RECIPE TO CREATE REAL RULES FOR DIVISIBILITY BY 7

BEFORE YOU READ THE TEXT I ASK YOU TO WATCH THIS VIDEO.

FOLLOW THESE STEPS

1 – Choose a sequence of 3 to 6 of Pascal’s multipliers: 546231546…

2 – Arrange the best way to perform the operations.

3 – Make use of the additive inverse mod 7 when needed

4 – Create the algorithm

5 – Apply it.

If necessary look at my last post.

First Example:

1 - Chosen sequence: 231

2 – N = abc; 2 . a + bc

3 – In this case the additive inverse mod 7 will be used.

4 – The algorithm: - ( (x) + bc ) mod 7; if N has more than one period the result of one period must be added to the next period till the last period is reached.

5 – Application

N = 154; - ( (2) + 54 ) mod 7 ≡ Ø; 7|Ø → 7|N

N = 4.641; - 4 mod 7 ≡ 3; ( 3 + (5) + 41 ) mod 7 ≡ Ø; 7|Ø  → 7|N

N = 823.424; - ( (2) + 23 ) mod 7 ≡ 3;  ( 3 + (1) + 24 ) mod 7 ≡ Ø; 7|Ø → 7|N

N = 6.453.243; - 6 mod 7 ≡ 1; - ( 1 + (1) + 53 ) mod 7 ≡ 1; ( 1 + (4) + 43 ) mod 7 ≡ 6; 7ł6; 7łN*

*In this case 6 is the remainder of N divided by 7. Observe that I don’t apply the additive inverse mod 7 to the last result. In this way, if N is not divisible by 7, the last result will give us the remainder of the division of N by 7.

N = 38.453.952; - 38 mod 7 ≡ 4; - ( 4 + (1) + 53 ) mod 7 ≡ 4; ( 4 + (4) + 52 ) mod 7 ≡ 4; 4 is the remainder of the division of N by 7.

 

Second example:

1 – Chosen sequence: 1546

2 – N = a.bcd; (6 . cd + a) mod 7; N is reduced to a’b

3 – Additive inverse mod 7 is not necessary as a link.

4 – The algorithm: - (cd  mod 7 + a ) mod 7 ≡ a’; it is applied from right to left.

5 – Application:

N = 154 → Ø154; - (54 + Ø) mod 7 ≡ 2 = a’ → a’b = 21; 7|21 and 7|N

N = 4.641; - ( 41 mod 7 + 4 ) mod 7 ≡ 5 = a’ → a’b = 56; 7|56 and 7|N

N = 823.424; ( - 24 mod 7 + 3 ) mod 7 ≡ Ø = a’ → a’b = Ø4; ( - 4 mod 7 + 8) mod 7 ≡ 4 = a’

→ a’b = 42; 7|42 and 7 |N

N = 6.453.243; ( - 43 mod 8 + 3) mod 7 ≡ 2 = a’ → a’b = 22; ( - 22 mod 7 + 4 ) mod 7 ≡ 3 = a’

→ a’b = 35; ( - 35 mod 7 + Ø ) mod 7 ≡ Ø = a’ → a’b = Ø6; 7ł6 and 7łN*

*Coincidentally the result was the same of the anterior rule but generally this rule does not result in the remainder.

N = 38.453.952; ( - 52 mod 7 + 3 ) mod 7 ≡ Ø = a’ → a’b = Ø9; ( - 9 mod 7 + 4 ) mod 7 ≡ 2 = a’

→ a’b = 25; ( - 25 mod 7 + 3 ) mod 7 ≡ 6 = a’ → a’b = 68; 68 mod 7 ≡ 5; 7ł5 and 7łN

Changing what must be changed this rule may be applied to divisibility by 11 and 13.

