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- 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

# Magic number 9

**Most people remember from school: A number is divisible by 9 if and only if the sum of its digits is divisible by 9.**

*by Franka Miriam Brueckler*

For example, 256482 is divisible by 9 because 2 + 5 + 6 + 4 + 8 + 2 = 27 is divisible by 9, while 1234 is not divisible by 9 because 1 + 2 + 3 + 4 = 10 is not a multiple of 9. A number of mathemagical tricks belonging to the class of “number magic” are based on this property; in some books on this type of magic, they are mentioned as tricks based on ‘digital roots’ (the sum of the digits of a number, repeatedly calculated until one obtains a one digit number, is called the digital root of the number; e.g., the digital root of 256482 is 9 because the sum of its digits was 27, and 2 + 7 = 9). Here we present five of such tricks. They have all been described elsewhere, but we shall also give the corresponding mathematical explanations.

**The missing digit**

The performer asks the spectator to think of any (positive integer) number – the larger, the better. Now, the spectator is asked to add its digits and subtract this sum from the original number. Finally, the spectator is asked to cross one of the non-zero digits of the result and read the remaining digits aloud, in any order. The performer will always be able to guess the crossed digit!

How? While the spectator is reading the digits, the performer adds them mentally. When the spectator is done, the only thing the performer has to do is find the difference of his sum to the next multiple of 9 – that is the missing digit!

Don’t believe it? First an example, then the explanation! Say, the spectator has chosen 467896. The sum of its digits is 40, and 467896 – 40 = 467856. Say the spectator crossed the 6. When he reads the other digits, the sum 30 is obtained. The difference to the next multiple of 9, and that is 36, is 6, so 6 is the crossed digit!

What happened is the following: The reason why the abovementioned criterion for divisibility by 9 is true is that the difference of a positive integer and the sum of its digits is always a multiple of 9. The proof of this fact is not hard and can be found elsewhere. This means that when the spectator subtracts the sum of digits from the original number, he gets a new number whose sum of digits is a multiple of 9. When he crosses out a non-zero digit *d* the remaining digits add up to a partial sum *s* with the property that *s* + *d* is a multiple of 9. As the order in addition is irrelevant (addition is commutative), it is not important in which order the digits are read – the performer is able to determine *s* and thus easily the missing digit *d*. Note that one must require to cross a non-zero digit, because both for *d* = 9 and *d* = 0 one would obtain *s* divisible by 9, so the performer could not be sure if *d* is 0 or 9.

**Another missing digit**

A simple variation on the previous trick is the following: The spectator chooses a positive integer and then takes its digits and writes them in any other order (“permutes them”) to obtain another number. He is to subtract the smaller from the larger, cross any non-zero digit of the result and reads the other digits in any order. The performer can guess the crossed out digit by determining the difference of the sum of the heard digits to the next multiple of 9. The explanation is essentially the same as above, after one notices that tow numbers with the same digits written in different orders have the same remainder when divided by 9.

**20 matches**

To perform this simple trick, suitable as an introduction to the “9-magic” for young audience, you need 20 matches (or coins, or other small items that can serve as counters). These are put on a table, and the performer turns his back to the table. The instructions given to the spectator are as follows: “Remove some matches, at least one, from the table and put them aside. You may not take more than 10. Now, count the remaining ones – there are obviously at least 10 left. Sum the digits of this number, e.g. if there are 12 left, you calculate 1 + 2 = 3. Remove that many matches from the table and put them aside. Now, of the remaining matches on the table take a few and hold them in your hand.” Now the performer turns around and instanteneously announces the number of matches held in the hand by the spectator.

The explanation is elementary, de facto it is the same as before. The first step was to ensure that there are between 10 and 19 matches on the table, which after subtracting the corresponding sum of digits always leaves 9 matches after the second step so the spectator has just to calculate the difference of the number of visible matches to 9.

**Magical 19**

This is one of the rare number magic tricks that are to be performed with a pack of cards. One takes a standard pack of cards (52 of them) and asks the spectator to shuffle the cards and then cut the pack approximately in half and keep one of the ‘halves’ himself, while the other ‘half’ is returned to the magician. Now, the spectator is instructed to count his cards, but not announce the number. Obviously, as the pack has been cut approximately in half, this number has two digits. The spectator is asked to sum these two digits to obtain a single digit number. E.g., if the spectator had 24 cards, he gets 6. Now, he is instructed to look at the card that has this position from the bottom (in our example: the 6th card counted from the bottom of his pack). The spectator has to remember this card, but leave it at the original position. Now, he returns the pack to tbe performer, who says: “Number 19 is in some cultures considered to be magic. Let us see if it is … I will count the cards from the top of your pack: one, two, …, nineteen. And this nineteenth card is – your card!”

The trick works automatically, since the difference of the number *n* of cards and the sum of the digits of this number is a multiple of 9 that is smaller than *n*. As the pack of 52 cards was cut approximately in half, *n* is between 20 and 29 cards and the subtraction of the sum *s* of the digits will always yield 18: *s* + 18 = *n*. Thus, to arrive at the position *s* from the bottom we can also count of 19 from the top.

If one wants to make sure tha the trick works, before determining *s* the spectator could be given an additional instruction: “If you have less than 20 cards, add 10 to the number. If you have between 20 and 29 cards, leave the number as it is. If you have more than 30 cards, subtract 10. Now, sum the digits, …”.

**And the results is … 8**

Probably the most attractive of the five tricks presented here is the following. The performer first has to decide how he wants to disclose the final result, which is – as the subtitle suggest – always 8: By writing it on a paper put in an envelope, by having a text handy of which he remembers the 8th word and at the end of the trick asks the spectator to find the word corresponding in order to his result, by preparing a box with exactly 8 items of some sort, …

The spectator is asked to write down his telephone number (or the serial number of a banknote, his birthday as a sequence of digits without dots, …). Now, this number is to be multiplied by 8. The spectator now has to look at all three numbers mentioned: the telephone number, the 8, and the product. He is further instructed to add all digits he sees. If the result has more than one digit, the sum of its digits is to be determined, if necessary repeatedly, until a single digit result is obtained. And now in the most dramatic way the performer can think of he shows that he knows it is and 8!

While the explanations of the previous four tricks are basically the same, the explanation of this one is slightly more advanced: Let *k* be the remainder of the starting number *n* if it would have been divided by 9, so *n* = 9*m*+ *k* for some *m*. Multiplication by 8 gives 8*n* = 72*m* + 8*k*, so the remainder of 8*n* when divided by 9 is the same as when one divides 8*k* by 9. The number 8 obviously has the remainder 8 when divided by 9. Now, the sum of all digits of a number, calculated repeatedly until a single digit is obtained, is exactly the remainder of the division of the number by 8. This means that the number the spectator calculates is the same as the remainder of the division of *k* + 8*k* + 8 = 9*k* + 8, and this remainder is obviously 8.

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