop at Khol's regularly and this is the one place where I recieve great Service and Product. I would love to be a part of a Retailer where I can provide Great Service and Product to our Customers. |

What is 29^2? 4 Answers841 Mental math. You could do it mentally. 29 is close to 30. So the upper bound is 30*30 = 900. You would have to subtract 30 and 29. 900- 59 = 841. or, take 29 x 30 to get 870. Take away one 29, get 841. Same idea, but more elegant. You've multipled 29 thirty times, you only need 29 of them, so take one away. Show More Responses Easiest way I can think of is to mentally calculate (30-1)(30-1) and then mentally foil the answers. You end up getting 900 - 30 - 30 + 1 = 841 which is pretty easy to do in your head. |

YPhone files a suit against XCell for damages associated with stealing a proprietary technology, how would you evaluate the damages? 3 AnswersThere are a variety of ways to approach it, but basically you need to describe a few methods for the valuation of lost sales due to XCell stealing market share from YPhone. One approach is to look at expected sales by coming up with a historical data model. Another approach could be model the situation as discontinuous innovation and computational reconstruct what sales should have looked like in the absence of the competitor. Finally, you could create a comparison model based on other similar technologies and their sales figures. Moreover, the damages could result from more than just sales. Damages could come from the costs associated with a need to change marketing tactics, and the costs stemming from filing the lawsuit itself. were you contacted via phone/email by Cornerstone to let you know you didn't go through to the second round? if so, how long after the interview were you contacted? Good question. I called the company on a Monday morning to follow up approximately one week after my phone interview. Although at the time there was no one available who could inform me about my status, I was allowed to leave a phone message. A few days later, I got a call in response to my phone message informing me that they were sorry to say there would not be a position available for me. Since it had already been a week and a half since my phone interview, I was already fairly certain that this was the case. Grateful to be informed, I thanked them for the interview opportunity and wished them well. I want to note that the employee from the firm who called me was very respectful and invited me to go ahead and apply for a full time position in the fall if I was still interested. Overall it was a positive experience, and I am happy to have had the chance to interview with them. |

### QA Automation Engineer at BitTorrent was asked...

You have 2 identical crystal orbs and a 100 storey tall building. How do you determine which floor the orbs will shatter at, and what is the minimum number of tests you need to execute? 5 AnswersOne test. If they shatter, they will shatter at the bottom floor. =)D If this is supposed to be a "what's the worst case" sort of question -- Drop first at the 50th floor. If it shatters, work you way up from the first floor. If it does not, drop from the 75th floor. If it shatters, work up from 50th floor. It it does not, drop from the 88th floor. If it shatters, work up from the 75th floor. It it does not, drop from the 94th floor, etc. You can get it done in about 4-5 tests, I think. Start by dropping one from the 33rd and one from the 66th. If both shatter, it's a floor below 33. If 33 is intact and 66 shatters, then it's a floor between 33 and 66. If both remain intact, then you'll check floors above 66. For your next test, divide the range you need to investigate into 3. So, for example, if the 66th breaks and the 33rd does not, then you have 33 possible floors. 33 divided by 3 is 11, so go up 11 floors from 33 and down eleven floors from 66 and drop them. Same idea as last time- you're narrowing it down to 1/3rd of all of the possible floors. You'll know the correct floor at which they will break when the number of possible floors is exactly 1 percent of the number of original floors (in other words, when there is only 1 floor out of 100 that could be the lowest shattering floor). Since each test reduces the possible floors by 66% (leaving 1/3 of the previous floors remaining), you'll have one floor left when (1/3)^n = .01, where n is the number of trials. This occurs at log base 1/3 of .01, which wolfram alpha says is 4.19, so it would take at least 5 trials to get this narrowed down to a single floor. Show More Responses The problem here is that you only have two crystal orbs. If you had closer to 100, a strategy of beginning on the 50th floor and working up or down would work great. Unfortunately, if it breaks at 50, and then breaks at 25, you're out of orbs. You'll have to be cautious. You can start on 50. If it shatters at 50, though, you'll have to go to the second floor (assuming you can simply place the orb on the ground when on the first floor) and work your way up one at a time--because you've only got one orb left. If it doesn't shatter at 50, go to 75. If it shatters at 75, you have to go to 51 and work up one at a time, as, again, you only have one orb left. If it doesn't shatter at 75, go to 88, etc. The minimum number of tests is 2. Shatters at 50, shatters at 2. If you only have two orbs, use one to test and the second to confirm. starting from the bottom make your way to the top on steps of 3 floors. This is slightly better than n/3 (n/3 + 3) where n is 100 in this problem. For example: let's say the orbs break on the 48 floor. On increments of 3 you get to the 48 floor in 16 steps. Once the orb breaks go to floors down and throw the second orb. You know that the orb didn't break on the 45th floor. If it breaks on the 48th floor, it won't break on the 46th, or the 47th floor, when you throw the orb from the 48th floor again it will be confirmed. let's say the orbs break on the 49th floor. you know 48th checks, 51st will break the 1st floor. go to 49th if it breaks you can confirm 49th is the limit. let's say they break on the 50th floor. 48th checks, 51st doesn't. go to 49th, checks, 50 doesn't therefore 50th is the limit. |