This is a question we ask when a conversation gets too heavy, especially while drinking:
There are 8 teddy bears. They all have the same weight, except for one that has a diamond inside, thus weighing more. You have a balance scale. How do you find the teddy bear with the diamond by using the scale only twice?
The first person to get the correct answer wins a copy of Podcast Maker. Post your answer as a comment.
Comments
Congratulations Andrew. Check your email for your winning. As Amancay says, you provided a very thorough and complete solution.
Something fun to think about is how to solve the problem if there are 9 bears.
Weigh them? Stuff that, just rip all there stuffing out until you find the diamond, much more fun!
Buy new bears with your new found wealth, if you really care! :-)
That would be my answer after a few pints anyway!
growl figures someone would beat me to the punch...
congrats andrew... and way to explain it so thoroughly that there is no question whether or not you hit it on the head!
Take six bears, place three on each side of the scale.
If the two groups are of equal weights then the diamond must be one of the two remaining bears: balance the two remaining bears and you'll know which has the diamond (the heavier one).
If they are of different weights, then take two of the bears from the heavy side and weight them. If they are of equal weights, then the remaining bear from the original heavy side has the diamond. If they are of different weights, then the heavy bear has the diamond.
or 100 bears?
Ahh, just quit futzin about.
I see that somebody has already won, but why not just take a knife to all of the bears until you find the diamond. (Yes, I know, so violent.)
Ron B
with 9, you seperate them into 3 bunches of 3...
weighing two groups to find a difference or not, then deducing to toss out the even groups, or the lighter group & the one left out of the weigh in...
then you set one on each side, leaving one out... again looking for a difference or a balance, tossiing out the balacing bears, or the lighter & the left-out one, determining the final bear!
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