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There's a certain kind of egg about which you wonder: What is the highest floor of a 36-story building from which you can drop an egg without it breaking? All eggs of this kind are identical, so you can conduct experiments. Unfortunately, you only have 2 eggs. Fortunately, if an egg survives a drop without breaking, it is as good as new--that is, you can then conduct another dropping experiment with it. What is the smallest number of drops that is sure to determine the answer to your wonderings?

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Posted · Report post

Am thinking

you could always determine which floor with twelve drops? starting with the third floor and skipping every three floors

*** For anyone having spoiler issues, if you type between the tags

[/spoiler ]without quoted text in the leading tag it will work (also leave out the space)***

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8 drops.

Drop from 8th, 15th, 20th, 26th, 30th, 33rd, 35th and finally the 36th (to see if it even breaks at the 36th) floors

Whenever the first egg breaks, start dropping second egg from the next lower floor. The max number of drops will be 8

[/spoiler ]

Still having problems with the spoiler!

Edited by DeeGee
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Posted · Report post

8 drops.

Drop from 8th, 15th, 20th, 26th, 30th, 33rd, 35th and finally the 36th (to see if it even breaks at the 36th) floors

Whenever the first egg breaks, start dropping second egg from the next lower floor. The max number of drops will be 8

[/spoiler ]

Still having problems with the spoiler!

Its correct.......

its goes by formula

q(q+1)/2>=36

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Posted · Report post

Its correct.......

its goes by formula

q(q+1)/2>=36

Hmm somehow I don't see this... if the egg breaks above the fifth floor, how can you determine that dropping the first from the 8th and the second from the 7th?

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Have to agree with the last post - how would you know if it was floor 5 or 6?

Here's the real answer

If you throw an egg out of the 36th floor with the pointy bit at the front and a slight spin, it will end up hitting the ground 'tip' first. The egg won't break as long as that happens - go try it!

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What is the smallest number of drops that is sure to determine the answer to your wonderings?

Best Case Scenario: 2

Drop one from one height, it breaks. Drop the 2nd from the floor right below it, it survives. There's your answer.

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Hmm somehow I don't see this... if the egg breaks above the fifth floor, how can you determine that dropping the first from the 8th and the second from the 7th?

If it breaks on the 8th floor you then try the 1st floor and the 2nd etc until you find where it breaks.

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If it breaks on the 8th floor you then try the 1st floor and the 2nd etc until you find where it breaks.

LOL my bad, yep it still makes 8 drops :thumbsup: bad brain day :P

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Drop your first egg from the following floors until (or if) it breaks:

6th, 12, 18, 24, 30, 35

Worst case comes if the egg breaks on either the 30th or 35th floor.

If it breaks on the 30th floor, then take the second egg and start at 26 and go through 28, for a total of 8. You can skip 25, since you know 24 worked. No need to go past 28 because you know 30 didn't work.

If it breaks on the 35th floor, then take the second egg and start at 32 and go to 33, for a total of 8.

All other breaks result in less than 8.

Good puzzle. Thanks.

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Posted · Report post

Drop your first egg from the following floors until (or if) it breaks:

6th, 12, 18, 24, 30, 35

Worst case comes if the egg breaks on either the 30th or 35th floor.

If it breaks on the 30th floor, then take the second egg and start at 26 and go through 28, for a total of 8. You can skip 25, since you know 24 worked. No need to go past 28 because you know 30 didn't work.

If it breaks on the 35th floor, then take the second egg and start at 32 and go to 33, for a total of 8.

All other breaks result in less than 8.

Good puzzle. Thanks.

This does not work. Looking at your 35th floor example, you test the 35th floor on your 6th attempt. If the egg breaks there, then it must be either 31,32,33 or 34.

If you use your 2nd egg to start at 32, you're not attempting to break the egg at 31. Its possible that if you were to try it at 31, it would still break and the 30th floor would be the highest.

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Have to agree with the last post - how would you know if it was floor 5 or 6?

Here's the real answer

If you throw an egg out of the 36th floor with the pointy bit at the front and a slight spin, it will end up hitting the ground 'tip' first. The egg won't break as long as that happens - go try it!

Thats only If Gravity wasn't pulling it down over 100 miles per hour, praying that kinetic laws will allow it to stay on the sidewalk without breaking, and then hoping that it was spinning?

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Posted (edited) · Report post

8 drops.

Drop from 8th, 15th, 20th, 26th, 30th, 33rd, 35th and finally the 36th (to see if it even breaks at the 36th) floors

Whenever the first egg breaks, start dropping second egg from the next lower floor. The max number of drops will be 8

[/spoiler ]

Still having problems with the spoiler!

shouldnt the numbers be 8, 15, 21, 26, 30, 33, 35, 36?

if it was 20th, then if the egg didnt break on the 25th level, you wouldnt know until you tried 9 times.

8 15 20 26 21 22 23 24 25

Edited by flamedudex
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You can skip 25, since you know 24 worked.

?????

what kind of logic is this?

The Answer is 8.. but here's why..

You can determine the answer logically by working backwards incrementing the spacing between drops by 1 floor each time until you reach the ground.

Drop final egg on 36th floor -1=35-2=33-3=30-4=26-5=21-6=15-7=8-8=0 Therefore start on the 8th floor.. If the egg breaks start at floor 1 and work your way up until you get to 7.. if you do the max drops would be 8.. if the egg does not break on 8 then move up to 15.. If the egg breaks go to the next floor above the highest floor that didn't break and work your way up.. rinse and repeat

The formula x(x+1)=yz y=the number of eggs, z=number of floors, solve for x (the number of drops)... therefore x(x+1)=2*36... x(x+1)=72... 8*9=72.. x=8

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