You have 9 billiard balls. They are all the same weight, save one. You want to find out which one is heavier. All you have is a scale that determines the heavier side. You can weigh any number of balls at any time. What is the least amount of weighings you can do to find the heaviest ball?
weasel7711 0 Junior Poster in Training
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Jump to PostCool problem. You can do it in two weighings.
Edit - I explained how, but after thinking about it, I took down the explanation in case other people want to puzzle over it themselves.
As a generalization, if you have n balls and one of them is heavy, how many …
Jump to PostHahaha, well done mate.
Easy when you know how eh? :icon_cheesygrin:
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VernonDozier 2,218 Posting Expert Featured Poster
You have 9 billiard balls. They are all the same weight, save one. You want to find out which one is heavier. All you have is a scale that determines the heavier side. You can weigh any number of balls at any time. What is the least amount of weighings you can do to find the heaviest ball?
Cool problem. You can do it in two weighings.
Edit - I explained how, but after thinking about it, I took down the explanation in case other people want to puzzle over it themselves.
As a generalization, if you have n balls and one of them is heavy, how many weighings does it take?
chriswellings 0 Junior Poster in Training
correct me if im wrong (and might be the case) but if you got 9 billiard balls all of the same weight wouldnt the answer be 0 : none would be hevier as they weigh the same??
VernonDozier 2,218 Posting Expert Featured Poster
correct me if im wrong (and might be the case) but if you got 9 billiard balls all of the same weight wouldnt the answer be 0 : none would be hevier as they weigh the same??
The original problem states that there is one ball that is heavier.
You have 9 billiard balls. They are all the same weight, save one. You want to find out which one is heavier.
thoughtcoder commented: Using red is discriminatory against the color blind. +0
chriswellings 0 Junior Poster in Training
lol... i read the q over and over mabee the question shoulda been worded a little different say they all weight the same excep one??
Thinka 40 Posting Whiz Team Colleague
Cool problem. You can do it in two weighings.
Edit - I explained how, but after thinking about it, I took down the explanation in case other people want to puzzle over it themselves.
As a generalization, if you have n balls and one of them is heavy, how many weighings does it take?
I've been thinking about it, and I can't work it out. Pray tell, what be the answer?
VernonDozier 2,218 Posting Expert Featured Poster
I've been thinking about it, and I can't work it out. Pray tell, what be the answer?
First, separate the balls into three groups of three. Pick two of the three groups. Put one of the groups of three on one side of the scale and another group of three on the other side. Weigh them. If they weigh the same, the heavy ball is in the group of three that wasn't on the scale. Pick that group of three for the next weighing. If one side of the scale is heavier than the other, the heavy ball is on the heavy side of the scale. Pick that group of three for the next weighing.
So after one weighing, you've narrowed it down to one of three balls. Randomly pick two of the three balls. Stick one ball on each side of the scale. Weigh them. If they weigh the same, the heavy ball is the ball that isn't on the scale. If one ball on the scale is heavier than the other, that's the heavy ball.
Thinka 40 Posting Whiz Team Colleague
Hahaha, well done mate.
Easy when you know how eh? :icon_cheesygrin:
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