Third example:

1 – Chosen sequence: 462

2 – N = abc; 6 . a₁b₁ + (x) + a₂b_{2 …}

3 – Additive inverse mod 7 is not necessary as a link

4 – The algorithm: ( - a₁b₁ mod 7 + (x) ) mod 7 + a₂b_{2 ….}

5 – Application:

N = 154; ( - 15 mod 7 + (1) ) mod 7 ≡ Ø; 7|Ø and 7|N

N = 4.641;   - [ ( (1) + 64 mod 7 ) + (2) ] mod 7 ≡ Ø; 7|Ø and 7|N

N = 823.424; [ ( - 82 mod 7 + (6) ] mod 7 ≡ 1; - [ ( 1 + 42 ) mod 7 + (1) ] mod 7 ≡ Ø; 7|Ø and 7|N

N = 6.453.243; - [ ( (5) + 45 ) mod 7 + (6) ] mod 7 ≡ 5; - [ ( 5 + 24 ) mod 7 ) + 6 ] mod 7 ≡ 5; 7ł5 and 7łN

N = 38.453.952; ( - 3 mod 7 + 2 ) mod 7 ≡ 6; [ -( 6 + 45 mod 7 ) + (6) ] mod 7 ≡ 4;  [ - ( 4 + 95 mod 7) + (4) ] mod 7 ≡ 3; 7ł3 and 7łN

Changing what must be changed this rule may be applied to divisibility by 11 and 13.

The remainder is the value of the digit that forms with the final result a two-digit number multiple of 7.

Fourth example:

1 - Chosen sequence: 315

2 – N = abc; ab + c + (y)

3 – Additive inverse mod 7 is necessary

4 – The algorithm: - ( ab + c + (y) ) mod 7 …

5 – Application:

N = 154; - ( 15 + 4 + (2) ) mod 7 ≡ Ø; 7|Ø and  7|N

N = 4.641; - ( 4 + (2) ) mod 7 ≡ 1; - ( 1 + 64 + 1 + (4) ) mod 7 ≡ Ø; 7|Ø and 7|N

N = 823.424; - ( 82 + 3 + (5) ) mod 7 ≡ 4; - ( 4 + 42 + 4 + (2) ) mod 7 ≡ 3; 7ł3 and 7łN

N = 6.453.243; - ( 6 + (3) ) mod 7 ≡ 5; - ( 5 + 45 + 3 + (5) ) mod 7 ≡ 5;

- ( 5 + 24 + 3 + (5) ) mod 7 ≡ 5; 7ł5 and 7łN

N = 38.453.952; - ( 3 + 8 + (4) ) mod 7 ≡ 6; - ( 6 + 45 + 3 + (5) ) mod 7 ≡ 4; - ( 4 + 95 + 2 + (1) ) ≡ 3

Changing what must be changed this rule may be applied to divisibility by 13.

The remainder is the value of the digit that forms with the final result a two-digit number multiple of 7.

Fifth Example:

1 - Chosen sequence: 546231

2 – N = abcdef; 5 . a + 6 . bc + 2 . d + ef

3 – Additive inverse mod 7 is not necessary

4 – The algorithm: ( - bc mod 7) + a + (y) + (x) + ef

5 – Application:

N = 154; - 54 mod 7 + 1 + 4 = 7; 7|7 and 7|N

N = 4.641; - 4 mod 7 + (5) + 41 = 49; 7|49 and 7|N

N = 823.424; - 23 mod 7 + 8 + (4) + (1) + 24 = 42; 7|42 and 7|N

N = 6.453.243; 6 + (- 53 mod 7) + 4 + (2) + (4) + 43 = 62; 7ł62 and 7łN;

remainder = 62 mod 7 ≡ 6

N = 38.453.952; 38 + (- 53 mod 7) + 4 + (2) + (4) + 52 = 103; 7ł103 and 7łN

remainder = 103 mod 7 ≡ 5

This rule demands more attention and I included it because it uses the whole sequence of Pascal’s multipliers.

I think that five examples are enough to demonstrate that the creation of real rules for divisibility by 7 became very simple and accurate using the knowledge presented in my anterior post. As the creation of the rules follows the sequence of Pascal’s remainders ( I call them multipliers) that was proved by himself I consider that additional proves are not necessary. The examples presented confirm the validity of my reasoning.

