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# Always the same end

**While most number magic tricks have quite elementary background, there are exceptions. One of the most puzzling ones is the trick we shall describe here. ***by Franka Miriam Brueckler*

While most number magic tricks have quite elementary background, there are exceptions. One of the most puzzling ones is the trick we shall describe here. Let us do this by an example dialog between the performer (P) and the spectator (S). The trick is best performed in front of a blackboard on which the spectator can write his numbers, while the performer is turned away or has his eyes bound. The routine goes like this:

P: “Choose a positive integer and write it on the blackboard. It may be a large one, you do not have to be afraid of complicated calculations. In fact, the larger, the better!”

S: Writes a number, e.g. 5783956075658.

P: “Now, some of its digits are odd, and some are even. Remeber that 0 is an even number! Next to your number first write how many even digits it has, and then next to that how many odd. Finally, as a third number in a row, write the total number of digits you had in your chosen number.” (Note that this is best demonstrated beforehand on an example to avoid misunderstandings!)

S: Writes 5, 8, 13 next to his number.

P: “Now, cross or delete your starting number in order to avoid confusion. Look at your last three numbers and consider them as digits of just one number. If, for example, you had written 2-5-7, you consider it as number 257.”

S: Looks at his number 5813.

P: “This is your new number. Repeat the previous procedure for it: write down the number of even digits, of add and all digits, cross your number and take the last three numbers as a new number. Repeat until you get the same number twice in a row.”

S: Calculates 5813 >>> 1 3 4 = 134 >>> 1 2 3 = 123 >>> 1 2 3 = 123. As he has gotten twice the same number, he tells the performer he is done.

P: “Your last number is 123!”

Now, how on earth could the performer guess that? It is simple and yet complicated. The simple part is: This is one of the number tricks that always, provided the spectator makes no mistakes, end in the same number. The result is always 123. The complicated part is: Why?

First note that if at some point you get 123 it will start repeating itself because it has 1 even and 2 odd digits, 3 in total.

Let us first imagine we perform the procedure with a 3-digit number. It can have 0, 1, 2 or 3 even digits, so for a 3-digit number the instructions result in one of the 4 numbers: 033, 123, 213, 303. These four then lead to 123 in the next step. So, we have proven that for any 3-digit number the procedure ends with 123. Consequently, if we can prove that the procedure always sooner or later results in a 3-digit number, no matter with how many digits we have started, we have proven our point.

Now, let the starting number have less than 100 digits. In practice this will always be true. If the number has only one digit, it is even or odd, so the next number idf either 101 or 011. In both cases this leads to 123. If the number of digits is *n* between 10 and 99, the number of even digits *e* and the number of the odd ones *o* are at most *n* and their sum is *n*. Thus the next number in the algorithm is *eon* which has at most 6 digits.

If it has 4 digits, the next number is then *eo*4 with *e* + *o* = 4, so obviously *eo*4 has 3 digits. It *eon* has 5 or 6 digits the same argument applies. Thus, is we start with a number with less than 100 digits, we have proven that the instructions result in 123. To prove the general case, one would use mathematical induction (on the number of digits of the number of digits :-)).

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