This post is inspired by the response to my previous article, The Knights Templar & The Circular Number Nine , which generated comments about number 9 far beyond the original context. And numbers are fun!

Two tricks

- Write down a 3 or 4 digit number. Make it random.

- Change the order of the digits to create another number.

- Subtract the smaller number from the larger one.

- Draw a circle around one digit of the result (although not a zero), and keep it secret.

- Change the order of the digits of your answer and write them down, except for the one you circled (removed).

-> And I can tell you which number you circled (removed).

But how is it done?

This is a trick that can be quite fun, but it is not difficult to do. It is based on a couple of straightforward principles.

If you...

- form a number consisting of two digits or more (no upper limit).

- create a new number by rearranging the digits of the first one, randomly or in any way you want.

- subtract the lesser number from the larger one.

Then the result will always be a number that is divisible by 9!

Examples:

(i) 81-18=63 = 9x7 / divisible by 9

(ii) 84-48=36 = 9x4 / divisible by 9

(iii) 213-132=81 = 9x9 / divisible by 9

(iv) 6297-2967=3330 = 9x370 / divisible by 9

(v) 59643125-43619255=16023870 = 9x1780430 / divisible by 9

Further, as already stated above:

A number is divisible by 9 if, and only if, the sum of its digits is divisible by 9.

Let us now look at an example.

Assume you choose: 59643125 (without showing it to me).

Then you create a second number by rearranging the digits of the first: 43619255.

You subtract the lesser one from the larger one:

59643125-43619255=16023870.

Now you are "drawing a circle" around , say, the "3" – that is, you remove that number and keep it secret.

Then you tell me the other digits, in any order you wish: 7106802.

I quickly calculate: 7+1+0+6+8+0+2=24. I know you have removed something, with which this should be a number divisible by 9. The closest number higher than 24 that is divisible by 9 is 27. So I set 27-24=3. This was the digit you withheld from me!

Since, due to certain properties of the number 9, we can continue to add digits until we have only one left, without changing the result, I could have been continuing with the sum I got, 24, one step further. 2+4=6. And 9-6=3. The same result!

There is another similar trick, which can be done like this:

- Ask someone to choose a number, consisting of at least 2 digits, without telling you which.

- Tell him to rearrange the digits, so he gets two other numbers.

- Now he should add the three numbers (the original one, and the two he created by rearranging the digits of the first one), and multiply the result by either 3, 6, 9, 12, or any other multiple of 3.

- From the result, ask him to remove one digit, except a 9.

- Now you ask him to tell you the number he still has, and you can tell him which digit he removed.

An example:

Assume he chooses 1732, and he forms two other numbers by the same digits, 3172 and 2137.

These are added: 1732+3172+2137=7041

He multiplies this by 3: 3x7041=21123

He removes a 2, and still has 2113, which he tells you.

You add 2+1+1+3=7, 9-7=2, which was the digit he removed.

One of the two principles behind this is that, if you add three numbers, each one formed by the same digits, the sum of these three numbers is always divisible by 3. And if you multiply the sum by 3, or any multiple of 3, the result is always divisible by 9.

The second one is the same as above: A number is divisible by 9 if, and only if, the sum of its digits is divisible by 9.

(This material has previously been published in TMA/Meriondho Leo and in my e-book “Numericon”.)

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