There are other characteristics of Pascal’s multipliers that may be explored to create other rules that work perfectly well. That will be the matter for the next post.

THE MOST DIFFICULT ELEMENTARY PROBLEM

PLEASE, WATCH THIS VIDEO BEFORE YOU READ THE TEXT

 

CHALLENGE TO THE MATHEMATICAL ESTABLISHMENT

In the year 2005 I created, naming it as a method, the first real rule for divisibility by 7. I tried to reach the Mathematical Establishment sending emails to Number Theorists, Math Departments, specialized sites, Math Doctors, experts and universities of many countries of the world. They did not recognize the result of my research, but I am sure that I assumed a completely new approach towards the matter that works better than all the research efforts produced by any expert (and how hard they tried!!). 

Back in 2.005 I could not explain why my rule worked, but I did not give up and continued studying the matter. Now I know why my rule works and much more.

I learned and can teach anyone how to create various real rules for divisibility by 7. Now you can create your own rule!!!

My rules are based on the general criterion for divisibility of N by any integer that was created and proved by the French Mathematician Blaise Pascal in the year 1.654. His general criterion is very useful to test divisibility but its application is very slow. My work turned operational Pascal’s criterion regarding divisibility by 7.

I consider that the creation of a rule for divisibility by 7 is the most difficult elementary problem faced by the Mathematical Establishment. Before 2.005 there were no real rules for divisibility by 7 and a book quotation says: “In the Talmud, 100a + b is stated to be divisible by 7 if 2a+b is divisible by 7”. (History Of The Theory Of Numbers – Leonard Eugene Dickson - page 337). This elementary problem persisted for almost two millennia; Talmud was compiled between 200-500 CE!!!

On the seven years that followed the creation of my first real rule for divisibility by 7 I kept watching if the experts would create something better, but nothing new was revealed.  All they have been doing is to present methods, tests, tricks, shortcuts that only work quickly when applied to 3-4digit numbers. Frequently they do not explain why their tricks work.

If you want to confirm if there is a valid rule for divisibility by 7 type on the Google search bar the words: “divisibility by 7” and “no rule” or “divisibility by 7” and cumbersome. If you type “divisibility by 7” and tricks, methods, shortcuts or test then you will see how the experts are in the need of a real rule for divisibility by 7 .

I think that a real rule for divisibility by 7 must be quicker than the division of a large number by 7 and applied mostly through mental work. To apply the real rules that I will present it is not necessary to use pencil because the process happens inside your brain; all you need is to know the multiplication table 7 and a little bit of Modular Arithmetic.

I challenge anyone to prove that:

1) Before my real rule for divisibility by 7 it was not created any mean to verify if a large number is divisible by 7 quicker than performing the division by 7;

2) My real rules (the first and the new ones) were not created by me;

3) My real rules don’t work properly.

Please, do not criticize the language or the Math notation. I am not a Mathematician and all I got was created using my language and Math notation. The vast majority of Number Theorists have tried to solve the problem of creating a real rule for divisibility by 7, but failed. Please, do not assume the role of the fox portrayed in Aesop’s fable “The Fox and the Grapes”. Some experts have tried to diminish the merit of creating a rule for divisibility by 7.

I don’t like the word trick related to numbers; with numbers there are not tricks. Some experts call tricks something they are not able to explain. But if something works there is always an explanation when the objects are numbers.

From now on I will present why my first real rule for divisibility by 7 works.

Starting with the digit one in the center

(2) 1 (4)*

(2) and (4) were inferred because both (2) and 1 and 1 and (4) form a two-digit number multiple of 7.

Observe that 7|(2)1 and 7|1(4), and that 2 . 1 = 2; 3 . 1 = 2 + 1; 4 . 1 = 4 and 5 . 1 = 1 + 4

If you want the product of 1 . 1 just keep the 1 the way it is.

If you want the product of 6 . 1 just apply  – 1 mod 7 ≡ 6.

