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:

10

^{0}/7 → r = 1; 10^{1}/7 → r = 3; 10^{2}/7 → r = 2; 10^{3}/7 → r = 6; 10^{4}/7 → r = 4; 10^{5}/7 → r = 5;
10

^{6}/7 → r = 1; 10^{7}/7 → r = 3; …
From 10

^{6}/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.

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