Answer To A Puzzle

You are given 6 balls and a weighing scale. 2 balls are defective and they weigh less than the other 4. You are only given 2 weighs (you can only use the scale twice; taking off the balls don't count). Find out which are the defective balls.

Remember, it's TWO defective balls, no typo.

Aaaaaaaand I've got great news for you, reader! I'm going to reveal the answer to this puzzle!!

Warning: If you're not really into abstract thinking, be prepared to get really lost in your thoughts....

(Press the "read more" button at the bottom of the post).


The first step you must take is to put 3 balls on each side of the scale. You might be faced with 2 scenarios, as the scale can either be imbalanced or balanced. I will start with the imbalanced scale first. I'll number the steps to prove that it can be done in two weighs (2 steps; you can use the scale twice), and so you know what's going on.

Imbalancedor Scenario 1
This tells you one thing: both defective balls have to be on the same side, otherwise it would appear as a balanced scale (one on each side).

1. Your first step (as shown above) would be to put 3 balls on each side of the scale. If it turns out imbalanced, remove all 3 balls on the heavier side, because you can bet on your life that there are no defective balls there.
2. Move one ball to the empty side (you removed 3 balls from there in the first step), hold one in your hand, and leave one the scale untouched. [If you're following the illustration above, move the black (shaded) ball over the empty side, hold the peach ball and leave the unshaded one where it is.]
Balanced: Both balls are on the scale
Imbalanced: The one on the lighter end of the scale is defective. The other one is in your hand.

There! The first scenario completed!!

And things get really messy now........refer to the diagrams as often as possible to avoid getting confused.

Balanced
This type means that there is one defective ball on each side of the scale.

1. Remove one ball from each side.

There are two possibilities after step one: the scale can either stay balanced or become imbalanced. I'll tackle the balanced one first.

Scenario 2(a)

Balanced = Both defective balls have to be EITHER
a) One on each scale
b) Both removed, and are on the ground now

Step 2(a): Switch ONE ball from ONLY ONE side of the scale. [Illustration: Switch the blue ball on the scale with the blue ball on the ground.]

After switch
2. Imbalanced (the side you switched got LIGHTER): You replaced a normal ball for a defective one. BOTH DEFECTIVES WERE ON THE GROUND. Story ends here for this scenario.

2(a) Imbalanced (the side you switched got heavier): You switched the defective ball with a normal one. The pink one is normal.

2(a) Balanced : Defective ball is pink.

[How do I know?? Because both defectives have to either be on each scale OR both on the ground.
1. If both were on the ground, the scale would have become imbalanced right after swapping the blue balls, because that's what happens when you switch a heavier ball with a lighter one.
If the ball you swapped was defective itself, then BOTH defectives MUST have been on the scale (one on the other side) and the 2 balls on the ground MUST be normal. You would have caused the scale to become unbalanced, because that's what happens if you swap a lighter ball for a heavier one.] More on this in scenario 2(b).

Still in step 2(a): You now know the type/condition of at least ONE ball from ONE side. In every scenario (except the one where "the story ended"), remove ONE BALL FROM EACH SIDE to find the other side's defective. Please rely on your imagination and common sense for this one.
(Eg: Remove the defective and another ball from the other side. If the scale is balanced, you know that you're holding both defectives. If scale is unbalanced, you're not stupid enough to not be able to tell which side is heavier or lighter.)

Scenario 2(b) [right afterstep 1]

The scale became imbalanced because the type of balls you swapped were different from those on the ground. Whatever the case, the lighter side has one defective, the heavier side has 2 normals. Remember, both defectives CANNOT be on the lighter side or you would've got scenario 1.

Step 2(b): This step is NOT related to step 2(a), so no switching of balls is involved. Remove both on the heavier side and shift one over to the now empty side OR remove one ball from each side.

Ta-daaaahhh!!! Now I hope you were smart/not blur enough to digest all that. Happy moping.