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MMM #28 Winner

The winner for this edition of MMM is Sue VanHattum. I thought this was a pretty fun problem. I didn’t arrive at the correct answer until a couple of faulty tries. Thanks for everyone who submitted!

We got some really cool explanations, and a sweet diagram from Vladimir Filimonov, which is shown below.

Hope you guys enjoyed this round of MMM, next week is Sol’s turn over at wildaboutmath.com. Be prepared!

The Answer by Nate Burchell

Suppose you have only one MacBook, because if and when the first one breaks, you will have only one. If you have only one MacBook, you will need to increase by consecutive heights in order to be certain of the maximum height from which a MacBook can be dropped. Prove this by supposing that you test at non-consecutive heights. If a MacBook survives a drop from the fourth floor and breaks when dropped from the sixth floor, we have no way of determining (since we have broken our one-and-only MacBook) if the fifth floor drop would have been survived or not by the MacBook. Therefore, we must proceed by consecutive heights when we have only one MacBook.

With two MacBooks, we can optimistically skip a number of floors. In case of breakage, we resort to the simpler procedure for one MacBook.

The method for finding the maximum height with two MacBooks and n drops needs to acknowledge that if the first drop breaks one of the MacBooks, we have only n-1 drops to check the remaining heights. Therefore, there should not be more than n-1 heights left to check. Drop the first MacBook from the nth floor. If it does break, we have n-1 drops left to check (beginning at floor 1 and moving up through consecutive floors) the n-1 heights that we skipped. If it does not break, we have n-1 drops left, so we drop from (n)+(n-1)=2n-1, knowing that in case it breaks on the second drop, we still have n-2 drops with the remaining MacBook to work through the n-2 heights between the nth floor and the (2n-1)th floor. This continues and we get n+(n-1)+(n-2)+(n-3)+…+3+2+1=(n(n+1))/2 total floors tested by n drops.

With n=8 drops, we could test a maximum of 8*9/2=36 floors. With 7 drops, we could only test (at most) 7*8/2=28 floors, so we will need at least eight, and eight is sufficient.