$2016$ coins are placed on a table with $50$ coins heads up and the remaining coins tails up.
Suppose you are blindfolded, and the only thing you can do is flip some of the coins. Explain how you can separate the $2016$ coins into two groups such that each group has equal number of head.
This is the question one of my friends give me. I don't know if it is true or false. I can't prove it either way.
Best Answer
Divide the coins into two groups of 50 and 1966 coins, then flip all the coins in the first group.
After division, the first group will have $50-x$ heads and the second one $x$ heads.
After flipping, both groups will have $x$ heads.