Super Strong Eggs -- III

For super strong eggs II we started from n, then n-1, n-2, n-3 and so on.

For this one, since this has three eggs, we do it somehow like powers of two, like this: n, n-1, n-1-2, n-1-2-3, so on.

n=100, so the first egg drops at 16,44,65,80,90,96,99,100

If the egg breaks at 16th floor, using the method to solve super strong eggs II, we need 7 drops 44th floor, 9 drops 65th floor, 9 drops All others, 9 drops 100th floor, 8 drops

So the answer is $\boxed{9}$

You may ask why will all the other floors require 9 drops, and this is because the numbers we minus.

First think if it breaks at 99th floor. We dropped 7 times. We still have two more levels, 97 and 98. Drop at those 2 floors. Then think if it breaks at 96th floor. We dropped 6 times. We have 5 levels. Using the method to solve super strong eggs II, we need to drop 3 times. 6+3=9 Then think if it breaks at 90th floor. We dropped 5 times. We have 9 levels. Using the method to solve super strong eggs II, we need to drop 4 times. 5+4=9

See the pattern? For all the levels, there are nine drops since we minus triangular numbers. And we're done.

Well this is the way I went about solving this. Keeping the principle of reducing balance in mind at higher level floors i worked it backwards. Starting with 2 eggs broken leaving floors 1-7 to try out step by step with the third egg, Floor 8 is where I start and work upwards to find out which floor my first egg drop should be. It turns out to be floor 36.

A) First egg drop from floor 36--- if egg breaks -- drop from 8 ( if breaks check 1-7 step by step) --if not-- drop from 15( if breaks chk 9-14 step by step) --- if not-- 21( if breaks chk 16-20)---If not-- 26 (if breaks chk 22-25)--if not --30( if breaks chk 27-29)--if not drop from 33 (if breaks chk 31, 32) If not chk 34, 35.

If first egg does not break when dropped from 36-----

B)Drop the first egg again from floor 64--- If breaks drop from floor 43(if breaks chk 37-42) -- if not -- drop from 49 ( if breaks chk 44-48) -- if not -- drop from 54(if breaks chk 50-53) -- if not -- 58 ( if breaks chk 55-57)-- if not -- 61( if breaks chk 59, 60.) If not chk 62,63.

If first egg didnt break when dropped from 64------

C) Drop first egg from 85--- If breaks drop second egg from 70(if breaks chk 65-69)-- if not drop from 75 (if breaks chk 71-74)-- if not -- 79( if breaks chk 76-78) if not-- 82( if breaks 80,81) If not chk 83, 84.

If first egg didnt break when dropped from floor 85----

D) Drop first egg from floor 100 (If doesnt break-- experiment over)-- If breaks -- drop second egg from floor 90( if breaks chk 86-89) -- if not -- 94( if breaks chk 91-93) -- if not -- 97 ( if breaks chk 95, 96) if not check 98, 99.

9 drops total in any and every scenario. I am sure some recursive formula could be whipped out given any number of floors and eggs by a good computer scientist.

let f(n)= sum of n+ve integers frm 1 to n, let x be some +ve integer, then the best way to split is to follow this, f(x)+f(x-1)+f(x-2)....+f(1) > 100 then the minimum value of n is 8. so be dropping the another egg at the points of the function f(8), v end up with 9 drops

another solution: if u solved the other 2 problems, u definitely know that the answer will be less than 12 and greater than 7, so it can only be 8,9,10,11. as u have 3 chances easy solve :p

PS:I not so good at explaining