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Two rows of pebbles
In this trick we use a number of smaller pebbles and the performer asks a spectator to choose any odd number of pebbles
by Franka Miriam Brueckler
For this trick you need a number of smaller items, e.g. pebbles (or matchsticks, coins, marbles, toothpicks, …). Each of them we illustrate by a ✿. The performer, mathemagician, asks a spectator to choose any odd number of pebbles to perform the trick, but the chosen number of pebbles is not disclosed to the mathemagician, who during the whole trick remains turned so that he can’t see the table.
As soon as the spectator has chosen the number of pebbles to work with and has removed the rest, the mathemagician asks him to arrange the pebbles in two rows so that the bottom row contains one pebble more than the top one. For example, if the spectator chose to work with 17 pebbles, he would arrange them like this:
✿ ✿ ✿ ✿ ✿ ✿ ✿ ✿
✿ ✿ ✿ ✿ ✿ ✿ ✿ ✿ ✿
Now the mathemagician asks the spectator to name a (positive integer) number smaller than the number of pebbles in the top row. In our example, the spectator would choose a number smaller than 8, say 6. The mathemagician instructs him to remove that many pebbles from the top row. In our example, this leaves the spectator with:
✿ ✿ ✿ ✿ ✿ ✿ ✿ ✿ ✿
Then the mathemagicians asks the spectator to remove from the bottom row as many pebbles as there are left in the top row. In our example, the spectator removes 2 pebbles from the bottom row and obtains situation:
✿ ✿ ✿ ✿ ✿ ✿ ✿
Finally, the mathemagician instructs the spectator to remove all the pebbles that still remain in the first row. After that, the mathemagician “guesses” (correctly, of course!) the remaining number of pebbles. In the described example, the last step would be removing the three pebbles from the top row and the mathemagician would announce that there are 7 pebbles left.
The background of the trick is simple, basic algebra. We encourage the reader to try to work it out by him- or herself. For those who prefer to have the full description of the trick, or want to check if their solution is correct, the following paragraph gives the full explanation.
First note that any odd number of pebbles can be divided in two rows differing by 1 in the number of pebbles. So, if we denote by n the number of pebbles in the top row (at the beginning of the trick), the bottom row contains n + 1 pebbles (in the described example, n = 8). Denote by m the number named by the spectator (in our example, m= 6). The first instruction leaves the spectator with n – m pebbles in the top row, and the bottom one still containsn + 1 of them. Then the spectator removes as many pebbles from the bottom row as there are in the top row, i.e.removes n – m pebbles from the bottom row, so the bottom row now contains n + 1 – (n – m) = m + 1 pebbles. Eliminating the top row one eliminates the dependence of the number of pebbles on the table on the first chosen number unknown to the mathemagician and leaves m + 1 pebbles on the table. Since the mathemagician knows the value of m, it is easy for him to “guess” the number of pebbles remaining on the table. That’s all folks!
This trick is described (without explanation of the mathematical background) under the name "Coin rows" in Oliver Ho's book Amazing Math Magic (Sterling Publ. Co., New York, 2002).