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Problem in the dark

Problem:

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?

Solution:

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.

  • Leading house

  • Co-leading house


The National Centres of Competence in Research (NCCRs) are a funding scheme of the Swiss National Science Foundation

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