You are studying the effects of gravity on clay spheres. You conjecture that they will shatter... but at what height? You want to find out the smallest integral height in meters from which the clay will fall and shatter.

Unfortunately, you only have four identical clay spheres, at least until the company that makes them starts returning your calls. Also, you only have enough time for 8 tests, during which the general area will be cleared of people. Last time someone did such an experiment, an egg... well, it was messy. Up to what height can you test the effects of gravity on the clay?

(In reply to re: Possibly right solution, partial expl. by tomarken)

Maybe I'm taking an overly simplistic approach. 

Let #of Trials allowed = t, #of spheres = s, then the maximum you can test is (t-s)*2^s + s^2

If there were only one sphere, then you could only test up to 8 by dropping at 1, 2, 3, 4, 5, 6, 7, and 8.

If there were only two spheres, you could drop at 2, 4, 6, 8, 10, 12, 14, and 15 knowing that any breakage would allow you to fully test the interval immediately below the break.

If there were three spheres, you could drop at 4, 8, 12, 16, 20, 24, 26, and 27; same logic.  As you run out of spheres OR out of trials, you can't increment as much.

This formula gets you 8, 16, 24, 32, 40, 44, 46, 47 with 4 spheres (ie, a max. of 47.)

tomarken, I'm trying to follow the logic of how you drop first at 24m and if it breaks, how you would narrow down to the exact meter.  24 (break) then..? let's say it was 13:  12 > 18 (break) > 15 (break) > 13 (break) so that's 4 breaks and you haven't tested 14m yet.

Thanks!

Posted by John on 2006-05-04 15:08:15