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Sun 27 Apr, 2008 12:45 am
Suppose you had 8 billiard balls, and one of them was slightly heavier, but the only way to tell was by putting it on a scale against another. What's the fewest number of times you'd have to use the scale to find the heavier ball?
suppose 8 balls out of which 7 one pound balls and one 1.1 pound ball.
total wieght is 8.1.now....
pan 1 pan2
4 4.1
remove 1 ball from each pan
if both are equal...dump pans and weigh these removed 2 balls to find out heavier balls else remove 1 ball from each of the pans.check for same. until last 2 balls in pan tp find out haevier ball.
so min number of chances is 3
2 (and you can add one more ball to the mix and the answer's the same)
No. It can be done with exactly two weighings every time.
Actually, the only ball that has a different weight on any billiard table is the cue ball. No weighing is required.
G'day Metro17. Markr is correct!
1. First weigh 3 v 3
2. If one pan is heaver; 1 v 1
Obviously, if the scales are balanced; it must be a ball not on the scales.
You could do the same with nine balls.
The maximum number of times would be 1 v 1 seven times.
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