### Coins on a table

There is a table on which a number of coins are placed. You also know that there are as many coins with Head up as many coins with Tail up. Now you have to divide the coins (number of coins is even) into two equal piles such that number of coins with Heads up and Tails up in either piles be the same. The catch is you are blind folded and you cannot determine the sides (for sure) if you are blinded

1. This can be solved pretty easily. Since we know that there are equal number of heads and tails on the table we can stack them up in a single column. Now divide the column in two equal halves. Invert the upper half and keep it on the table. This would make the number of coins with heads equal in both the halves.

The trick here is that the table has equal number of heads and tails.
Let the total number of coins be '2n'.

Let the first half of the column contain 'x' heads. so the number of coins with tails up is 'n-x'.
Also we know that equal number of heads and tails would mean 'n' heads and 'n' tails.
Thus the number of heads in second half of the column is 'n-x', and the number of coins with tails up is 'n-(n-x)' i.e. 'x'. This is just the reverse of the first half.

Thus we invert the first half to make the number of heads and tails equal in both the halves. :)

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