# Problem in the dark

You enter a room. On a table there are some counters, each of which has

a red side and a blue side. Twenty of these counters have their blue side

facing up while the others have red facing up.

Your task is to separate the counters into two piles, not necessarily of

the same size, but so that each pile contains exactly the same number of

counters with blue side up. To do this, you can turn over as many counters

as you like.

Here’s the catch: the room is in total darkness.

How would you separate the counters into two piles so that each pile

contains the same number of counters with blue side up?

Divide the counters into two piles: one consisting of a randomly selected

20 counters and the other containing the rest. Now turn over each counter

in the pile of 20.

Suppose that the pile of 20 contains n blue counters. Then, since there

are 20 blue counters in total, the other pile has 20 - n blue counters. Since

the pile of 20 has n blue counters, it must have 20 - n red ones. By turning

each counter over, we are left with a pile having n red counters and 20 - n

blue ones - exactly the same number of blues as in the other pile.