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# 1089

By on 01/02/2016

This is one of the mathemagical classics! The trick itself is a simple example of ‘number magic’, but not only that it can be used to teach some maths in school, the background has even been extended to a research paperabout what happens when one uses numbers of arbitrary length in arbitrary bases!

#### Franka Miriam Brueckler

 The effect of the trick is that the performer is able to guess the result of a calculation done in secret by the spectator. There are many ways that have been suggested for making the announcement of the number 1089, which will always be the result: besides simply stating it, the classics are putting a note with the number in a sealed envelope before the calculation even starts or (the personal favorite of the author of this text) to use a book and at the end of the trick say something like “Your result must have at least 3 digits, am I right? OK, then first look at the digits before the last two and find the corresponding page in the book. Then, on that page find the line corresponing to the second to last digit. Finally, in that line, find the word corresponding to the last digit. Your word is …” (Obviously, when choosing the book, one must make sure that the 8th line on page 10 has 9 or more words ;-)) If you do not already know the trick, here is how it works: The spectator is instructed to choose a 3-digit number with first and last digit having a difference of 2 or more. Then he is to reverse the number. The smaller of the two numbers is to be subtracted from the larger to obtain a third number. This one is reversed again, and the obtained fourth number is added to the third. And the result is … 1089!

It has been suggested by other authors to use this trick to teach the basics of the decimal system. This trick, along with most others belonging to mathematical magic, is great to implement discovery learning in a class! Here is the essential explanation of the background: If you start with number abc = 100a + 10b + c, the reversed number is cba = 100c + 10b + a. Let e.g. c < a. Subtraction results in 100(a – c) + ca. As c is less than a, the subtraction carries a digit over to the second to last place, and then further to the first position (e.g. 672 – 276 = 396 = 100(6-2-1) + 90 + (10+2-6)). Thus the third number in the procedure has three digits: first ac – 1, second 9, and last 10 + ca. In other words, the third number is 100(a-c-1) + 90 + (10+c-a). Reversing this gives us 100(10+c-a) + 90 + (a-c-1), and the final addition results in 1089.

Note that the trick also works if one requires just that the first and third digit of the starting number are not the same, provided the spectator is instructed that if the result of the subtraction has only two digits, then one should consider it a 3-digit number by placing a 0 in front of it.