Problem in the dark - Solution

19 Nov 2018

This is the solution to the monthly maths problem from RTSdécouverte

Problem in the dark

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.

You can read the original problem here: Problem in the dark

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