Find the marble

Logic Level 2

You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.

The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it.Find minimum number of moves required to get the defective marble.


The answer is 3.

Karandeep Singh Ludhar by karandeep singh ludhar , 21, India

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1 solution

Aaaaaa Bbbbbb
Sep 5, 2015

Divide 12 marbles into 4 groups of 3. Compare two groups by scale. If they are equal, continue to compare one of these two group with the one of two remaining groups. If they are not equal then the group has just been selected that has defective marble in accordance with its weight is less or more than normal group. By only one time comparing two marbles of this group to detect defective marble. So: A n s = 3 Ans = \boxed{3}

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I have one concern here. Suppose the first two groups are equal, and then the next measurement of one of these groups with one of the remaining groups yields an equal measurement as well. This tells us that the 4th group contains the different marble, but we don't know whether the different marble will be lighter or heavier. Thus when we compare two marbles from this group, if they are unequal then we won't know which is the different marble, and thus we would need a 4th measurement in this worst-case scenario. I still think the answer is indeed 3 3 measurements, but we might need a different strategy. (Mine is quite complicated, so I was hoping someone would come up with a simpler method first.)

Brian Charlesworth - 5 years, 9 months ago

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Agree. You've reminded me of the possibility that the outcast marble may be either lighter or heavier than the others.

Nhi Huỳnh Ngọc Phúc - 5 years, 9 months ago

Sir please, see this

karandeep singh ludhar - 5 years, 9 months ago

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Yes, I've seen that solution before and it is roughly what I had in mind as well. The answer is indeed 3 , 3, but I was hoping that someone had a simpler approach to suggest. The solution posted above has one flaw which I outlined in my previous comment.

Brian Charlesworth - 5 years, 9 months ago

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