I have gathered in my house three friends (Alberto, Benito and Carlos) whom I plan to surprise with a magic trick.

I put three objects on the table: A ring, a pen and a matchbox. I also leave a plate with 24 hazelnuts. I give Alberto a hazelnut from the plate, I give Benito two and Carlos I give three. Finally I propose that you keep one of the three objects that are on the table each without me seeing it (hazelnuts do not count). For this I leave the room for a moment.

Once the objects have been stored, I come back and propose the following: Without me seeing it, the person who took the ring should take as many hazelnuts as I gave it. The person who has the pen must take twice the hazelnuts I gave him and the person who has the matchbox must take four times the number of hazelnuts that I gave him without me seeing him. To give you more emotion, I tell you that everyone eats their hazelnuts. To do this, I leave the room again.

Upon returning, I see that there are 6 hazelnuts left on the plate ...

**Who took the matchbox?**

#### Solution

The solution to this problem is purely mathematical. Suppose we call Alberto "A", Benito "B" and Carlos "C" and the objects on the table are "a" the ring, "b" the pen and "c" the matchbox. We are going to make a table with all the possible person - object and hazelnut combinations delivered by me and hazelnuts taken by them:

Case | ABC | Hazelnuts Meals | Total hazelnuts | remaining hazelnuts | ||

Alberto | Benedict | Charles | ||||

1 | abc | 1 + 1 = 2 | 2 + 4 = 6 | 3 + 12 = 15 | 23 | 1 |

2 | acb | 1 + 1 = 2 | 2 + 8 = 10 | 3 + 6 = 9 | 21 | 3 |

3 | bac | 1 + 2 = 3 | 2 + 2 = 4 | 3 + 12 = 15 | 22 | 2 |

4 | bca | 1 + 2 = 3 | 2 + 8 = 10 | 3 + 3 = 6 | 19 | 5 |

5 | cab | 1 + 4 = 5 | 2 + 2 = 4 | 3 + 6 = 9 | 18 | 6 |

6 | cba | 1 + 4 = 5 | 2 + 4 = 6 | 3 + 3 = 6 | 17 | 7 |

So we can deduce that if there are 6 hazelnuts left, case 5 is met and **Alberto has the matchbox**. We also know that Benito has the ring and Carlos the pen.