Question

You have 8 coins which are all the same weight, except for one which is slightly heavier than the others (you don't know which coin is heavier). you also have an old‐style balance, which allows you to weigh two piles of coins to see which one is heavier (or if they are of equal weight). what is the fewest number of weighings that you can make which will tell you which coin is the heavier one?

Answer 1

A wise man has committed a capital crime and is to be put to death. The king decides to see how wise the man is. He gives to the man 12 pills which are identical in size, shape, color, smell, etc. However, they are not all exactly the same. One of them has a different weight. The wise man is presented with a balance and informed that all the pills are deadly poison except for the one which has a different weight. The wise man can make exactly three weighings and then must take the pill of his choice. Remember that the only safe pill has a different weight, but the wise man doesn't know whether it is heavier or lighter. Most people seem to think that the thing to do is weigh six pills against six pills, but if you think about it, this would yield you no information concerning the whereabouts of the only safe pill. So that the following plan can be followed, let us number the pills from 1 to 12. For the first weighing let us put on the left pan pills 1,2,3,4 and on the right pan pills 5,6,7,8. There are two possibilities. Either they balance, or they don't. If they balance, then the good pill is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the good pill is either 11 or 12. Weigh pill 1 against 11. If they balance, the good pill is number 12. If they do not balance, then 11 is the good pill. If 1,2 vs 9,10 do not balance, then the good pill is either 9 or 10. Again, weigh 1 against 9. If they balance, the good pill is number 10, otherwise it is number 9. That was the easy part. What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these pills could be the safe pill. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the good pill is either 3 or 4. Weigh 4 against 9, a known bad pill. If they balance then the good pill is 3, otherwise it is 4. Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a good, heavy pill, or 1 is a good, light pill. For the third weighing, weigh 7 against 8. Whichever side is heavy is the good pill. If they balance, then 1 is the good pill. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a good heavy pill or 2 is a light good pill. Weigh 5 against 6. The heavier one is the good pill. If they balance, then 2 is a good light pill. I think that one could write all of this out in a nice flow chart, but I'm not sure that the flow chart would show up correctly over e-mail.

Answer 2

Answer:

The fewest number of weighings is 2.

Step-by-step explanation:

The total number of coins is 8.

The coins can be piled in 3 groups.

Group A and B containing 3 coins each and group C containing 2 coins.

Weighing (1)

In first weighing Group A and Group B should be weighed.

Here arises two cases.

Either the coins should balance each other or the coins will not balance each other and one side will be more heavier.

Case 1

Both the group A and B of coins balance each other.

That means the group C contains the heavier coin.

Weighing (2)

Now the 2 coins in group C should be weighed individually, in this way one side will be more heavier than other.

So, the heavier coin will be determined.

So the total number of weighing is 2 in this case.

Case 2

The group A and B do not balance each other.

One side on the balance will be heavier, it means the heavier side contains the heavier coin.

Weighing (2)

The coins on the heavier side should again be weighed individually.

There are three coins in the group.

Again two cases will appear here.

Either the two coins will balance each other or the two coins will not balance each other.

i) The two coins balance each other.

When two coins will balance each other, the third will definitely be the heavier coin.

ii) The two coins do not balance each other.

When two coins will not balance each other, the coin on the heavier side of the balance is the heavier coin.

Final Answer

The total number of weighings done in this case is 2.

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