there r 8 coins of which one weighs less. u shud find it by using a weight scale only twice?

Answer 1

The answer from Valdis has given some good clues.

I will modify it as follows:

Let ABCDEFGH be the coins. Take ABC and DEF in the balance. If ABC is say lighter than DEF, you know the coin is one of ABC. Now take A and B in the two pans. If A or B is light, you will know which one. If they balance C is the coin.


If the two sets ABC and DEF balance, you know GH is the set that contains the light coin. Just balance them against each other.

Answer 2

8? Usually, the problem is stated as '9', and you're given a *balance* scale rather than weigh scale. This is solved as follows:

1: Test 2 sets of 3 coins against each other on either side of the scale. After this, you know it's one of 3 coins - either the light 3, or the *untested* 3 if the 2 sets balance.

2) Test 2 of the 3 coins. Again, either one is light, or if they balance, it's the 3rd one....

Figuring out how to do *8* is left as an exersize for the reader. :)

Answer 3

this cannot be done - (i have a hat and i WILL eat if you prove me wrong)