# Brain teaser Interview Questions

interview questions shared by candidates

## Brain teaser Interview Questions

You have a birthday cake and have exactly 3 slices to cut it into 8 equal pieces. How do you do it? 39 AnswersCut in half, stack, cut in half, stack, cut in half. All you have to worry about is the 45 degree rotation of one of the 4 pieces after the second cut. Blade can be kept in place, like a paper cutter, as to minimize the margin for error. Slice it horizontally across the middle creating two equal halves top and bottom. Then simple two slice cross from above like normal. Cut each slice into 3 slices. Then eat one of them. Show More Responses This is really an easy one. First cut into half across the top, then cut the halves in half also across the top (you now have 4 equal pieces) then cut across the middle = 8. No, Jason and Sharon, you will only wind up with 6 slices. z, too many stacks. You need to cut in half, then make another cut - to get four pieces. NOW you stack these four pieces and make the last third cut - and you get 8 pieces. A rather easy lateral thinking question. Alina's got it. The stacking seems to be the "right" answer. But this is a stupid question. Who stacks cake? The frosting from the bottom slice would meld with the stacked slice, thus making the cake inseparable. I wouldn't get the job because I would swear at the interviewer for asking a dumb question. What good is an answer to this question if it wrecks the cake? Assuming the cake is square: slice 1: cut horizontally to create 2 equal pieces slice 2: cut vertically to create 4 equal pieces slice 3: line up all 4 pieces of cake side by side and cut horizontally to create 8 equal pieces. don't stack, it will ruin the frosting. With a knife Alina would be penalized for not being able to count past 6. But then, she could get a job at another company where they appreciate people who say someone's idea is wrong, then put forth the same idea and take credit for it. A new solution for you: who says slices have to be a straight line and not circular? I would cut a concentric circle in the middle (would have to calculate the radius compared to the whole), and then slice an X with the remaining 2 cuts. It will look like a target. If done correctly the sizes will be the same, it says nothing about having the same shape! I guess it's much easier if you think of two planes: first cut in half, then to quarters. That's the easy part. Now look at the cake from the side, and cut it across... Each quarter is cut into two and all pieces are equal. I would consider the interviewer's emphasis on "equal pieces." While my first thought was to bisect across each of the three dimensions, half of the pieces would end up with less frosting than the other half. Stacking the pieces would result in frosting transfer, which would also screw up some of the pieces. Therefore, I'd go with lining the pieces up and have a large knife on hand for each bisection. Hey, people get crazy about their cake slices. Show More Responses I would take a different approach to this. First of all the question asks for the cake to be cut into 8 equal pieces, not 8 identical or 8 of the exact same size. Equal doesn't always have to mean the same, just equivalent to. First I would find out who I am cutting the cake for, if its 4 old ladies and 4 young guys, equal pieces would not mean that they all needed to be the same. If everybody wanted the exact same amount of cake I would figure out a way to give everybody 12.5% of the entire cake volume, but if some wanted larger or smaller pieces I would come up with a way to satisfy each individuals desire. If you know how to cut an arbitrary shape in half, you have the solution. After every round, plan the cuts for each individual piece. Then align them so the proposed cuts are in one straight line. Make a cut. This way you can cut any cake into 2^n equal size pieces with n cuts. An interesting question is, if you start with one connected piece, will you always be able to end up with connected pieces. Think of cutting letter S in the middle, like this: $. You end up with two equal figures that are not connected (or, in other words, with 4 pieces). It's also easy to design a cake that can be split into 8 pieces with one straight cut. I LOVE Andrew's answer! Eat one of the freakin' pieces. That's the corporate way in America anyway. Mike is either a socialist, or works in non-profit, or government. I thought "slice them long ways" but then someone misses out on the freaking frosting, which is the best part. UNLESS, it's a layer cake. Cut 'em all in thirds and give the extra to the birthday boy/girl to take home. First I'd yell at whoever cut the cake incorrectly to start with. He's ruining the party. Then I'd squish the 3 pieces of cake together and re-cut the cake into the required 8 pieces. Boom. This is as easy as pie. Viewing the cake from the top, make 1 cut vertically down the middle of the cake and another horizontally. Viewing the cake from the side, make your third cut horizontally through the middle; QED three slices and 8 pieces of cake with a beamsplitter and prisms it could be done in 1 (with a laser) That's an oddly presented question that is understood 2 ways: - 3 cuts allowed to cut one cake in 8 pieces. Which yields cut in 4 parts, then split those in 1/2 again with the last cut, either by stacking, realigning the slices or making a round cut. - 3 pieces of cake must be re-cut to make 8 equal parts. Which is an impossibility unless one piece is 2/3 smaller than the other 2. That yields different answers like cutting in 3, and eat one to leave 8 pieces. They key to answering any of the brain teaser questions is to ask a few critical questions before even attempting to answer. I would start with: Is the original cake round or square? If square, line up all three pieces and recut to be 4 equal widths of cake. If round, was the original cake cut in 6 or 8 pieces? If 6 pieces you have 1/4 of a cake = easy to redivide into 8 equal slices. If 8 pieces, you have 1/3 of a cake and a little math needs to be applied to create 8 equal slices. First slice a strip off each of the 3 slices to create a fourth slice. then divide each in 1/2 to make 8 equal slices. Cut each slice into 8 small slices. Then give each person 3 small slices. Show More Responses There are some posting above who seem to not have a good grasp of numeric's. The answer is not that difficult. First, presume the pieces are not equal size (nothing states they are). Second, presume two pieces are of equal size and the third piece is twice that size. Third, cut vertically (the most usual manner in which to cut cake) the 2 equal pieces (we now have 5 pieces -4 the same size and one larger piece). Fourth, cut the large piece in half, then those two pieces in half again. Fifth, voila, one now has eight equal pieces of cake. Hmmm, I always assumed they meant three knife cuts by