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# Magical mathematics: The Integers

By on 01/02/2016

Choose a three-digit number and write it twice in succession. For example, if you chose 761, then you should write down 761 761. The game begins by dividing your six-digit number by 7. The remainder, that is, whatever is left after the division, is your lucky number. This will be one of the numbers 0, 1, 2,3 ,4 ,5, 6, since these are the only possible remainders on division by 7. Now write your number and the remainder on a postcard and send it to the editor of this newspaper (Die Welt). By return post you will receive as many 100-euro notes as indicated by your lucky number.

If you are unfortunate enough to have ended up with zero as your lucky number, you are in good company, since the same fate will have befallen all of your fellow readers. (If such were not the case, the publisher would never have agreed to print this article.)

The reason for this phenomenon rests in a well-hidden property of the set of whole numbers, or integers. Namely, placing a three-digit number next to itself is equivalent to multiplying it by 1001, and since 1001 is divisible by 7, the six-digit number will be divisible by 7 as well.

This idea can be packaged as a little magic trick for one’s private use; one can replace the promise of 100-euro notes by predicting the remainder.

Indeed, it happens frequently that a mathematical fact somehow finds its way into a magician’s hat. One simply has to find mathematical results that contradict everyday experience and that also have their basis hidden in the depths of some theory.

Here is a piece of advice: Magic is like perfume: the packaging is at least as important as the contents. No one should be suggesting that the chosen three-digit number is to be multiplied by 1001; such a multiplication is equivalent to writing the number twice in succession, but then the whole trick would fall flat. Those looking for a variant from dividing by 7 can substitute 11 or 13, since 1001 has these numbers as factors as well. It will just make the calculation of the remainder a bit more difficult.

Is there a reason that we have to write down precisely a three-digit number? Could we achieve a similar result with two or four digits?

Let us consider a two-digit number n, written in the form xy. If we write the number twice in succession, then we obtain the four-digit number xyxy, which is equivalent to multiplying the original number by 101. But 101 is a prime number, and so the divisors of xyxy are the divisors of xy together with 101. Since in performing this magic trick we know nothing about the number xy, we can say only that there will be zero remainder on division by 101. But asking for division by 101 gives the trick away, or at least strongly suggests what is at work, and furthermore, dividing by 101 may be too difficult for your friends and acquaintances. We conclude, then, that starting with a two-digit number is not such a good idea.

With four-digit numbers we are dealing with multiplication by 10001. This number is not prime, since 10001 = 73·137, with both factors being prime. Therefore, if you write down a four-digit number twice to form an eight-digit number, it is guaranteed that it is divisible by 73 and 137. But who is eager to divide by 73?

Since the number 100 001 has only the prime divisors 11 and 9091, both inconvenient divisors, five-digit numbers are not optimal for our magic trick. And so it goes. We again find small divisors with 1 000 000 001 (it is divisible by 7). But do we really want to begin our magic act with, “choose a nine-digit number and write it down twice to form an eighteen-digit number”? My recommendation is that you stick with the original trick.

Here is a table of prime factors for the first several numbers of the form 10 …01:

 Number Prime Decomposition 101 101 1001 7 ·11 ·13 10001 73 ·137 100001 11 ·9091 1000001 101 ·9901 10000001 11 ·909091 100000001 17 ·5882353 1000000001 7 ·11 ·13 ·19·52579 10000000001 101 ·3541 ·27961 100000000001 11 ·11 ·23 ·4093 ·8779 1000000000001 73 ·137 ·99990001