Its cool to balance the eggs

Logic Level 3

There are total of 9 eggs out of which 8 have identical weights and one has different weight. Find the minimum number of balances needed to figure out the egg with different weight.

Note : You can use your observation and visualization to recognize which egg you have placed at what position.

2 3 4 5

2 solutions

. .
Feb 20, 2021

$\sqrt { 9 }$ is $\boxed { 3 }$ , so the answer is 3.

Brian Charlesworth
Jun 5, 2015

Assuming that it is not known whether the different egg is heavier or lighter, there is a strategy for finding the "odd" egg in just $3$ weighings, which is presented in the first comment in the following link . Although the strategy presented in this link involves numbering the eggs, it can be done without numbering by just placing the eggs in a row and keeping track visually which eggs are being weighed in each step. It is somewhat convoluted, but it is nevertheless legitimate.

I have to say, the problem could be revised to make it more clear that it has to work in all scenarios. If it were the most convenient way, you could do it it 2 steps which is the minimum times it could take to balance it. That messed me up on this problem and I don't want it to happen to anyone else. But otherwise, good problem!

Colin Carmody - 6 years ago

If we divide the 8 identical eggs in two groups, each group having 4 eggs then both the group would balance each other. The left out egg would be thew odd one out. Can we reason in this way?

@Brian Charlesworth

Soumo Mukherjee - 5 years, 11 months ago

Since we don't know which of the eggs has a different weight, (and even if its weight is less than or greater than that of the others), we would not know which $8$ eggs to divide into two groups of $4.$ In a question like this we need to find the minimum number of weighings that will guarantee the identification of the different egg, i.e., we have to consider all possible scenarios, not just the extreme ones.

There are various strategies that involve $3$ weighings. The one in the link I've provided is complicated, but the following is a simpler one. Divide the $9$ eggs into $3$ groups of $3.$ Weigh two of these groups against each other. If they balance, then the "odd" egg is in the remaining group. If this is the case, then choose two eggs from this group to weigh against each other. If they balance, then the remaining egg is the odd egg. If they don't balance, then choose one of them to weigh against the remaining egg. If these balance then the other egg (from the first weighing) is the odd egg, and if they don't balance then the egg that has been weighed twice is the odd egg. So at worst we've used three weighings to determine the odd egg.

Now if the first two groups of three eggs don't balance, then we choose one of them to weigh against the remaining group. If these balance then its the other group (of the ones weighed first) that has the odd egg, and if they don't balance then its the group that has been weighed twice that must contain the odd egg. Either way, we have isolated the group containing the odd egg in two weighings, and we will also know whether this group, and hence the odd egg, if heavier or lighter than the others, (since the scale presumably tells us which of the two items we're weighing is heavier than the other). So now that we've isolated the group with the odd egg, we choose two eggs from this group and compare their weights; if they are the same, then the remaining egg is the odd one, and if they are different, then whichever is heavier or lighter, (depending on the information we've gathered from the two group weighings), will be the odd egg. So again we have identified the odd egg in three weighings.

As we have now considered all possible scenarios, we can conclude that a minimum of three weighings is required to guarantee identification of the different egg.

Brian Charlesworth - 5 years, 11 months ago

This one was simpler. Thanks.

Soumo Mukherjee - 5 years, 11 months ago

That would be an extreme case. But we can include extreme cases when determining the minimum no. of trials. Can't we?

Soumo Mukherjee - 5 years, 11 months ago

Minimum number of balances is 2... Divide the 9 eggs into 3 groups of 3 eggs, put 3 eggs on one side of the scale, 3 on the other side, and leave 3 out for the first balancing. First balancing: If the scales lean towards one of its sides then we know what side the egg is with a different weight. If the scales stay balanced, then we know the egg is 1 of the 3 eggs that were left on the side. In any case, we now know the group of 3 where the egg is. From that group, take 2 eggs and put 1 of each on each side of the scales. Second balancing: If the scales lean towards one of its sides, then we have found the egg. If they remain balanced, the egg we were looking for was the one left out. Indeed a very nice problem, but sadly the "correct answer" was incorrect.

Ruben Wastiels - 4 years, 4 months ago

We aren't told whether the different egg is heavier or lighter, so on the first balancing you describe where the two sets differ in weight, we have no way of knowing whether the different egg is in the heavier or lighter set. In this case, we would have to take one of these two sets and weigh it against the remaining set. If these two balance then the different egg is in the other of the first two sets, and if they do balance then the different egg is in the set that has been involved in both weighings. At this point we will now know from the results of the first two weighings whether the different egg is either heavier or lighter, but we will need a third weighing to determine which egg among the three in this final set is different. So without at first knowing whether the different egg is heavier or lighter we cannot guarantee that 2 weighing will be sufficient, but we can do so in 3 weighings, hence the answer.

Brian Charlesworth - 4 years, 4 months ago

Its cool to balance the eggs

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