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Thursday, December 15, 2005

PLEASE WATCH THIS VIDEO BEFORE YOU READ THE TEXT
 
DIVISIBILITY TESTS

To find a quick method for divisibility by 7 is the most difficult problem related to Elementary Mathematics.

My name is Silvio Moura Velho and, although I am not a Mathematician, I've been studying divisibility by 7 for a long time. I have a different approach towards this matter. I think that elementary problems must be solved through simple reasoning. The methods I created are based mainly on mental calculation. One of them is so simple that it is almost as quick as the test of divisibility by 9 and indicates the remainder if the number is not a multiple of 7. Paper and pencil are not necessary; the process takes place only inside your brain.
I have created the site: www.7and13divisibility.com (inactive) to show my work. My project is to include, according to my approach, divisibility tests to be applied to the prime numbers from 11 to 47. I will present my methods in the order of my preference; first I will show the one I like most. The first time I made it public was in 06/11/2005. I hope you like it.
"MOURA VELHO METHOD"1
divisibility by 7


This method is based entirely on mental calculation. Forget paper and pencil!
Look at this practical example and follow the explanation.

Let n = 36,554

36(3),(3)554(2)

3+6+3 = 12; 12 to 14 = "2" ; "2" +3+5+5+4+2 = 21

then 7 divides n exactly.

The digits between parenthesis and their neighbors form a multiple of 7; they are mentally inserted while the addition is processed.
No insertion is needed regarding the tens.
The digit 2 (between quotation marks) is a "link" that must be added to the next sum. It is the difference between the sum and the next upper multiple of 7. If the sum is a multiple of 7 then the link is 0 (zero). Mathematicians would represent it like this: -12 mod 7 = 2; I prefer to express this difference in a simpler way: "12 to 14 = 2".

If the last sum is a multiple of 7 then the same happens to the given number.

Other examples:

1) Let n = 266,453

(4)266(3),(1)453(5)

4+2+6+6+3 = 21; "0"+1+4+5+3+5 = 18

then 266,453 is not a multiple of 7.

2) Let n = 3,126,495

3(5),(2)126(3),(1)495(6)

3+5 = 8; 8 to 14 = "6"; "6"+2+1+2+6+3 = 20; 20 to 21 = "1"; "1"+1+4+9+5+6 = 26

then 3,126,495 is not a multiple of 7.

THE REMAINDER

Rules of divisibility usually are applied to determine if a number is or is not a multiple of another. I like methods that determine the remainder.
If you would like to know the remainder when the given number is not a multiple of 7 then there is a final step:

Create a last link with the last sum; the digit that forms a multiple of 7 with this last link is the remainder.

The last sum of the example 1 is 18; 18 to 21 = 3; 3(5); then the remainder is 5.
Regarding example 2 we have 26; 26 to 28 = 2; 2(1); then the remainder is 1.


Copyright 2005 by Silvio Moura Velho, all rights reserved

1 comment:

ciadedominiopublico said...

Very good your blog, Mr. Silvio!

Alê