the word "slice" Kind of interesting to me that others assumed the cake was given to you in 3 parts as defined by the word "slice" I'd say that you should cut the cake horizontally using the knife as a measuring device to find the exact center of the circle, then cut vertically using the same method, then take each quarter , using the knife as a straight edge, build an alignment diagram that places each set of two quarter pieces point to point along an axis that defines their center lines, and cut all 4 quarter slices with a single cut of the knife ( defined by the word slice ). Put it all in a blender. Pour each of the resultant mixture onto a plate or into a bowl. 1. assemble 3 sliced cakes into a big cake (original shape) 2. cut it half (don't care about the indentation) , you would get 2 piece of cake 3. cut it half again, you would get 4 pieces of cake 4. cut all of them half again, that's finish. To those who think it means you start with 3 pieces ("slices") of cake, READ IT AGAIN. It says (emphasis added): “You have a birthday cake and have exactly 3 slices to cut IT into 8 equal pieces." Get it? "... to cut IT [the cake] into 8 equal pieces."... There is NO WAY it means to cut 3 slices of cake, otherwise it would say "...exactly 3 slices to cut into...". And Mike: "equal doesn't mean same, just equivalent". That's the funniest (and stupidest) thing I've heard all day. Anyway, since 2^3 = 8, you have to stack. Assuming a round cake: Cut (or "slice") 1 creates 2 semi-circles. Stack them. Cut 2 creates 4 quarter-circles. Stack then Cut 3 creates 8 1/8th circles. is the 3 slices equal in size? Great answer and explanation here: http://www.programmerinterview.com/index.php/puzzles/birthday-cake-8-pieces/ Make sure the guest of honor has Blown out candles 1st! (was not specified but hey so were many other things) If the birthday boy/girl is under the age of 10, I am not too sure you want to be messing with their cake!! Cakes come in many different sizes and shapes...ESPECIALLY Birthday Cakes!!! ACK They even come in characters and shapes you can NEVER get into equal pieces but, back to the solution! Will use 2 shapes: Round and Square! Cut 1: Parallel to cutting board and horizontal to create 2 layers of equal depth; Cuts 2 and 3: Perpendicular to cutting board once then rotate 90 degrees and repeat! Now give it to the Kid in the high chair to for quality control/assurance! Like Andrew, I would eat one piece and then cut the 2 in four equal pieces. Remember 1 whole cake, 3 slices with a knife = 8 equal pieces Place cake flat on table. Grab a knife big enough to cut the cake horizontally. 1st cut - Cut the cake horizontally leaving the cake flat on table as if the cake still in one single piece. Now you have 2 cakes instead of one. 2nd and 3rd cut - cut through the cake vertically in the form of a cross. Now you have 8 equal pieces of cake. As if you had cut 2 cakes in 4 pieces each. Show More Responses I would state that I only eat cakes in the shape of circle. then 3 equal cuts across the middle. think of it like a pizza... or a pie.... or a pizza pie. Remember these questions are made to have you think outside the box. Not all cakes are square. cut the diagonal portion then centre line of cake Cut each of the 3 slices into 8 equal parts which makes the slice count 3*8 = 24, Divide the 24 between 8 people 3 each. Cut 2 slices into 4 pieces each, cut the third one into 8 pieces. Separate 2 of the 3 slices, dividing 2 slices for 8 people is easy (each slice in 4) Then with the left piece I divide it in 8 and offer one to each person. One more: I could put all the 3 pieces together and re slice into 8 equal pieces. |

An array of 99 elements contains integers from 1 to 100 with one missing element. Find the missing element. 22 Answers100 1. calculate the sum of elements in array say SUM 2. sum of numbers 1 to 100 is(n* (n+1))/2 = 5050 when n==100 3. missing element is (5050-SUM) 100 Show More Responses The parameters of the question do not allow you to determine what element is missing. Either more information should be supplied, or all answers are equally correct. How could an array size of 99 elements contain 1 - 100? Should either be integers 1-99 or 2-100 , in either case there is no missing element. All indices are accounted for. Sum them and then subtract them from 5050. In general, if an array of size n - 1 elements has unique elements from 1 to n, then the missing element can be found by subtracting the sum of the elements in the array from sum(1 ... n) = n * (n + 1) / 2. Alternately, one could use a boolean array of length n with all values set to false and then for each value, set array[val - 1] to true. To find the missing value, scan through the array and find the index which is set to false. Return index + 1. This requires O(n) memory and two passes over an O(n) array (instead of constant memory and one pass), but has the advantage of actually allowing you to verify whether or not the input was well formed. Admittedly, this question is poorly posed; however, the answer they are looking for refers to the syntax/nomenclature of some (not all) programming languages to index arrays starting at “0.” As such the 1-100 stored values would be in entries 0-99 of the array. Read the question. Here are the steps to solve it: 1) find the sum of integers 1 to 100 2) subtract the sum of the 99 members of your set 3) the result is your missing element! Very satisfying! Sort array. While loop with an index variable with condition of next element being 1 greater than previous element. When loop breaks, return the value of the index. Doing the expected sum and subtracting the actual gives the run time of O(2n), however a bucket sort will almost always do it in less time (somewhere between O(n) and O(2n)): 1. create a 101-int (or boolean) array (to have a 100-index) 2. traverse original and for each int, assign value in bucket array to 1 or true. 3.After first traversal, traverse created array starting at one, and when value is false, print it. 100 100 coz in array it initial value starts frm 0 to 100. or else 4 further clarification u can study array chapter in c or c++ 100 Show More Responses The question: "An array of 99 elements contains integers from 1 to 100 with one missing element. Find the missing element." The information states that the integer count is 1 to 100. I take this to be inclusive of all elements in the array so that the missing inters would be subjective to their arrangement or random. In other words, I do not have enough information to say which one. 1 I need more information. 1. Are the integers unique in this array? 