(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

You've15coins out of which 5 coins are turned with their HEAD facing upwards and the rest 10 have their TAILS facing upwards. Suppose you have beenblindfolded. Now you've got to divide these 15 coins in two groups such that each group has the same number of coins have theirHEADsfacing up. But youcannot overturnany coin. How can you do it?? It may seem practically impossible but it can be really done!

2. The attempt at a solution

I thought it might be a problem of probability. I found the probability of HEAD which is 1/15 i.e. 1/3 which means one out of three. But actually I think it is impossible as to the fact that the number of HEADs is odd and we can't divide it into two equal groups! But in the question it says that it can be done! I suppose there must be some trick in it. Please help me.

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# A difficult and tricky problem

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