# Spot the tumbler

There are 4 identical tumblers each filled with the same number of 1 kg balls. One of these four tumblers contains all defective underweight balls. You have a weighing scale which you can use to weigh the tumblers, but it can be used only once.

How can you figure out which tumbler contains the defective balls?

Note: The tumblers are not transparent, but it doesn't matter, because the underweight balls look and feel the same as the genuine balls anyway. Underweight ball can be between (0,1).

I am not the original author of this question. This question was asked in an interview to my friend and he agrees that this is not solvable until and unless there are restrictions put on the underweight quantity. I don't wish to modify the question now and have accepted the answer which concurs with this.

• This puzzle is likely impossible for an arbitrary number as that could include 1 ball in each tumbler. Apr 24, 2015 at 13:47
• Is putting all tumblers on the scale and removing balls while the tumblers are on the scale considered as one use or multiple uses?
– A.D.
Apr 24, 2015 at 13:59
• @kaine Are you sure Old's didn't reject that answer because of Alcona's comment? That's not about the number of balls available Apr 24, 2015 at 14:25
• @IvoBeckers, the "flat-bottomed beverage container" definition seems most likely. Even though 1kg balls aren't really a beverage ;-) Apr 24, 2015 at 16:53
• Going off of Alconja's answer, are we trying to find which tumbler originally contained the defective balls, or which tumbler ultimately contains the defective balls (or either)? Apr 24, 2015 at 18:51

I don't really see how this could be solved without further restrictions. Let's look at a simplified case where we have to pick x balls from one tumbler, y from another, with one of them containing the defective balls. We get the following:

• If x is defective, y okay then the weight range is between y and y + x.
• If x is okay, y defective then the weight range is between x and x + y.

No matter what you pick for x and y, if both are at least 1 then intervals will always overlap. If the scale shows a number within the overlapping part then there isn't a solution. As an example, for x = 2, y = 5, the first range is 5 to 7, the second range is 2 to 7. Getting anything above 5 and below 7 doesn't tell us anything.

• I think you are right. This problem usually specifies the amount the defective balls are underweight (like .1 kg). Apr 24, 2015 at 17:47
• I am very much inclined to consider this as the final answer as it concurs with the individual who was asked this question and he also said "not solvable without any further restrictions".
– Old
Apr 25, 2015 at 1:14

Since the weighing scale can only be used once, I feel free to disassemble and modify it.
Now, tie a ball from each tumbler to the 4 vertices of the weighing-scale, as shown in this picture: Now, pull up the structure and see how it tilts: the highest ball from the floor is the lightest one!

• This idea was definitly found not inside of a box. Apr 24, 2015 at 17:26
• ingenius.......
– JLee
Apr 24, 2015 at 18:19
• This is really good answer but unfortunately this needs modification to the weighing scale.
– Old
Apr 25, 2015 at 1:11

Take 4 balls out of one, 8 out of another, 12 out of a third, and 16 out of the last one and weigh all tumblers together. Based on how underweight the scale reads, you have your answer.

In this way if the scale is <4kg under your 4*N kg expected total, then the tumbler missing 4 balls is underweighted. If 4kg < scale < 8kg then its the tumbler missing 8 balls. If 8kg < scale < 12kg it's the tumbler missing 12 balls. Lastly, if 12kg < scale then it's the tumbler missing 16 balls.

• No it will not work as you can only weigh once.
– Old
Apr 24, 2015 at 3:32
• Let me add the explanation: basically take 4, 8, 12 and 16 balls out and we expect 4+8+12+16=40 right? But suppose the underweight balls is in tumbler from where we took out 4 balls so now the total weight will be less than 40 but greater than 8+12+16+"x" amount where "x" is some value i.e 36+x. So we know if the value is more than 36 then it is tumbler from where 4 balls were taken. Same goes for other tumblers where different numbers such as 4+12+16+x will indicate tumbler from where we will take out 8 balls.
– Old
Apr 24, 2015 at 4:03
• I was going to give (a variation of) this solution, but it fails under certain circumstances. Eg. How do you tell the difference between 8 missing 1kg balls + 12 missing 0.5kg balls and 8 missing 0.25kg balls + 12 missing 1 kg balls? Apr 24, 2015 at 4:15
• @Old - your explanation doesn't work... Taking your second example, you say that 4+12+16+x will be between 32 and 40, which indicates 8, but if the weight is, say, 36, then your x would be 4 (8*0.5kg bad balls). However, 36kg could also be 4+8+16+y, with y=8 (12*0.66kg bad balls). So if you weigh your tumblers once and get 36kg, you still don't know... Apr 24, 2015 at 4:31
• Unless you know how much underweight the underweighted balls are, this question is unsolvable. Because in that case you need to first deduce the weight of the underweight balls and then deduce the tumbler containing them. This can't be done with one use of the scale. My answer below works if you know how much underweight the balls are (e.g .95 kg) Apr 24, 2015 at 15:16

Place all four tumblers on the weighing scale, but on there sides. By tilting and rolling off tumbler by tumbler, you would be able to find the defective one. Assume by chance it were the first, a weight of 3kg would be displayed upon rolling off. If the decimal value after first, continue. If second, 2kg displayed, if third 1kg and if last, it wont be a whole number. I guess this way you are only using the weighing scale once.