please solve this riddle

[Jonmo1]

If you mean weight by using a scale, you can get the wight of the lightest ball by using a scale only once. All other steps involve COMPARING weight by just determining which is heavier, not needing to know how much each weighs...

Doms approach is the best.

4 sets of 3.

Put balls 1-4 in a box
put balls 5-8 in a box

Lift each one and an average person should be able to tell which one is lighter, or if they are the same. You have not used a scale.

If the same, then the lighter ball is in the group 9-12, otherwise, the lighter ball is in either group 1-4 or 5-9.

Take the group with the lighter ball in it, put 2 balls in 1 box, the other to in another box.

Pick them up and determine which is lighter. Still have not used a scale.

Then pick up each ball of the lighter group of 2, and you can tell which is lighter.

Then use the scale for that lightest ball to get it's weight.

 

[Andrew Fergus]

Unless I'm misunderstanding something, where in the question did it say the odd ball was lighter?

 

[klb]

Note that the question says you can only weigh all balls 3 times and it does not say that you can only use the scale 3 times. My preference is to use a digital scale - not a balance scale.

Divide balls into 3 groups of 4
Weigh each group.
One group will have a different weight - either higher or lower than the other two.
Weigh each ball in the odd group. Read the weight on the scale to determine the weight for each and you now know the weight of the ball which is different.

You have weighed 8 of the balls only once and the other four a maximum of twice.

 

[Jonmo1]

Unless I'm misunderstanding something, where in the question did it say the odd ball was lighter?

Lighter / Heavier, doesn't matter. The same logic applies.

Note that the question says you can only weigh all balls 3 times and it does not say that you can only use the scale 3 times

If you can weigh EACH BALL 3 times, then it's not a very good riddle.

You can just weigh each one a single time (one at a time) to dermine which ball is different...

 

[klb]

The riddle lies in how it is interpreted or misinterpreted. Then again it is not my riddle and not likely to happen in real life.

I like your idea of weighing each ball. Keep it simple. Potentially you only use the scale three times if one of the first three is the odd ball - lol
With my "simple" solution you use the scale a max of seven times and none of the balls are weighed three times.

 

[Andrew Fergus]

Lighter / Heavier, doesn't matter. The same logic applies.

Yes it does. If you weigh the lighter set of 4 balls a 2nd time, then you have wasted one of the maximum of three weighs if the different ball was heavier. That's the point - you don't know if it is heavier or lighter. If a relative scale with two set of four balls tips to the left, was one of the balls on the left side lighter, or one on the right side heavier? You cannot tell with a single weigh - by pre-measuring relative weights of balls in boxes, you have used one (or more) of your maximum number of weighs.

I believe the OP has been a little unclear in stating the problem. I found the original problem stated correctly here : http://256.com/gray/teasers/

• There are 12 balls.
• 11 of the balls weigh the same.
• 1 of the balls is either heavier or lighter than the rest.
• You have an unmarked balance scale.
• Using the scale only 3 times, determine which ball is different and whether it is heavier or lighter than the rest.

I think you'll find the solution I worked out provides the correct answer.

Andrew

 

[cornflakegirl]

I did find an interesting extension to the problem: can you devise a solution whereby you weigh the same sets of balls each time, regardless of the results you obtain?