2. Do I have enough information to find the sum of the integers in the array (or some aggregation)? If sum is available, then, the answer is 5050-sum{integers}. Bucket Sort works and summation works. I think both are good, practical and clever solutions. I think sorting the array then searching may be unnecessary computation. Another interesting method which may be faster. SIMD computers may do this particularly quickly: Do a bitwise operation on all the elements: Result = Array[0] xor Array[1] xor ... Array[98] xor 1 xor 2 xor ... xor 100 Result = Missing number. Explanation: When you xor 2 identical numbers your result = 0. For example, 5 xor 5 -> 101 xor 101 = 000. (5 in decimal is 101 in binary). Knowing that "xoring" 2 identical numbers results in zero is useful. Now we apply this useful info to the problem. Array is Identical to a list of 1,2,3,...,100 except for one number. In other words 1,2,3,...,100 duplicates all of array's elements and adds one extra element that is missing in Array. Therefore, we now have 2 instances of each element in the Array in addition to one extra element in 1,2,3,...,100. We can see when you xor two duplicate numbers you get zero. Because we have pairs for all numbers in Array and one extra number we are essentially "xoring" the missing number with zero. When we xor the missing number with zero we get the missing number. (For example, 6 xor 0 -> 110 xor 000 = 110) The question states that one (not two or three or n) element ("value") from 1 to 100 is missing. There are 99 elements ("values") in the array. The question implies that the data is well-formed because it states that only element is missing. It doesn't ask you to find the missing value(s), but the one (singular) missing element. With the problem constrained, the solution falls out. Subtracting from 5050 is an elegant solution, but not obvious as to why it works. The array of booleans is more obvious, but doesn't scale well. I agree with one of the answers in this thread...5050-sum(elements) = missing item. Other approach that crossed my mind is something similar to binary search. Check the index of 50th element: if A(50) == 50, the missing element > 50, else if A(50) > 50, missing element <50. Do this iteratively. The number of comparisons would be log 100 = 7. 0 100 Add 1-100 to a hash of 100 elements. Then compare each element with the hash.. Answer in o(n) |

How far apart is the hour and minute hand a 3:15? 6 Answers7.5 degrees 360 degrees on the clock. Each number is 360/12 = 30 degrees apart from one another. at 3:15, the minute hand is at 3 while the hour hand is 15/60 = 1/4 away from 3. So (1/4) * 30 = 30/4 = 7 + 2/4 = 7.5 degrees The previous two answers are incorrect. If it is 3:15, the hour hand and minute hand are in the same orientation. Therefore they would be 0 degrees apart. Show More Responses No matter what time of the day, the minute and hour hand are always touching in the center of the clock face. Is that 3:15 a.m or p.m.? 180 |

What is your idea of reasonable turn around time for a project? 3 AnswersDepends on the clients ability to provide assets, and the scope of work. Reasonable would be the projected, based on original scope and estimated labor hours. The word reasonable in itself indicates the use of reason to determine the outcome. Just |

How would you handle a situation where you have a client who interrupt your existing interaction with a client? 6 AnswersI request the existing client to step aside momentarily which I then assess all my available resources to assist this client. Should it then become heated I attain a manager to resolve the situation. I politely tell the client who interrupted to please standby and that I will be able to answer their questions/handle their issues shortly. If they need immediate attention I will call a member of the team to assist. I apologize and ask him or her to please step in line and I’ll be right with them. If they interrupt I say understand completely and look for a manager. If there isn’t one available, I’d ask for the source of their urgency. If it’s just them being difficult without an interest in telling me their issue I’d tell them I’ll be with them shortly, I just have to help another customer. 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Q2: A pnp transistor with its base connected to a voltage source, the V source is connected to a +10V source. The emitter of the transistor is connected to a resistance, and then to the same +10V source. The collector side is connected to a capacitor, which is not charged at t=0-. Given the graph of Vsource = 10 V stepping up at t = 0 to further, draw the graph of Vout. Vout is between the point of collector and capacitor. 3 AnswersANS: Vout should be constantly -10V until t=0, and will hit V=0 V linearly from V=-10 V after t=0. Hi, Can you explain why it linearly increases? Are you assuming that Collector is tied to -10V? The pnp transistor is completely cutoff for the given biasing. The only way the capacitor is going to charge is through leakage currents. It is very slow and takes a lot of time. Please advise me if my analysis is correct. I just made it back from Lutron's HQ and I was asked for the same question. My first approach will be identifying a PNP BJT, and elaborating all 4 BJT operating regions. Before t = 0, since q = 0, by using Q = CV, we can tell that the voltage across the capacitor is 0. Hence, Vo = -10V before t = 0. Recall the capacitor's current equation: I = C*(dv/dt), we can solve for the slope of changing voltage -> dv/dt = I/C. Here, I is simply the BJT's collector current, which can be found by looking at the BJT's emitter current. Given that the (beta) parameter is infinite, we see the base current to be 0. Now, at this point, we need to look for I_E. Since the R is given to be 9.3k, and VEE = 10V, it is natural to assume V_EB = 0.7V, and thus the voltage across R = 9.3V. Therefore, I_E = 9.3/9300 = 1mA. Voila, we now values for all currents: I_E = I_C = 1mA, and I_B = 0A. Plug the I_C value into the equation: I_C/C = dv/dt (C = 1uF). We know that the slope of the voltage change is 1000V/second, or 1 Volt per millisecond. Now, we know the capacitor voltage raises at 1V/ms from -10V, but we also need to know where is the upper limit. Looking back to the BJT basics about operating regions and BJT's 2-diode model, it is not hard to identify that this PNP BJT must operate in "Saturation" region (NOT IN "ACTIVE" REGION!). The boundary of that region is V_BC <= 0.7V (I hope everybody is able to solve for this). Hence, 0.7V will be the upper limit for capacitor voltage. At this point, you will have a flat line at Vo = -10V before t = 0. and raises at 1V/ms for 10.7ms and hit Vo = 0.7V. From t = 10.7ms and on, the Vo stays at 0.7V. |

### Systems Analyst II at AT&T was asked...

How would you respond to a customer who encountered a problem with a product or service. 1 AnswerFirst, try and reproduce the problem and if the problem could be reproduced acknowledge there was an issue, give thanks for bringing this to the company's attention, and apologize for the customer's encounter with the issue. Offer possible solutions or workarounds while the team worked to correct the issue. |