If you substitute the one in the center by any other digit you will observe that always the value of the digit on the left is the double of the value of the central digit mod 7 and that always the value of the digit on the right is the quadruple of the central digit mod 7.

In my first rule (x) will be used to represent the digit that forms a multiple of 7 with the digits of the hundreds of N and (y) will be used to represent the digit that forms a multiple of 7 with the digits of the ones of N.

Now it is known that (x) and (y) are not digits that belong to N, and that they are easily inferred by anyone who knows the multiplication table of 7.

Pascal’s general criterion of divisibility

Pascal’s general criterion of divisibility is based on the remainders produced by the division of powers of 10 by the divisor that must be tested.

Regarding 7 the sequence of remainders is:

10⁰/7 → r = 1; 10¹/7 → r = 3; 10²/7 → r = 2; 10³/7 → r = 6; 10⁴/7 → r = 4; 10⁵/7 → r = 5;

10⁶/7 → r = 1; 10⁷/7 → r = 3; …

From 10⁶/7 on the sequence repeats itself infinitely.

To simplify communication I will call Pascal’s remainders as multipliers.

To test if an integer is divisible by 7 it is necessary to multiply the value of each digit of N by the value of each multiplier of the sequence from right to left; if the sum of the products obtained is a multiple of 7 then N is also a multiple of 7, otherwise it is not.

According to Pascal’s criterion the ones must be multiplied by 1, the tens must be multiplied by 3, the hundreds must be multiplied by 2 and so on. But I think I discovered something new!

When N is a multiple of 7, if the order of the sequence is maintained it is possible to apply any other multiplier to the ones; the sum of the products will be a multiple of 7, and if N is not a multiple of 7, the sum of the products mod 7 will be equivalent to the multiplication of the real remainder of N/7 by the multiplier applied to the ones mod 7.

The sequence of Pascal’s multipliers is a geometric progression mod 7 common ratio 3 from right to left. This explains why it is indifferent to start the application of the criterion with another multiplier instead of 1; if you start the application with the multiplier 6, for example, the whole progression is multiplied by 6 and the second above mentioned consequence is explained.

To simplify, as the extension of N is not important to this demonstration, let us apply the multipliers to a three-digt number multiple of 7 from left to right:

N = 154; 1 . 4 + 3 . 5 + 2 . 1 = 4 + 15 + 2 = 21; 7|21 and N; multipliers 1, 3 and 2

                 3 . 4 + 2 . 5 + 6 . 1 = 12 + 10 + 3 = 28; 7|28 and N; multipliers 3, 2 and 6

                 2 . 4 + 6 . 5 + 4 . 1 = 8 + 30 + 4 = 42; 7|42 and N; multipliers 2, 6 and 4

                5 . 4 + 1 . 5 + 3 . 1 = 20 + 5 + 3 = 28; 7|28 and N; 5, 1 and 3 *

*This is the sequence of multipliers applied to my first real rule for divisibility by 7: 5, 1 and 3 from right to left.

THE ALGORITHM

Let N = abc;

 - ( (x) + a + b + c + (y) ) mod 7

Remember: ( (x) + a ) mod 7 ≡ 3 . a mod 7; b = 1 . b and ( c + (y) ) mod 7 ≡ 5 . c mod 7

Applying the algorithm, without performing the multiplication, automatically “a” is multiplied by 3, “b” is multiplied by 1 and “c” is multiplied by 5. Note that 3, 1 and 5 form a sequence of Pascal’s multipliers regarding divisibility by 7.

N = 382.473

The algorithm may be applied from right to left or vice versa; the result will be the same. The result of one period must be added to the next period. It is indifferent if the inverse additive mod 7 is applied or not to the last period.

It must be applied successively to each period of N. I prefer to apply the algorithm from left to right.

