There are nine balls. One is slightly lighter than the rest; the difference is small enough that you can't tell just by picking them up. Using a basic two-sided scale, what's the minimum number of times you'd need to weigh balls to guarantee you find the light one?

Question

Mathematics

Kade Aguilar

28 August, 22:58

There are nine balls. One is slightly lighter than the rest; the difference is small enough that you can't tell just by picking them up. Using a basic two-sided scale, what's the minimum number of times you'd need to weigh balls to guarantee you find the light one?

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Maverick Gonzales · Answer 1

Just my thoughts:

this is assuming the rest of the balls are of equal weight.

put 4 and 4 on either side of the scale, if by some chance they end up equal then that means the one ball not on the scale is the lighter. (can switch out with another ball to confirm) and that will be 2 weighs by just luck.

systematically, the more likely case is that the lighter ball will be in the 8 you put on the scale, and it's lightness will cause one side to be higher. this means that the lighter ball is on the higher side.

then take the four balls from the higher side and put two and two back on the scale. again higher sides means lighter. then take those two and put one and one. you will then find the lightest ball.

this method will take 3 weighs