### Manager at Amazon was asked...

If you had 5,623 participants in a tournament, how many games would need to be played to determine the winner 61 AnswersBefore I could figure that out, I'd need to know whether the # of participants represents the number of individuals on larger teams, or the number of teams A specific numerical answer can be given, but there are multiple ways the tournament can be setup, for example, are there play-in games, byes, etc. I would think the question is being given to a manager to see how they think and process, and then come up with a specific numerical answer, as opposed to just a math problem. It can be just one game. A huge mock battle. Show More Responses 5,622. Assuming it is a single elimination tournament. All teams lose one game except the champs. It's always # of teams - 1 if each team plays until it loses in one(team)-on-one(team) contests, the answer is ln(5623)/ln(2) Assuming it is a simple process of elimination, it takes 5622 losers to get 1 winner from 5623 participants. So, it would require 5622 games. Assuming it is a simple process of elimination, it takes 5622 losers to get 1 winner from 5623 participants. So, it would require 5622 games. if it's a 1:1 type draw, then # rounds = (5623)**x where x is base 2. a good answer is between 12 and 13 rounds. 2**12 = 4096 so I'd draw up 13 rounds and give out 1,527 byes. There is no true answer as the question is very open ended. The interviewer is probably looking at task delegation, management and creativity skills. One One. Obviously. One game, all players participate. If participants equal number of teams involved, think power of 2. Show More Responses The interviewer is not looking for the right answer because there can be many. What he/she is looking for is your logical approach in solving the answer. So you could start by probing more is first I would like to understand if 5,623 participants represent the number of team or individuals. Then ask the next logical question based on the answer. Everyone who didn't ask a follow up question except Mike is right. The question says, "if YOU had ...". This requires no follow up questions, because YOU should decide how YOU are going to operate YOUR tournament. Why would Mike or any of the others think it's someone else's job to organize his/her tournament. Oh just a follow up. My tournament would be held in the top of a hardly dormant volcano. Everyone would get a backpack full of grenades and the first one out of the crater without dying wins. That makes 1 game. Also, I think most participants who got out of the volcano alive would consider themselves winners, but only one would get to keep the gold plated dancing chiquita banana. Yipee!!!! I'm right too! Take that Mike! it'd be one game if it was a battle to the DEATH I agree with Nancy there is no strict answer to this question it is all about problem solving. First thing to do is to get more information, if it is not forthcoming then make assumptions, as an interviewer I would not be impressed if the candidate didn't ask for more information, although I probably would not supply any more. Then looking for a logical (and humane) answer which is substantiated with appropriate reasoning. Ie number of people on a team, game being played, what is required to win a match, are there several games in a match? knockout style tornamant sounds like a good approach. I agree with Mike. Just show the interviewer how you think and how you will tackle the problem in a colloborative environment 1 I agree with the very first response. Many of you are perhaps making the assumption that this is a one-on-one tournament such as singles tennis. Isn't it possible the question could refer to a soccer or basketball tournament where there are multiple players on each team? That would certainly bring the number of games to be played down considerably. I'd give an approximate answer, stating my assumptions. The question asked is not the number of rounds (2 people per game: log base 2, giving approximately 13 rounds with everyone playing at least one game ) - it's the number of games. So, if two people per game, then it's the sum of 5623/2 + 5623/4 + 5623/8 + 5623/16 + ... The limit of this is not something I know off the top of my head, but it's less than 5623. Also, interestingly, you need to account for the original number being odd. That could be accommodated in a number of ways, none of them straightforward. 5622 ... if based on elimination between 2. Show More Responses I personally agree with most of you. If you read the posts here, you see all types of answers. Some say "1". Short and simple rules to a simple game. Others have posted all types of formulas and methods to figure out a process. The answers here are all a good example of different minds using different means to find an end. Those different answers are what a good interviewer would be looking for. If the job needs a person that is logical and takes time to plan things out, or perhaps someone that needs to think fast on their feet. That kind of question could come in handy for any kind of interview in my opinion. 1 I think the interviewer is asking you to ask for more information, ask three qualifieng questions to be exact in order to show you have the probing skills to fully understand a customers situation or company problem and have the ability to ask the appropriate questions to or to get the help to solve. Question 1) How many players are allowed per round? Question 2) Is there a time restriction on these rounds? (Daylight or Night as well) Question 3) Are there going to be different classes for the golfers? Are we taking handicaps into consideration? I guess there could be more questions but chances are the interviewer would stop you after the third question. The correct answer is: "I didn't come here to play bulls**t games for 8 hours, you Mac-slinging hipster. Ask me a real question." 5622....................................assuming single elimination 5623 is a prime number. Good luck dividing into even teams. Also, there is no use of the word "minimum" in the original question. This question is a good example of a problem with no absolute answer. If I were to ask this of an interview candidate (which I wouldn't because I think subjective questions are mostly a waste of time for everyone involved), I would look for someone who can: A) Ask questions to pin down a few details. B) Formulate options. C) Suggest options with recommendation and take feedback. D) Execute (pretty hard to demonstrate in the 10 minutes max I'd give this). p.s. I'd hire chapped. Good answer! Depends on the numbers of players per team. 'Excuse me, I'm just waiting for excel to open and my math wiz buddy in accounts to pick up to verify my calculation. I'll get back to you in two minutes with the answer.' 'Excuse me, I'm just waiting for excel to open and my math wiz buddy in accounts to pick up to verify my calculation. I'll get back to you in two minutes with the answer.' I would try and be creative and put my suggestions in front of them while I give them a reason for all the options that I choose. For instance, I'd say I would create 5 levels for each game as adding more levels makes the game more challenging and interesting. I wouldn't want to set up too many games as it would require a lot of overhead using up a lot of resources for organizing large number of games. Hence, in each round I would eliminate 20 participants. That would make 100 players getting eliminated after every game. After every 10 games, I would allow all the eliminated contestants to battle it out and 15 can re-enter the game as the eliminated ones would get a chance to observe, learn, refresh and get a second chance... (The interviewer might stop me eventually before it gets too long). Even though my answer was too long, I think I would show them how my logical thinking works. They would see that I am thinking aloud and in the end all that matters is how we approach the problem rather than giving them a vague answer with no reasoning. Show More Responses Oh and I agree with Toasty, I would hire chapped :) I like the way they think. Oh and I agree with Toasty, I would hire chapped :) I like the way they think. i dont know how everybody else thinks butI divided by two with an extra game for when the number is odd and came up with 5627. the question was straight forward, " How many games".... I would have been startled as well, just reading it. Congrats on keeping your calm! My intuitive answer would have probably been: Game theory - 1 or rather none - when it's a battle to death everyone loses, even the winner (last man standing, howling at the moon). I would have likened it to the company and customer situation - in good company EVERYONE is a winner (win-win scenario). Good luck with your endeavors! 