2 Eggs 100 Floors Puzzle

2 Eggs 100 Floors Puzzle - 18 july

-> You are given 2 eggs.
-> You have access to a 100-storey building.
-> Eggs can be very hard or very fragile means it may break if dropped from the first floor or may not even break if dropped from 100 th floor.Both eggs are identical.
-> You need to figure out the highest floor of a 100-storey building an egg can be dropped without breaking.
-> Now the question is how many drops you need to make. You are allowed to break 2 eggs in the process


For Solution: Click Here

55 comments:

  1. from 100th till first floor it will not break.

    however if we are allowed to break two eggs in the process, we can through both of them.

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    1. whoever said we actually had to drop it out the window? couldn't you technically just drop it on a pillow on the 100th floor? and say that you have to drop it out the window. dropping the egg doesn't actually break it. it's just the egg hitting the ground after you drop it that breaks it. and it asks how many drops YOU need to make. couldn't you just make your friend do it? and-ok i'll stop.

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    2. This question is flawed. It says the eggs could be of a different hardness to one another. therefore the floors they can handle will be different and it didnt state which egg was to be measured. Now the second question was how many eggs you would need to use, not how many you are allowed. obviosly you need as many as it takes, dropping them from the 100th floor, untill one doesnt break, to say its floor 100.

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    3. They are identicle, toughness included.

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  2. Just one drop should be sufficient -- from first floor only. If it do not break, this is the hard egg. Else the other egg is the hard one.

    (Presuming we are given two eggs -- one hard and one soft; hard egg WILL NOT break even if dropped from 100th floor; soft egg WILL break even if dropped from first floor.)

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  3. Hey nkaini ....please read the puzzle again

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  4. 14 drops at max. first drop the egg 1 from 14th floor if it breaks then with the egg 2 try each floor from 1 to 13.
    if egg 1 doesn't break from floor 14, next try would be from 14+13 = 27th floor. if it breaks, try 15th to 26th with egg 2.
    if egg 1 doesn't break from floor 27, next try would be from 27+12 = 39th floor. if it breaks, try 28th to 38th with egg 2.
    So, keep trying like this for 50th, 50+11=61th, 61+10=71th, 71+9 = 80th, 80+8=88th, 88+7 = 95th, 95+6 = 101th >100(total).

    So, answer is basically sum of n numbers == 100.
    i.e. sum of 1 to 13 is 98 and sum of 1 to 14 is 105. So, to cover the 100th floor we need 14 drops.

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    1. way to totally over think it lol...

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    2. Right you start at the bottom, if it doesnt break you go up one floor, as simple as that...

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    3. Waldo scholtz, the problem is that, "Eggs may break if dropped from the first floor or may not even break if dropped from 100 th floor."
      that means even you drop it at the 1st floor, the possiblility of the egg breaking is still there, with 100th floor, 2 eggs is not enough.

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    4. this does not make sense i am going to check the solution

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    5. Great answer! This is easier to understand than the admin solution. Well explained!

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    6. 14 is not the best option!
      If us go with 10 floors each time, it will be 5 drops
      plus if egg does break, for exaple from 20th floor,then we will go and drop it from 11th floor, then 13th, 15th and so on, so in sum it gives us about 10 drops, if anywhere in the process egg breaks, u know that max floor limit is 1 less then current.

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  5. A correction in the above post.. sum of 1 to 13 is 91*

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  6. ==>> 14 <<===

    Let x be the answer we want, the number of drops required.

    So if the first egg breaks maximum we can have x-1 drops and so we must always put the first egg from height x. So we have determined that for a given x we must drop the first ball from x height. And now if the first drop of the first egg doesn’t breaks we can have x-2 drops for the second egg if the first egg breaks in the second drop.

    Taking an example, lets say 16 is my answer. That I need 16 drops to find out the answer. Lets see whether we can find out the height in 16 drops. First we drop from height 16,and if it breaks we try all floors from 1 to 15.If the egg don’t break then we have left 15 drops, so we will drop it from 16+15+1 =32nd floor. The reason being if it breaks at 32nd floor we can try all the floors from 17 to 31 in 14 drops (total of 16 drops). Now if it did not break then we have left 13 drops. and we can figure out whether we can find out whether we can figure out the floor in 16 drops.

