Number the coins from 1 to 10 and the weighings are as follows:

1,2,3,4,5 : 6,7,8,9,10

If they balance, then we expect that there is either 2, 1, or 0 fake coins on each side.

1,2,6 : 3,4,5

If they balance, it means that either there is either 1 or 0 fake coins in 1,2. There is either 1 or 0 fake coins in 3,4,5.

3,4,5,6 : 7,8,9,10

Assume 6 to be a fake coin. There must be a fake coin in 7,8,9,10 to balance it. But according to the previous weighing there must be at least one fake coin in 3,4,5 to balance the 6 on the left side. So there must be at least two fake coins in 7,8,9,10 to balance the two fake coins on the left side, which makes at least 3 fakes in 6,7,8,9,10. This is inconsistent with the original statement that the number of fake coins must be at most 2 in 6,7,8,9,10. Therefore 6 cannot be a fake coin. (Note, I’m using the term fake coin as the coin in minority.)

Now that we know 6 is a real coin, we know that there is 0 or 1 fake coins in 3,4,5 and 0 or 1 fake coins in 7,8,9,10. Therefore according to the first weighing, 1,2 must be real coins. According to the second weighing, 3,4,5 must also be all real coins. And now we realize by the third weighing that 7,8,9,10 must also be real coins.

Therefore the only way all three of these weighings would balance is if all of the coins were of the same weight.

]]>My soln does not cover 2) , but works for 1) and 3).

]]>Actually in this problem there are 3 cases :

1) 10 either real or fake.

2) 4 real and 6 fake or 6 real and 4 fake.

3) 2 real and 8 fake or 2 fake and 8 real.

In fact there are 6 cases but symmetric cases can be handled by the same solution.

The procedure I gave works for 1) and 2).

But fails for 3) in which case whatever you said can arise as a possibility.

Thanks again. You can try to mend the proof ðŸ™‚

]]>in the second time, there may be in each sidr one real coin and oae fake/

the thurd time we do not know uf noth sides are real or fake. ]]>

First we divide the 10 coins into 5 and 5.

Weigh them.If do not balance we are done.

If not we remove one coin(*) from one of the piles of 5.

And divide it into 2 and 2.

Weigh them.If they do not balance we are done.

If not we remove one coin from one of the piles of 2.

And weigh it with the coin(*) removed before the weighing.

If they balance we can be sure all of them are either real ones or fake ones.

If they do not the conclusion is one must be real and another fake.

**note: when pan balances with equal no. of coins it means there are exactly same no. of

fake and real coins on both sides.But when it does not it means they are unequal no. of real

and as well as fake coins on both the pans.

the second problem, 100 coins is the same as 50 or 1000 0r 13.

divide to three parts, n,n, m

such that 2n>m>n

weight n aginst n. if there is one side heavier divude to 2, all coins in heavy side now are real.

else weight m against n and (m-n)coins from second n.

please tell me the first solution

]]>Oops, I meant the classical question where you know that fake coins weigh less.

]]>I think the latter is actually a bit much to ask from people.

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