1. I think, this question is for a management position. The size of team is given as a too big random number. One cannot control a team of 5627. Divide them in a measure size, e.g size of 10 or 20 (any measureble size). With 10, there will ve 563, which can be further divided into 10, leading to apx 57. that can be futher divided into 10, leaving it to 6 teams. proformance/ goals can be set and can be evaluated later. 2. or do a marathon. None. Our PC world demands everyone gets a trophy. ********************* Keep in mind the position, Nathaniel is right. YOU are organizing the tournament, make a rational decision and describe it. Your answer should be formulated to convey a skill. For example I might suggest something like this: 1 on 1 round robin, 5,623 players, SUM(1+2+3+...+5621+5622) games Quick and simple, shows some knowledge of algorithms but not very practical. The 1st participant plays 5622 others, the 2nd plays 5621, until the 5622nd plays 1 (5623rd participant). Notice you don't add 5623. Participant with the most wins is the champion. Supposed it is a single elimination. should be 5626 games because the first row would be 5623/2 = 2811.5 which means 1 person must be going to the next round to compete therefore we will have 2812 contestants then the second row would be 2812/2 = 1406 the third row would be 1406/2 = 703(odd number which means the one of them is going to the next round without a fight. ect... I believe I would have responded with "Is that how many applicants there are for this job?" Followed by "One game, one victor." Before answering you should read the interview report this question is linked from, where the question is explained in more detail -- """If you had 5,623 participants in a tournament, and each participant had to play games until he/she one or lost, and every game had a winner and loser, how many games would have to be played in order to determine the winner of the tournament""" So it's pretty unambiguous: Participants are individuals; the tournament is single elimination; games involve just two parties (a winner and a loser.) The question doesn't ask about the number of rounds involved, nor about timeframe. Just the sheer number of games. So we're left with an answer of 5622 games (because every game has one and only one loser and 5623 - 1 participants need to lose for there to be a single participant left as the winner, so that's how many games there must be.) Show More Responses Assuming in each game "n" people participate and there is just one winner in a game. If N is the total number of people (in this case 5623), then the approximate number of games would be: log(N)/log(g) -1 After each round, you would have half the number that started the previous round; except if it were an odd number it would he half + 1. So 13 rounds. 2812 1 1406 2 703 3 352 4 176 5 88 6 44 7 22 8 11 9 6 10 3 11 2 12 1 13 It is far simpler than you guys are making it out to be. In ALL single elimination tournaments there is one less game than the number of participants. Because in every game 1 team gets eliminated. And at the end 1 team has to be left standing. This will be a detailed explanation. Since they're asking for a tournament, that means one on one matches, and eliminations of 'participants' or players. With such a large number doing a round robin style tournament would not be very efficient, as every player would have to play every other player, ((N-1)^2)/2 =15,803,442 matches. I would first start to get more details of the tournament. If it were up to me to design the tournament and easily determine the number of matches, I would go with single elimination bracket because since its based on power of 2. Contrary to what what Bob InNorCal did, you dont start halving the at the beginning with 5623, because its not power of 2. You will get to a point where there wont be even numbers, and BYEs will have to be given, it would be unfair to give byes out at the end or middle of the tournament, players would complain that others got BYEs and they didnt. Detailed Explaination: In a single elimination bracket, the brackets end with 1 match between 2 players, then 2 matches and 4 player, and eventually for this tournament, 4096 matches between 8192 players. But since we dont have 8192 players, there will have to be BYEs which wont count as matches. In this case we'll have to use a 8192 single bracket, with the first 4096 players spread every other position then the last 1527 players spread evenly throughout the bracket and the remaining 2569 positions are BYEs (4096 + 1527 + 2569 = 8192). Here are how the rounds look like: A - 1527 matches, (4096 matches was suppose to happen, but 2569 BYEs are no counted) B - 2048 matches, C - 1024 matches, D - 512 matches, E - 256 matches, F - 128 matches, G - 64 matches, H - 32 matches, I - 16 matches, J - 8 matches, K - 4 matches L - 2 matches, M - 1 match From round B-M, its (2^12)-1 = 4095, so with Round A, its 4095+1527=5622. The calculated answer 5622 is one match less than the number of participants. However, just because its easy with single elimination to determine the number of matches, does not mean that it is what they initially asked, conversing with the interviewer for more details of the tournament is important. If the interviewer said it was a double elimination tournament then a player would have to lose twice in the bracket to be eliminated from the tournament. Depending on the number of players and the placement of the BYEs, then calculating the number of matches in a true double elimination maybe difficult. Also in Double Elimination, the winner maybe won by someone without a loss or with 1 loss, the winner without a loss means one less game. it is assumed that the competition is a head to head knockout competition like wimbledon. the only correct answer is 5,622. the quickest and smartest way to get that answer is to see that in a knockout tournament every body loses once and only once, except the winner. in every match, one person loses. therefore the number of matches required equals the number of players minus 1. PaulO, Jordan, madhur, simplebrain and key2success you are all hired! How many games are played? Well all of them of course, how else do you find the winner. It need only one game to find the winner If it is your game you can appoint a winner without playing any games. Then 0 is a possible answer. 1 is also a viable answer, if it is a game of life and death it is possible that no one lives then there are no winners on the first round. Such as a game that on the first round were everybody is exposed to a nuclear blast. Even those exposed to the fall out might be considered losers. A game like chess where you eliminate draw contestant then 1 round to 13 rounds would be required. You could divide by 2 to get the rounds or use logarithms to convert 2**13 which is greater than 5,623 to a log equation such as log(5,623)/log(2). Potentially everybody can loose on the first round because of the draw rule and the person with the bye on the first round can loose because they have no one to play to win. You can change the rules so that infinite rounds are required to determine a winner. One way to do this is to increase the elimination rounds so that it approaches infinity. As this goes on your interviewers glaze over and fall to sleep and when they wake up they decide not to give you the job. This can be poker tournament (as too many players). From 5-7 people - one wins. This can keep number of games to reasonable limit. Depends on how many rounds there are to determine a winner. Logical answer here could be 5623 Show More Responses 5,623 - each person plays, person with the highest score wins. There would be 5613 head-to-head games cosisting of 12 rounds with one team receiving a bye in each round except the fourth round, the tenth round, and, of course, the twelfth round. Assuming the tournament is Mortal Kombat everyone knows that earth realm has to win because out realm is evil and evil... Sucks. No games necessary. What kind of tournament? There will be 2 games. The first game is a question: Would you like to play a game? Which is then followed by the second game - one involving thrones. One 2 The simple answer is Number of Teams minus ONE. Simplified, a four team bracket plays 3 games. |