    Lets take the case with 16 as the answer

    1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
    1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
    1 + 13 45 .....
    1 + 12 58
    1 + 11 70
    1 + 10 81
    1 + 9 91
    1 + 8 100 We can easily do in the end as we have enough drops to accomplish the task


    Now finding out the optimal one we can see that we could have done it in either 15 or 14 drops only but how can we find the optimal one. From the above table we can see that the optimal one will be needing 0 linear trials in the last step.

    So we could write it as

    (1+p) + (1+(p-1))+ (1+(p-2)) + .........+ (1+0) >= 100.

    Let 1+p=q which is the answer we are looking for

    q (q+1)/2 >=100

    Solving for 100 you get q=14.
    So the answer is: 14
    Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (i.e. move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100)

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    Replies
    1. ahhhh...I see. Not really. acially i didn't get anything

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    2. your answer makes a lot of sense, but i don't think it fittingly answers the riddle. lets say you dropped it from the 14th floor and it broke, and then the second one broke when it was dropped from the 1st floor. then you lose both eggs; in which case the answer is 2. it could break on the 2nd floor; in which case the answer is 3. so there is no definite answer as far as mathematics goes.

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  7. start with 1st floor and go on up till it breaks...

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  8. This comment has been removed by the author.

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  9. The question is not well posed. Are we trying to minimize the expected value of drops or the maximum possible number of drops. These problems are not equivalent. It makes more sense to minimize the expected value. Also we need an assumption about the probability distribution of number of floors an egg can be dropped off without breaking. If it is for example overwhelmingly more likely that the egg breaks somewhere between 91-100 than between 1 and 10 minimizing the expectation will have different solutions.

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    Replies
    1. Plz chk the soln at

      http://solution-dailybrainteaser.blogspot.com/2011/07/2-eggs-100-floors-puzzle.html

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  10. how does one account for wear and tear on the eggs? after you drop it once it it no longer an intact egg, and would break probably break more easily. @L@

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  11. just 1 from the 100th floor if it breaks it is the soft one if it does not, then it is the hard one..

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  12. Bunch of idiots its only one according to the puzzle defying the fact the egg becomes vulnerable with each fall we would start from floor 1 and proceed to the next floors unti the egg breaks
    -quoted by a Havard graduate

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  13. Drop an egg at each floor 10, 20, 30, ..., 100 until it breaks. Then drop the second one at (breaking floor-9) going up single floors until it also breaks.

    You could generalize this to drop every log_b (100) floors and then drop the second from the last safe floor up until the breaking point.

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  14. once an egg cracks you can't drop it again - & you can only break two eggs in the process - the solution doesn't make sense.

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  15. 0 eggs need to be dropped, the highest floor an egg can be dropped from = 100

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  16. the 100 floor its the highest

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  17. couldn't it be 6?

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  18. in fact you only need one egg...you start on the 1st floor and drop the egg if it breaks than you got your answer, if it doesnt you go and pick up your egg and go to the 2nd floor...repet proces until egg breaks

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    1. But in the worst case, you would drop an egg 100 times, which is not the minimum number of drops required to find the correct floor.

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  19. one egg is boiled egg

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    1. yes correct......lol.....

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  20. K i thought u were all over thinking the question but maybe im not understanding it right, it can fall 100 stories before it breaks because it can fall all it wants. It breaks on the ground so you can go as high as u want with it and (with no obstacles) it wont break till it comes in contact with the ground. Am I right? Or am I misunderstanding the question it would take one drop....? lol

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  21. It's one. More amusingly, you can take it to the 100th floor, and drop it an inch from the floor, because the floor is now 100 floors up, you can drop the egg really clsoe to the floor and it will not break...

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  22. Very easy. Since the eggs are identical they will both break at the same floor. Start from the first floor and see if the egg breaks, if so this is the maximum floor, if not then continue on to the next floor until the egg breaks, this is the maximum floor on which the egg breaks.

