- 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

# Inductive Magic

This trick, as well as most of the other mathematically based magic tricks, belongs to the class of so-called parlour tricks, which means that it needs simple and easily obtainable equipment and is suited for performances in smaller groups of people and without a stage.

#### By Franka Brückler

## So it begins…

For this trick you need a number of smaller items, e.g. marbles (or matchsticks, coins, pebbles, …), and it is advisable to have a piece of paper and a pen (or a small portable blackboard and chalk). It is performed by a “mathemagician” and one participator. From the beginning to the end of the trick, the mathemagician has to be positioned so that he does not see what the participator is doing. In the following description of the routine, in the example illustrations we represent each marble by ✿.

Disclaimer: The author of this article is not the author of the trick, but has picked it up somewhere several years ago and is unable to find or recall the reference (or even if it was a book or a website, or even if it only contained the description or also the explanation of the trick).

## The routine

You will be doing some actions with the marbles on the table, and will have to calculate as you perform the actions. The calculations will be simple, but if you want, you can use the paper and pen. Until the end, you shouldn’t give me any information on your actions and calculations.

Say, we have 9 marbles on the table:

✿ ✿ ✿ ✿ ✿ ✿ ✿ ✿ ✿

First, divide the marbles in two heaps. Then multiply the numbers of marbles in each heap and remember, or write down, this number.

Say, the participator decides to divide the 9 marbles in one heap with 3, and the other with 6 marbles:

Heap #1 | Heap #2 | Score |
---|---|---|

✿ ✿ ✿ | ✿ ✿ ✿ ✿ ✿ ✿ | 3 ⋅ 6 = 18 |

Now, subdivide one of the two heaps in two, multiply the numbers in the two new heaps and add to your last number.

Say the participator decides to divide the second heap with 6 marbles into heaps of 2 and 4 marbles each, so he has to add 2 ⋅ 4 = 8 to his last number (18):

Heap #1 | Heap #2 | Heap #3 | Score |
---|---|---|---|

✿ ✿ ✿ | ✿ ✿ | ✿ ✿ ✿ ✿ | 18 + 8 = 26 |

Now continue with the process of dividing one of the heaps on the table in two, adding the product of the two numbers in these two new heaps to your last total, until all your heaps contain only one marble.

Each of the imaginary actions of the participator (including the first two) are represented by a row in the table below. In each row, the heap that is subdivided to obtain the next row is coloured in red.

Heap #1 | Heap #2 | Heap #3 | Heap #4 | Heap #5 | Heap #6 | Heap #7 | Heap #8 | Heap #9 | Score |
---|---|---|---|---|---|---|---|---|---|

✿ ✿ ✿ | ✿ ✿ ✿ ✿ ✿ ✿ | 18 | |||||||

✿ ✿ ✿ | ✿ ✿ | ✿ ✿ ✿ ✿ | 26 | ||||||

✿ ✿ ✿ | ✿ | ✿ | ✿ ✿ ✿ ✿ | 27 | |||||

✿ ✿ ✿ | ✿ | ✿ | ✿ | ✿ ✿ ✿ | 30 | ||||

✿ ✿ | ✿ | ✿ | ✿ | ✿ | ✿ ✿ ✿ | 32 | |||

✿ | ✿ | ✿ | ✿ | ✿ | ✿ | ✿ ✿ ✿ | 33 | ||

✿ | ✿ | ✿ | ✿ | ✿ | ✿ | ✿ | ✿ ✿ | 35 | |

✿ | ✿ | ✿ | ✿ | ✿ | ✿ | ✿ | ✿ | ✿ | 36 |

When the participator says he has finished subdividing and calculating, although his actions have been quite arbitrary, the mathemagician surprises the public by knowing the final result of the calculations. How is this possible?

## Mathematical induction explains it all

Surprisingly, or for a mathematician not so, the only thing the final result depends upon is the starting number of marbles, so this is all the mathemagician has to know. When starting with 9 marbles, one always ends up with the final score being 36. If you do not believe it, try it. And then try to prove it 😉 (this is also a note for the mathematics teachers who are reading this article – the trick, as are many of the mathemagical class, is very suitable both for performance in the classroom and as a starting-point for a problem-based or even discovery-style lesson).

Although it is much more interesting to discover why this trick works, and how to calculate the correct final score for other starting numbers of marbles (besides the possibility of trusting the author that the result depends only on that number and performing the trick once yourself to discover the final score experimentally :-)), we present the secret for the “lazy” ones among the readers. And those who want to do it themselves can just finish reading here!

The maths behind the trick is as follows. Denote the starting number of marbles by ♥. For ♥ = 1 we cannot do any subdivison, so the final result is 0. For ♥ = 2 the only possible subdivision is into two heaps of size 1, so the final score is 1 ⋅ 1 = 1. This is the basis for the mathematical induction proof of the general formula. One can try to make a conjecture of the formula by performing the trick with several different starting numbers of marbles. For ♥ = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, … one experimentally obtains the following corresponding final scores: 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, … Can you guess the formula? The next paragraph starts with its statement.

If one starts with ♥ marbles, the final score is always ♥ ⋅ (♥ – 1) / 2. We prove this conjecture by mathematical induction. The basis has already been proven. Suppose the formula is true for all positive integers smaller than ♥. If we divide the ♥ marbles in two heaps of, say, *m* and *n* marbles, obviously *m* and *n* are smaller than ♥ and *m* +*n* = ♥. Also, the product *mn* is the starting number in the calculation of the final score. By induction hypothesis, sequential subdivisions of the heap with *m* marbles will result in a subscore of *m* ⋅ (*m* – 1)/2, and sequential subdivisions of the heap with *n* marbles will result in a subscore of *n* ⋅ (*n* – 1)/2. Consequently, the total score is*m*(*m* – 1)/2 + *n*(*n* – 1)/2 + *mn* = (*m^2* – *m* + *n^2* – *n* + 2*mn*)/2 = ((*m* + *n*)^2 – (*m* + *n*))/2 = (♥^2 – ♥)/2 = ♥ ⋅ (♥ – 1) / 2. So, since we proved both the basis and the inductive step, we conclude that for every positive integer ♥ as the starting number of marbles the trick routine ends with the final score ♥ ⋅ (♥ – 1) / 2. The end.

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