### ASIC Verification Engineer at Zoran was asked...

You have 2 pieces of rope, each of which burns from one end to the other in 30 minutes (no matter which end is lit). If different pieces touch, the flame will transfer from one to the other. You cannot assume any rope properties that were not stated. Given only 1 match, can you time 45 minutes? 49 AnswersTake one rope (Rope A), place it down as a circle. Light match and start burning rope A at the tips that are touching. When the rope completely burns out, 15 minutes will have passed (since both ends are burning and being consumed at once). Hold the second rope (Rope B) straight and place one end so that it will immediately catch fire when the two burning points from (Rope A) finally touch and are just about to burn out. Thus 15 minutes on Rope A + 30 minutes on Rope B gives you 45 mins. How about this: Fold the first rope double so the ends touch. Lay it down and lay the second rope so it touches the fold of the first rope. Light the ends of the first rope. After 15 minutes the second rope should ignite. Once second rope finishes burning it is 45 minutes. Same principle as above, I just don't want to sit there for 15 minutes in order to light the second rope.... :-) Make a T. Simple Show More Responses Light both ropes at the same time with the match: ------------* () *---------- || then place the two ropes next to each other with the burning ends opposite each other: this way one of the ropes burns left to right, while the other is burning right to left. ----------* *---------- In 15 min the two burning ends will be next to each other. -----* *----- Great Puzzle, thanks! ** You cannot assume any rope properties that were not stated Burn like this *-------- ===> After 30mins, Rope A finished burning, and both ends of Rope B start burning burn one rope, wait till it gets to the half way point, then you transfer the first one to the second one to initiate the other flame. wait till the end. 45 minutes are up You have 2 pieces of rope, each of which burns from one end to the other in 30 minutes (no matter which end is lit). If different pieces touch, the flame will transfer from one to the other at the point at which the burn rate consumes the first rope to its point of contact with the second rope. The only thing we know about the second rope is that it will burn in 30 minutes if ignited from one end. There is no assurance the second rope will ignite, or burn at the stated rate, if an end is not made to point of contact. do you want to buy saints jerseys,vikings jerseys? depends on what the ropes are made of @eaasy you can assume that the second rope with burn at the same rate if lit from the middle, as the rope burns a set amount of time from either end. If Point A to Point B is the same as Point B to Point A, igniting from the center would cause the flame to finishing burning Point A AND point B at the same time. As for ignition, you have a point. There is no assurance that the first rope does not have a prolonged burn rate at the ends and an accelerated burn rate in the center. Therefore the second rope could have ignited before the first rope finishing burning (at any point before the 30 minute limit) which would make timing 45 minutes improbable. Lay one down, and have the second one touch the first rope perpendicularly at the midpoint. Light the first one, when it gets to the midpoint (15 minutes), the second will start burning. When the second one extinguishing you have 45 minutes. T. Simple, quick, walk away and do 45 minutes' work where you can still see light from the flame. Both ends will finish burning at the same time if environmental conditions are consistent. LAY THE TWO ROPES WHERE THE SECOND ROPE IS TOUCHING ONE END TO THE MIDDLE OF THE FIRST ROPE. IT SHOULD TAKE 15 MINUTES TO START THE SECOND ROPE AND THEN ANOTHER 30 TO COMPLETLY FINISH BURNING SO THIS IS 15+30=45 Show More Responses lay ropes end to end, to make one long length, when first length ignites the end of the second length, touch the last remaining end to it and let the flames meet somewhere in the middle. That way each flame will travel in the correct direction, are considered to be burning end to end, without ambiguously being lighted in the middle and hoping the direction will be as desired.... since it takes 30 minutes end to end, then to ignite both ends of the second rope will draw a 15 minute interval no matter what the burn rate is per divided section, thank you and I will accept $150 K per year and 30 days vacation time yearly as well... I had no idea that rope burning was so lucrative a career field... and I want my own rest room. it all depends on what kind of match you have. They said you have one match (no book or box) so if it's not a strike anywhere match then you're screwed. However, if it is then make the T formation... The vertical rope burns 30m then the horizontal burns both directions for 15. No loops, circles or folding required :-) burn A at the two ends will give you 15 min, and when finish A, then burn B next. Lay the map in a T formation, light two ends of the horizontal rope with the match. when the flames meet in the middle in 15 minutes they will light the second rope which is perpendicular to (and touching) the first rope. That rope will take 30 minutes to burn. Hence, 15 plus 30 = 45 minutes. burn A at the two ends will give you 15 min, and when finish A, then burn B next burn one rope, wait till it gets to the half way point, then you transfer the first one to the second one to initiate the other flame. wait till the end. 45 minutes are up Helpful Answer? Yes | No Lay the two ropes in a "T" shape. Burn the one rope at the bottom end of the "T". It will take 30 mins for the flame to go through first rope, and when it reach the other end, transfering the flame onto the second rope right at the middle with the flame move towards both directions, taking 15 mins to completely burn out. This is a silly riddle.... because we're not given the property that the rope can bend, the best formation for the two ropes is to lay them in the "+" formation that everyone got. However, no one mentioned burning the rope at the intersection of the two ropes, ropes will burn in 15 minutes. Also, there isn't anything saying you can't light the ropes more than once with one match (assuming you have the ability to light the match, as someone already pointed out as a significant problem), in that case, you could burn the ropes up in less than a minute. If the rope can bend, just bundle the ropes up into a ball and light the rope ball on fire... it take 1.3 seconds to burn the ball if you create the most efficient weave... true story, I was there. But of course, if you wanted to make the most efficient use of your time (e.g., if you're lazy), you can just crumple the ropes up together, drop the heap of rope on the ground, and throw the lit match on the pile and be done with riddle (and the ropes... and the match). I agree with a couple people here. In this you cant assume the rope burns uniformly. Therefore loop one rope and put the end of the other so it touches where the ends meet. so you have this C--- light the straight rope, when it burns it'll light the "C" which will burn in half the stated time (15 mins). Just a quick one for all the "T" people, you're wrong... The rope doesn't burn uniformly so you can't measure to choose time. If you could you may as well just cut one of the ropes... Show More Responses Solution: NO YOU CANNOT GUARANTEE 45min. Most likely solution to work: Straight followed by loop. ---O One straight, followed by a loop. Ignite straight first, have it ignite the loop. You ignite the straight, gives you 30 minutes. Loop ignites, both ends may burn at different speeds, but when it finishes it will be 30mn/2. The flame may end anywhere on the loop, not necessarily at the 1/2 way of the rope. Why wouldn't it work: Above, you're assuming the rope burns end to end. If the rope is really non-uniformly burning, you can devise a case where burning on both sides at the same time doesn't