Riddle me this: Let's say you have 2 phones and a 100 story building. What's the minimum # of drops you can do?

Final exams are a bitch.

Is the building 100 stories plus the ground floor, or 100 stories including the ground floor?

Also, does the building have a seventh and thirteenth floor, or are these skipped?

How high does common sense place lowest. This is a good test of how smart you are as a function of cellphone usage and technology First number that came to your mind when you read the OPs riddle?

Rtfm. Done.

49

binary search is best i can think of

50

bearly trying wrote:

binary search is best i can think of

You only have two misses so you can't do that.

Already answered this in the other thread which got deleted but my answer is 14.

Drop phone #1 at 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, ... until it breaks

Drop phone #2 at the floor above the previous successful drop, and each successive floor until it breaks

phone a friend

assuming what never mentioned.

that phone wont break from over 1 floor drop

wont break from certain min

Since you're lucky, you drop it from the right floor on the first try. However, you aren't certain unless you test the one above it. So you trudge up the stairs to the next floor and drop that sucker. It breaks. You're not really surprised because iPhones break if you look at them wrong. But hey, at least you determined the correct floor, and it only took two tries. To make sure it wasn't a fluke the first time, you go back down to try again. (I mean, you know it won't break anyways, so what's the harm? You chuck that sucker as hard as you can.) Much to your dismay, you realize you just tossed an old Motorola flip phone that wouldn't break even if a tank drove over it. Screw that third throw, you only needed the first two to answer the question anyways.

+1

14 drops for 100 floors

45 drops for 1000 floors

141 drops for 10,000 floors

by dropping phone 1 at floors

{n, 2n-1, 3n-2-1, 4n-3-2-1, ... , n^2-(n-1)-(n-2)...-1}

we can test up to 1+2+3+...+n floors

note that OP said we can assume the top floor fails, so we don't need to test it (though this doesn't change our answer for a 100 or 1000 story building).

to clarify,

n drops lets you test up to 1+2+3+...+n floors.

HardLoper wrote:

bearly trying wrote:

binary search is best i can think of

You only have two misses so you can't do that.

Already answered this in the other thread which got deleted but my answer is 14.

Drop phone #1 at 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, ... until it breaks

Drop phone #2 at the floor above the previous successful drop, and each successive floor until it breaks

So what happens when the phone beaks after being dropped from the 14th floor? Because it would.

HardLoper wrote:

bearly trying wrote:

binary search is best i can think of

You only have two misses so you can't do that.

Already answered this in the other thread which got deleted but my answer is 14.

Drop phone #1 at 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, ... until it breaks

Drop phone #2 at the floor above the previous successful drop, and each successive floor until it breaks

ah right, i didn't think of that

interesting problem though. If anyone knows some math maybe they can explain this for me though (kinda veers off topic)...

i checked out how the solution works and the idea is n + n-1 + n-2....+1 >=100

or n(n+1)/2 >=0

i was bored so i plugged it into quadratic formula just to see and you end up with a negative discriminant. How can we have a real solution to a problem with a non-real root? i don't know much about imaginary numbers or how to use them or anything but i feel like i am missing something here

thats n(n+1) >= 100 not zero in the second part*****

It should be real not imaginary

n(n+1)/2 = 100

n(n+1) = 200

n^2 + n = 200

n^2 + n - 200 = 0

Quadratic forumula solutions are 13.65 and -14.65. As knox harrington the same formula for 1000 and 10,000.

Your fancy math still doesn't explain to me what happens when the first phone breaks after the initial drop from the 14th floor.

14

HKop wrote:

Your fancy math still doesn't explain to me what happens when the first phone breaks after the initial drop from the 14th floor.

You go floor by floor from the first at that point. If it's an iPhone it breaks on floor 1 and you're done.