Explain that you can adjust even large numbers quickly, so that they’re evenly divisible by a chosen number from 1-10.
First, you ask for a number from 1-10. Then, someone from your audience is asked to create a large random number (about 15-20 digits long). You take a brief glance at the number, make some minor adjustments to it, and declare it is now even divisible by the chosen number. Checking with a calcaultor, you are proven absolutely correct!
How is this done?
There are actually two tricks involved here. One is used if they chose 7, and the other is used if they choose any other number from 1-10.
If 7 is chosen:
Have them write down 9 random digits, and then have them write down the same 9 digits immediately to the right, but they don’t have to be in the same order as the first 9. For example, if the person writes:
…then they would write the same digits again next to it, but in a jumbled order, we might get this 18-digit number:
Emphasizing there is no way you could have known that this would be the number generated, you take a quick glance at the number, ponder (for effect), and then look down and change the 2nd and the 11th numbers to a 7. In our example, the number would now read:
The number is now evenly divisible by 7!
By generating the number as desribed above, and replacing the 2nd and 11th digits with a 7, you will ALWAYS get a multiple of 7!
If possible, have the person use a number that means something to them, like a phone number, to generate the first 9 digits. In the US, you can have them use their 9 digit “Social Security” number.
If any number except 7 is chosen:
In this case a 16-digit (not an 18-digit) number is created, and there is a little more freedom.
Have the person create a random 16 digit number (any 16-digit number will do).
Again, they should use a handy number, such as their phone number, from which to obtain the digits. In the US, the paper currency always has an 8-digit serial number on it, so you could have the person take out two bills, and copy the serial numbers next to each other.
As an example, let’s use the following 16-digit number:
From here, there are two ways to go to make this 16-digit number divisible by the chosen number:
If the chosen number is 1, 2, 4, 5, 8 or 10:
Simply add “6240″ to the end of the number. It is now divisble by 1, 2, 4, 5, 8 and 10 (although you only emphasize the number they chose)! In our example, this would now look like:
Actually, just adding 240 to the end would also do it, but adding 4 digits makes it look like it took more thought!
If the chosen number is 3, 6 or 9:
You’ll need to figure the “digital root”. To get the digital root, you take the number add its digits, then adding the digits of numbers derived from it, etc., until the remaining number has only one digit.
In our example, we would add:
We then add 7+0 to get the number’s digital root: 7
If the digital root is even (0, 2, 4, 6, 8 ), then you subtract the number from 18, and append that number to the right of the 16-digit number.
If the digital root is odd (1, 3, 5, 7, 9), then you subtract the number from 9, and append that number to the right of the 16-digit number.
Remember: Even numbers are subtracted from 18, (the first even multiple of 9), and the odd numbers are subtract from 9 (an odd number).
The above rules insure that the resulting number will be divisible by 3, 6 AND 9.
In our example, the digital root is 7, so it’s odd. We subtract that from 9 (since it is odd), resulting in 2. So, we append a 2 to the right of the number, giving us:
You can now confidently say that this number is divisible by the chosen number (if it were 3, 6 or 9).
I know this seems like a lot of work, but with a little practice, you can get the whole routine to flow quite smoothly.
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