make the burn 1/2 the time: For example, the outside burns real fast and the core real slow: no matter where you ignite it, or how many times, it would burn in 30mn. For that case, you could split the rope at their diameter and maybe rig a solution. But likely there can be another counter example that can be built, more and more far fetched. Interviewer will want you to explain the thoughts, non-uniform burn, etc. So this is an opening to check your deductive powers. What's the purpose of such a simple question? To judge how you react and explain yourself? Fold each rope in half & mark them in the middle. Light 1 rope & when it reaches the middle light the other one. When the 2nd rope burns completely you'll be at 45 minutes. no comment Too easy... fold one of the ropes in half end to end with the full rope... viola! There are a couple of answers. It depends how you would interpret the question. "the flame will transfer from one to the other" Assuming no special materials you would have on the ropes. This phrase can be interpreted that THERE CAN ONLY BE ONE FLAME. Thus the flame moves from one to the other and does not multiply i.e you can have more than one rope or section of the rope burning .(If it could the solution is easy-just arrange both ropes end to end into a pentagon shape or do what some people have suggested above). for the 45 mins thing you do the T shape like people above suggested so 15+30 =45. If you assume you HAVE to take 45 mins this is the best way. Assume that an earlier finish is best, the Pentagon would be fast. -When I mean pentagon I mean the 5 pointed star thing you know from the occult.- If there was only one flame which could transfer from one rope to the other without creating multiple flames you would have to ask how fast does the flame transfer and what distance does the flame burn at before being transferred. The reason for this is, if the ropes are lied down next to each other TOUCHING each other then the flame would burn, transfer, burn, transfer etc. this would take 30 mins assuming that there is no time lapse on the transfer(hence my questions above). If you wanted to hit the 45mins like it said in the question just have both ropes lie next to each other but have the second rope start from the middle of the first. So after 15mins the second will burn and the flame from this point the remainder of the first rope(15mins) and the second rope will burn 30 mins. Sorry for the poor spelling . Cheers Take one rope, fold it over so one end touches the other, then cut the rope in half. Place one half of the rope at the end of the second whole rope so their ends are touching. Then light either end for 45 minutes of burn. I can't assume rope properties, but nothing says I can't cut it in half. Fold Rope A in half. Lay it down next to Rope B, which is not folded at all. Light Rope B. Reminds me of some of the stupid questions in the Corvirtus test. I took one recently and felt like a 3rd grader. I'm Schmart, I swer. Make a plus sign the the ropes. Light any one end. 15 mins to get get to center, 30 to burn the other rope. Show More Responses Must be a crew of techies answering this question. The obvious answer is to look at a clock. lay both pieces of rope down so that end of rope A touches the end rope rope B light the end of either rope, after 30min have passed the flame will transfer to the other rope,when the rope is half burn 45 total min will have passed. so yes you can time 45min. 45 minutes is easy. I would rather have an hour of light. Tie the ropes together, making one long rope. Light one end and get 1 hour. The Question is "Given only 1 match, can you time 45 minutes?". Option 1. You have a timing device and are allowed to use it - no prohibition stated in setup or question - of course you can "time 45 min" to the precision of your timing device. Option 2. We know: Ignition can occur at either end. From reaction start to reaction end of each rope takes 30min. We can not assume that the rope burns evenly, can be bent, lit up other than the ends or even transmit the flame to the outside of the rope (Think trapped chemical burn in a tube). Therefore we can NOT build a timing device. Put one rope in a straight line, bend the second rope into an O shape (with both ends touching one end of the first rope), when the first rope reaches its end, 30 minutes will have passed, and ignite the second rope. The second rope, which has both of its ends ignited at the same time, will meet at its center, timing 15 more minutes have passed. -> ///////////O Lay down both ropes evenly side by side. Slide one rope down 1/2 the length of the other then light one of the ends. Make a "T" with the ropes. 1 rope burns in 30 min, that rope is touching the middle of the second rope, and with two equal halfs(30/2) it should burn in 15 mins. ----------- -------------, then push them together First of all, the question is "Given 1 match, can you time 45 mins?"... well you can stand there with a match in your hand and time 45 mins with any stopwatch. Or is the question really can you time 45 mins using that match? Then using the match to create a shadow..... hem hem maybe not. Let assume the question was meant to be "can you burn both ropes in 45 mins?"... well it doesn't say anywhere that the match can only light one rope. I've lit many birthday cakes in my days and I promise that only one match can light a full array of candles. So if the ropes burn in 30 mins. and you light both of them with the one match then yes; the burning of the ropes should be done under 45 minutes. But then again.... maybe the question was "can you burn both ropes in EXACTLY 45 minutes?" Now the real answer I think is that the question need to be more clear :) Light one of the ropes from one side only. when it burns out completely (30 mins) light the second rope from BOTH sides. that will double the rate at which it burns, making it completely burnt out in 15 mins. Total 45 min Show More Responses ***CORRECTION*** didn't realize the "one match only" make a loop and a long rope running out of it (like a key figure) light the loose end of the standing rope. it will burn in 30 min. When it reaches the end it will ignite both sides of the other rope, doubling the rate at which it burns (i.e. it will burn to completion in 15 min only) that's a total of 45 min. Keep the first rope straight, and the two tips of second rope touching one end of first rope. Now lit the other end first rope. So first rope takes 30 min to burn and transmit fire to two ends of second rope which will take 15 min to burn. So total of 30+15= 45 min The answer is simple... Make a sundial with the match and the rope. Can Keep the ropes in this fashion----------》 《--------- And fire dem with opposite side burning. Now when both fire meet it wil be 15 mints...nw both have 15 mints remaining now bur first remain rope which will take 15 mints more and den burn second remaining rope which will take another 15 mints..so total 45 mints..isn't it simple ..;) The question does not state that the items in the question are required to provide the answer so... Take note of your start time. Take the first length of rope and lay it next to the match. Then, carefully inspect the second length of rope for any defects. If none are found, or even if there are defects present, place the rope on the ground in a straight line facing north to south. Place the first length of rope, which is tired to the match, approximately 5 feet to the north of the second length of rope. **Warning: This step requires quick action on your part** While the match is burning, tie the first length of rope to the match. Take the second length of rope and touch the north end to the match while it is still lit. Grasp the second length of rope in your left hand and spin 360 degrees counter clockwise and throw it as far as you can. Now go have a beer and keep an eye on your watch to see when the 45 minute timeframe has finished. |

