There are 8 billiard balls. One of them is slightly heavier than other seven. Only way to tell which one is heaviest is by putting them on scale against the others. What would be minimum number of times you have to use scale to find out heavier ball.
It's two - how?
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I'll give you an answer by the end of the weekend...and no, I am not going to google or ask for help.
You are awesome. I know you would figure it out on your own - you are too brilliant and stubborn to not figure it out! Let me know. =)
We have to assume the balls weigh less have identical weight. We would put three balls on each side of the scale. There are three possibilities. 1.)the scale balance out. 2.)right side is heavier. 3.)left is heavier. If the result is no. 1, the heavier ball isn't in the group of six. It's one of the two remaining balls. Put the two on the scale and you will find out which one is heavier. If the result is no.2 or no.3. The heavier side of the scale has the heavier ball. Pick any two from the heavier side. Weigh them. If the scale balance, the remaining ball that didn't get weighted is the one we are looking for. If the scale isn't balanced, the heavier ball will be on the heavier side of the scale.
BTW... not even my mom called me brilliant. But she did call me stubborn all the time;)
Did I get it right? Is there another solution?
I love these kinds of problems! Most I've seen before, but I have a horrible memory sometimes so I have to re-figure them out.
You are right about your answer - very nice job. Their also another solution that only takes two times...I bet you'll figure it out.
Took me awhile and it was frustrating. Here is another way. First step is put 3 balls on each side and weigh them. If the scale is balanced, weigh the remaining two balls and you will find the heavier one easily. That part is same as last solution. If the scale isn't balanced the first weigh-in, we know the 3 balls on lighter side and the other two balls that didn't get on the scale have identical weight (based on the original assumption that all balls are equal in weight except the heavier one).We also know the heavier ball is within the other group of three. Make two pairs of balls that consist one ball from the heavy side with one ball from the lighter side (or the two balls that didn't get weighted). That will leave one ball from the heavier group. If the scale is balanced second time, the one ball from the heavier group that didn't get weighted would be the heavier ball. If the scale isn't balanced, we know the pair of balls has the heavier ball and we know which ball was from the lighter side when we paired them. Eliminate the lighter ball from the pair and you get the heavier one. Is that the second solution. It's a bit redundant compare to first solution.
Tob: You are too smart for your own good. Can I borrow your brain?
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