- Snapshots of modern mathematics
- Diderot Mathematical Forum 2013: “Mathematics of Planet Earth”
- Pierre de Fermat and Andrew Wiles in Czech Republic stamps
- Stefan Banach (March 30, 1892 – August 8, 1945)
- Diderot Mathematical Forums
- Guessing the Numbers
- What is mathematics for Ehrhard Behrends
- What is mathematics for Krzysztof Ciesielski
- The Three Ducks Trick
- What is mathematics for Franka Brueckler

# Guessing the Numbers

For this trick, you need a number of small counters: coins, pebbles, matches, …

#### by Franka Miriam Brueckler

## The effect

You, the performer, ask a spectator to take the counters into his/her hands and distribute them in his/her hands so that you cannot see how many counters are in the left and how many in the right hand.

Then you ask the spectator to multiply the number of counters in his/her left hand by 4 and the number of counters in his/her right hand by 5. Finally, he/she is to add the obtained two products and tell you the result. Almost instanteneously you can tell how many counters are in the spectator’s left and how many in his/her right hand!

## How can that be?

The magic behind the trick is, of course, mathematics. To illustrate this, we shall first use an example, and then generalise. Say the number of counters was 9. This means that in the beginning the performer knows that the sum of the counters in the spectator’s left and right hand (let’s denote these numbers by *L* and *R*) is 9:

*L* + *R* = 9.

Secondly, the performer knows which numbers – coefficients – were named for multiplication of *L* and *R*. In the described case, the spectator had to calculate 4*L* and 5*R*. Finally, the spectator had to add these two products and name the result (let us denote this by ♥), so the performer also knows that

4*L* + 5*R* = ♥.

Note that all the coefficients in the obtained system of two linear equations with two unknows are known to the performer in advance. This means that he can solve the system beforehand. Substituting the first into the second equation he gets

♥ = 4*L* + 5*R* = 4(*L* + *R*) + *R* = 36 + *R*,

i.e. when he learns the number ♥ all he has to do is to subtract 36 from it and obtain the number *R* of counters in spectator’s right hand, and obviously the difference from this to 9 is the number *L*in the left hand.

Note that one has to choose different coefficients for multiplication: if they were equal we would get a dependent system with an infinite number of solutions.

Now, the best about this trick is that it can easily be adapted to different total numbers *N* of counters and different coefficients *A* and *B* for multiplication (as already said, the only restriction is *A ≠ B*). Generally the system of equations corresponding to the trick is

*L* + *R* = *N*,

*AL* + *BR* = ♥,

so substitution of the first into the second equation results in

*AN* + *(B – A) R* = ♥, i.e.

*R* = (♥ – *AN*)/(*B – A*).

Note that the simplest situation for the performer is when *B – A* is 1 or -1 (in that case *R* = |♥ – *AN*|).

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