A frog is at the bottom of a 30 meter well. Each day he summons enough energy for one 3 meter leap up the well. Exhausted, he then hangs there for the rest of the day. At night, while he is asleep, he slips 2 meters backwards. How many days does it take him to escape from the well? 49 Answerstry to answer this question as seriously as u can 28 I agree -- it's 28...because on that morning, he'll be at 27 metres and he can jump to the top in one bound. Show More Responses Answer: 28 Each day he makes it up another meter, and then on the twenty seventh day he can leap three meters and climb out. It's 29 - on the 28th day he can leap 3 meters and hang at the top (but he can't climb higher & out), and on the 29th day he'll leap and be out of the well at 31 meters. 27 At day 0, he jumps to 3m. At day 27, he jumps to 30m and gets out. 28. Each night he ends up/starts the next morning at the number of days he's been there (first night, he's at 1 foot, 2nd night he's at two feet). Hence, on the 28th day he jumps 3 feet to 30 feet. 28 days. At start of day 27, he jumps 3m to reach the top of the 30m well but has no energy left to climb out. At start of day 28, he jumps another 3m and entirely out of the well. Never....the frog would be dead by day 10 since nothing to eat or drink. 28 Saying 28 is too easy. What about the external (1) and internal factors (2): (1) there might be a heavy rain during the first night and the frog can easily float up or, ..., drown, at all; (2) the frog may decide it's sunday - let's have a rest and spare the energy for a bigger jump on the next day... 27 days Day 1 - It jumps 3 meters. 0 + 3 = 3. Then falls back 2 at night. 3 - 2 = 1 Day 2 - It jumps 3 meters. 1 + 3 = 4. Then falls back 2 at night. 4 - 2 = 2. ... Day 26 - It jumps 3 meters. 26 + 3 = 29. Then falls back 2 at night. 29 - 2 = 27. Day 27 - It jumps 3 meters. 27 + 3 = 30 This question is ambiguous whether if the frog is able to get out when it reaches the top or if it needs to exceed 30 meters to climb out. Assuming it doesn't die of starvation, the answer is 28 days.* start of day 1 (0 days elapsed): 0m --> 3m (then falls back 2m by start of day 2) start of day 2 (1 day elapsed): 1m --> 4m start of day 3 (2 days elapsed): 2m --> 5m ... start of day 28 (27 days elapsed): 27m --> 30m start of day 29 (28 days elapsed): 28m --> 31m In other words, 28 days will have elapsed before the frog can jump to a height exceeding 30m.* * This answer assumes the frog is not able to walk away after it hits 30m. I would assume it has no energy left to climb out based on the problem description. If the questioner disagrees with this assumption, then the answer is 27 days. Show More Responses 28 27 ..... he will jump 3 meters this day to get out!!! The math certainly says 27, assuming he only needs to get to 30m to actually get out. its easy to forget that, as pointed out by Paul, he can jump at the beginning of the day, therefore he can reach 3 meters in 0 days. I like the out-of-the-box notions presented by HB and nic. Maybe croaking could get him some help from someone. Maybe he could get up in a well bucket. Why does he want to leave? -- maybe he has everything he needs there and is safe from predators. Why does he slip down? Can he stop that? I wonder what they're looking for in a question like that. I wonder if it really helps them choose good candidates. I wonder who's going to bother reading this. No math necessary. Frog is dead after a few days. 27!! as he will be out on the same day. it will be 28 if he spends the night as well. and c'mon, frog is not based on a binary system, that he wont have enough energy after the last 3 meter, he sure will be motivated enough to take the 3.000001 meteres on the last day to get out of the damn well. 27 days It is 28 assuming reaching 30 ft gets him out of the well, people saying 27 are making the error of assuming there is a day zero, when counting days as with years there is no 0. 28. 27 days to get to go to sleep on level 27. Next day outa da hole. Most frogs have to surface for air. After a short amount of time they will die without air unless they are in aerated water and can absorb through skin. So ask what species is the frog. Day 1: 0 + 3 = 3 - 2 = 1, D2: 1 + 3 = 4 -2 = 2, D3: 2 +3 = 5 - 2 = 3....D28: 27 + 3 = 30 -2 = 28, D29: 28 + 3 = Eaten by the bird that has been waiting for him.... Show More Responses bllshit If Mr. Frog manages to make it up the wall another meter everyday then, on the 27th day, he can leap three meters and climb out, the answer would be 28 days... 28 days. As the frog slips 2 meters down every night by the 27th day he has climbed 27 meters. On 28th, the frog will start from the point of 27m which means start climbing 3 meters foward and this way he reaches his 30m to get out. 28days NB - he does 3m/day, but the result in the next morning b4 his next jump is 1m jump the previous day In the morning of day 26 b4 he jump, he has 5m left...meaning he has done 25m for 25days. On day 26, he jumps 3m, leaving him wit 2m to go but cos he sinks 2m overnight, the4 his resulting meters to go is 4m On day 27, he jumps 3m and left with 1m but sinking 2m more means the resulting meters to go is 3m On day 28, he complete his 30m and walk out from the well victorious provided no predator in the well and he didnt die of starvation 28 days... kermit the frog? How much water is in the well? Frogs need water to survive. What about food? 27th day How did the frog get in the well in the first place? What is motivating him to get up? How does he know he can get out? If the frog slips two meters every night for 27 nights he will be very sad and wont try anymore so he will never get out. Wait....I am this frog! x = number of days A = 30 meters x+2 = A x + 2 = 30 x=28 Show More Responses Thirty days as he is only moving one meter a day. What kind of Frog is it? Is the well full of Water? My first thought is that the frog would not survive, Food and air seem the best bet for Frog-a-cide. Of course if the question is mathematical only in nature then you would have to follow the logic above x=number of days y=30 meters x+2=y x+2=30 x=28 Hey guys ... Its simple don't break your head.... 30 feet well OK 1 day 3 feet jump and sleeping and falling down to 2 feet.. So the frog can only climb up 1 feet per day... so on 26th day it was in 26 th feet and jumped and it reached to 29th feet and sleeping and falling down to 2 feet down.. so on 27th day it was in 27th feet jumped and it reached to 30 feet and went out of the well.. bec after reaching the 30th feet why the hell does the frog gonna sleep again. So it took 27 days for the frog to come out of the 30 feet well. Question created by me... so answered it....lolzzzzzzzzzzz Answer: Day 28 Day 1 - It jumps 3 meters. 0 + 3 = 3. Then falls back 2 at night. 3 - 2 = 1 Day 2 - It jumps 3 meters. 1 + 3 = 4. Then falls back 2 at night. 4 - 2 = 2. ... Day 26 - It jumps 3 meters. 25 + 3 = 28. Then falls back 2 at night. 28 - 2 = 26. Day 27 - It jumps 3 meters. 26 + 3 = 29 Then falls back 2 at night. 29 - 2 = 27. Day 28 - It jumps 3 meters. 27 + 3 = 30 Note: Assume after the first leap that his hind legs are exactly three meters up the well. His hind legs must clear the well for him to escape. Answer: Day 29 Remember that on Day 1 the frog ends at 1 meter, Day 2 the frog ends at 2 meters, ... So on Day 27, the frog end at 27 meters. On Day 28, the frog goes up to 30 meters, then back down to 28 meters. On Day 29 the frog finally makes it to 31 meters (out of the well). This will help if you are still not convinced: int height = 30; int curPos = 0; int days = 0; while(curPosheight){ break; } curPos -=2; if(curPos>height){ break; } } System.out.println(days); } 28 ITS ONLY 10 DAYS. Because, if he waits 10 days, he will summon the strength to jump 30 meters in one jump to the top of the well. Ya'll can't think outside the box... Cool 28 28 Show More Responses I know this isn't the right answer but if u think VERY VERY logically, well, he doesn't. 28 30 days 39 Day 1 -> Total distance Covered = 3-2 = 1m Day 27 -> 1*27 = 27 Day 28 -> 27+3 = 30 and the frog and he escapes 28 will be the final answer |