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    1. Doh...no! Look above, the answer's 14 and it most definitely isn't easy! Unless u are a genious of course! Judging by your response, you're not! :-)

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  23. The minimum amount of drops is 14. Drop first egg from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... until it breaks (or doesn't at 100). when it breaks use the second egg to increment up by 1 from the previous until it breaks. worse case 14 drops.

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  24. The minimum amount of drops required to determine this value is 9 (if egg breaking range is in between 1st to 10 Floors) & the maximum amount would be 18 (if egg breaking range is in between 90th & 100 floors). So if we take average of them, its exactly 13.5 ~ 14 drops.

    So we can start from 14th floor to maintain uniformity if trying on probability basis in a total of 100 floors.

    Here's the example.
    Say If we start from say : 10th floor, then the total amount of drops to find highest dropping value without breaking an egg is 9 if it all the egg breaks @ 10th floor. In this range to identify the breaking of egg, actually we would not be dropping the last drop to identify the floor & instead use deducing formula to isolate the floor where the egg doesn't break. i.e starting from 1st floor & going all the way till 8th floor.

    And the drops keep on incrementing by 1 if the first egg doesn't break @ 10th floor & then we try again from say 20th floor. i.e

    1st attempt : 10th Floor ( Egg didnt break)
    2nd attempt : 20th Floor ( Egg breaks)
    & range in between 10th & 20th Floor.
    Resulting in a total of 10 drops.

    & So on...

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  25. A slight correction in to my comment :

    {
    In this range to identify the breaking of egg, actually we would not be dropping the last drop to identify the floor & instead use deducing formula to isolate the floor where the egg doesn't break. { i.e starting from 1st floor & going all the way till 8th floor.}

    {i.e we will be starting from 9th floor going down all the way till 2nd floor.}

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  26. egg in elevatator press button :)
    Or
    Level 1 drop egg catch it all safe
    Level 100 drop egg on bed or drop and catch it none smash can do both ;)

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  27. I can solve this by 25 trials, this is my logic:
    start with 4, if it breaks, do 2, if it breaks 1 is the answer, if not 3, is the answer, we covered 1234, if it did not break at 4 do 8, and in this way you check 100 by 25 combination

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  28. Simple answer - Point 1 - You are supposed to drop them without breaking them. That means = you have to drop the eggs without breaking them open. So, you have to drop them entirely.
    Point 2 - the highest floor. The 100th floor.
    Point 3 - you can break 2 eggs while trying to determine the highest floor. You are given only 2 eggs.
    So you drop them without breaking them from the 100th floor, and you get to break those two. So, you have found out that the highest floor from which you can drop 2 eggs without having broken them open before dropping them, from a 100 floor building.

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  29. Maximum 19 tries ... Drop the 1st Egg from 10th Floor and then 20th, 30th and so on.... Will see the Max scenario .... If the 1st Egg does not break till the 90th floor that means you have used 9 tries and if it finally breaks at the 100th floor means u have used 10 tries. So you can conclude that the Egg will break between 90th & 100th floor. Now try from the 91st floor and then increase gradually. if the egg breaks at the 99th floor then you would have used a maximum of 9 tries to determine with the 2nd egg which makes a total of 19 tries.

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  30. Started from the bottom now we here

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  31. the riddle states that both eggs are identical which should include everything including how hard or fragile the egg is.It also states that we are allowed to break 2 eggs in the process, so this means that both eggs are fragile.so i think the answer is 0

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  32. i was wrong. the real answer was much more retarded

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  33. Assuming all eggs are identical, we could start from 2nd floor. if it breaks, throw the egg from 1st floor. if this one also breaks ans = 0.
    if it doesnt break ans = 1. If it did not break from floor 2, then we throw it from floor 4, and use same logic.

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  34. it just need only 10 drops......from 19th,37,53,67,79,89,97,100 th floor.

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  35. 100th floor. You are inside therefore it does not matter which floor the eggs are dropped as they will always travel the same distance.

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  36. Allowed 2 make 2 drops..

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  37. How about we drop the eggs in a pan as a side for bacon? That sounds like the smartest choice.

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