To Find the Heavy Marble, Divide and Conquer Asymmetrically
Date: 01/14/2011 at 15:54:13 From: Dawn Subject: find the heaviest marble of eight in two weighings There are eight marbles of the exact same size and color. One of the eight is slightly heavier than the other seven. With a balance scale, how do I find the heavier marble in just TWO weighings? By weighing four and four, I can eliminate it down to two and two; and on the third weighing, to one and one -- but to eliminate seven or six in the first weighing is boggling my mind. So I can solve this easily in three weighings; but ... two?
Date: 01/14/2011 at 15:59:08 From: Doctor Ali Subject: Re: find the heaviest marble of eight in two weighings Hi Dawn! Thanks for writing to Dr. Math. We can set aside two of the marbles and put the remaining six on the scale -- three in one pan, three in the other. If they had equal weights, it would mean that the heavier marble is among the two marbles that weren't on the scale. But if one side goes down, then we know that the heavy marble is among those three. In that case, we set aside one of those and put the remaining two of those three on the scale.... Can you take it from here yourself? Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/
Date: 01/14/2011 at 23:53:24 From: Dawn Subject: Thank you (find the heaviest marble of eight in two weighings) Thank You! This was a bit like the one about the three story brick building with no windows, three light switches at the bottom of the first floor, and you get one chance to go upstairs one time, without a time limit -- but you must figure out in that one trip which light switch works the light bulb on the third floor :) If only I hadn't been traveling for two days before this question.... But thank you so much! I love smart people! Will check in again if I get stumped. Dawn
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