– ( (6) + 3 + 8 + 2 + (1) ) mod 7 ≡ 1; 1 + (1) + 4 + 7 + 3 + 5 = 21; 7|21 and N

N = 32.473; – ( 3 + 2 + (1) ) mod 7 ≡ 1; 1 + (1) + 4 + 7 + 3 + 5 = 21; 7|21 and N

N = 4.473; - ( 4 + (2) ) mod 7 ≡ 1; 1 + (1) + 4 + 7 + 3 + 5 = 21; 7|21 and N

Please note that you only infer (x)’s before the hundreds digits.

Application to a really large number

N = 695.246.594.226 – ( (5) + 6 + 9 + 5 + (6) ) ≡ 4; - ( 4 + (4) + 2 + 6 + (3) ) mod 7 ≡ 2;

- ( 2 + (3) + 5 + 9 + 4 + (2) ) mod 7 = 3; 3 + (4) + 2 + 2 + 6 + (3) = 20; 7ł20 and 7łN

The reason for the use of the inverse additive mod 7

Any integer divisible by 7 formed by two periods or more has the following characteristic: if we perform an alternating sum of its periods the final result will be a multiple of 7.

Observe that:

N = 155.554; 554 – 155 = 399; 7|399 and 7|N

N = 155.555.554; 554 + 155 – 555 = 154; 7|154 and 7|N

When my rule is applied using the inverse additive mod 7 each subtraction is followed by an addition. The difference is that my rule reduces each period of N to a simpler expression but if 7|N the results will be always equivalents mod 7.

It is interesting to observe that when the multipliers 3, 1 and 5 are applied to a period of N and the inverse additive mod 7 is used the effect is the same as transforming the multipliers 3, 1 and 5 respectively to 4, 6 and 2. In reality considering two periods the six Pascal’s multipliers are applied, from left to right, in this order: 4, 6, 2, 3, 1 and 5.

N = 155; (2) + 1 + 5 + 5 + (6) = 19; 19 mod 7 ≡ 5; - 5 mod 7 ≡ 2

4 . 1 + 6 . 5 + 2 . 5 = 4 + 30 + 10 = 44; 44 mod 7 ≡ 2

Blaise Pascal has already proved that his general criterion works, so if an algorithm is created and applied in a way that the sequence of multipliers (remainders) of Pascal is followed then the proof of the validity of the algorithm is already done.

Observe that in my approach it is possible to apply two multipliers to a pair of digits. If it is necessary to multiply a two-digit number by the sequences 3, 1 or 4, 6 it is enough to proceed this way:

If the sequence of multipliers is 3 and 1 the two-digit number remains unaltered because if n = ab; (3 . a + 1 . b) mod 7 ≡ ab mod 7.

Examples:

n = 54; ( 3 . 5 + 1 . 4) mod 7 ≡ 54 mod 7 ≡ 5

n = 83; ( 3 . 8 + 1 . 3) mod 7 ≡ 83 mod 7 ≡ 6

n = 23; ( 3 . 2 + 1 . 3) mod 7 ≡ 23 mod 7 ≡ 2

If the sequence of multipliers is 4 and 6 the two-digit number must be expressed by its modular additive inverse because if n = ab; (4 . a + 6 b) mod 7 = - ab mod 7

Examples:

n = 54; ( 4 . 5 + 6 . 4) mod 7 ≡ - 54 mod 7 ≡ 2 ≡ (6 . 54) mod 7

n = 83; ( 4 . 8 + 6 . 3 ) mod 7 ≡ - 83 mod 7 ≡ 1 ≡ ( 6 . 83 ) mod 7

n = 23; ( 4 . 2 + 6 . 3) mod 7 ≡ - 23 mod 7 ≡ 5 ≡ ( 6 . 23 ) mod 7

These last examples are based on the fact that for any number this equation is valid:

- N mod x ≡ [( x – 1) N ] mod x

                      x . N mod x – N mod x

- N mod x            ≡ Ø         – N mod x

In the next posts I will present other algorithms applying Pascal’s remainders in a different order. The use of the pairs of multipliers above mentioned will represent a shortcut (valid because explained) that will turn quicker the application of the new real rules.

 

 

 

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