OPPOSITION TO MY CLAIM

OTHER FALSE RULES

 

A professional mathematician sent me some web addresses to contest the claim that I created the first rule of divisibility by 7 of the History of Number Theory. Now it is time to analyze the non-rules of divisibility by 7 mentioned in those web addresses and some others that were not mentioned.

I repeat: I have been trying to create a real rule of divisibility by 7 since the beginning of the 1990’s and I saw a great variety of procedures that are not real rules of divisibility by 7 according to the definition of “divisibility rule” mentioned by Wikipedia:

http://en.wikipedia.org/wiki/Divisibility_rule

My search was not continuous. It was intermittent and sometimes I abandoned it. However, after approximately 15 years of search, in 2005, I created my first method that I made public through a web site that is inactive now. In the same year, naming it as “Moura Velho Method” I created the first rule of divisibility by 7 that fits into the definition of divisibility rule.

Studying to demonstrate the “Why it works” of my first rule I learned some details of Pascal’s theorem 2.5 that he demonstrated in 1654 and that introduces a criterion for divisibility by 7. Pascal tried intensely to create a real rule of divisibility by 7, but he failed. His criterion for divisibility by 7 applied to a six-digit number demands the multiplication, from left to right, of each digit of N respectively by 5,4,6,2,3 and 1; if the sum of the products obtained results in a number that is a multiple of 7 then the tested number is also a multiple of 7.

 

 

N = 389,613; SP = 5 . 3 + 4 . 8 + 6 . 9 + 2 . 6 + 3 . 1 + 1 . 3 = 15 + 32 + 54 + 12 + 3 + 3 = 119

7|119 and 7 |N; Pascal would repeat the application of his criterion to confirm that 7|N:

2 . 1 + 3 . 1 + 1 . 9 = 14.

Pascal’s criterion for divisibility by 7 does not fit into the definition of “divisibility rule” because its application is very slow. References to Pascal’s great efforts to create a rule of divisibility by 7 may be found in the “HISTORY OF BINARY AND OTHER NONDECIMAL NUMERATION” written by Prof. Anton Glaser that may be accessed through this web address:

http://www.eipiphiny.org/books/history-of-binary.pdf

Many experts mention the procedure created by Pascal but they rarely mention his name. One of the addresses sent to me by the professional mathematician shows the application of Pascal’s criterion for divisibility by 7 without mentioning the name of the famous French mathematician.

Pascal’s criterion is cited in one of the addresses sent to me by the professional mathematician initially mentioned. That is the reason why I started my analyses by Pascal’s criterion that certainly does not fit into the definition of divisibility rule.

The experts’ preferred procedure to test divisibility by 7 was analyzed in my previous post (a false rule) and it also does not fit into the definition of divisibility rule.

 

N = 389,613; 389,613 ─ 63 = 389,550; 389,550 ─ 1050 = 388,500; 388,500 ─ 10500 = 378,000;

378,000 ─ 168,000 = 210,000; 7|210,000 and 7|N

There is one method of divisibility by 7 that was created by Prof. Gustavo Toja. The author do not claim that his method is a rule of divisibility. It certainly is a mathematical rule used to test divisibility by 7 but it also does not fit into the definition of divisibility rule. Linked to the same method the names of Martin Gardner and Alexander Bogomolny were cited.

N = 389,613; 38   96   13 → 3   2   6 → 6    23; 6   5 → 56; 7|56 and 7|N

3 2 6 from (38 ─ 3) (98 ─ 96) and (13 ─ 7) → inversion to 6 23

6 5 from (6 ─ 0) and (28 ─ 23) → inversion to 56

It reaches the correct result but I consider it complicated. See it in details accessing:

 http://www.sciencenews.org/view/generic/id/6193/description/Divisibility_by_Seven

My best rule of divisibility by 7 (see my post) also works with pairs of digits and fits into the definition of divisibility rule. Compared to Toja’s method it is applied in a much simpler way and it is incomparably quicker. Conclusion: Toja’s method cannot be used to object my claim.

 

N = 389,613; ( ─ 13 mod 7 + 9) ≡ 3 → 3836; ( ─ 36 mod 7 + 3) mod 7 ≡ 2 → 28; 7|28 and 7|N

It is very easy to apply my best rule entirely through mental calculation.

Prof. Gustavo Toja is Brazilian like me but he has a degree in Mathematics: this explains why his work had the receptiveness it had among the experts.

Now is the time to address other procedures used to test the divisibility of a number by 7 that were not mentioned to contest my claim.

There is one procedure that performs the algebraic sum of the numbers formed by the blocks of three digits of a number alternating the signs plus and minus applied to each block. If the result is a multiple of 7 then the tested number is a multiple of 7.

N = 389,613; 613 ─ 389 = 224; sometimes the result is a three-digit number and the solution needs an additional work. If N is larger, the application of the procedure is even more time-consuming.

This procedure cannot be considered a rule of divisibility by 7 because its application is very slow and it does not fit into the definition of divisibility rule.

Better than this procedure is my simplest rule (see the post). It is simpler and quicker than the procedure under analysis with one advantage: if 7ƗN, its result in modulo 7 is equivalent to the remainder of N divided by 7.

N = 389,613; ( ─ 38 mod 7 + 61 ) mod 7 ≡ 2 the tens

( ─ 9 mod 7 + 3) mod 7 ≡ 1 the ones → 21; 7|21 and 7|N

There is a procedure that consists of successively multiplying by 3 the first digit of N and adding the result to the next digit. The first digit is eliminated and the new first digit is submitted to the same procedure until the penultimate digit is reached. If the last two-digit number is a multiple of 7 then N is also a multiple of 7. Otherwise, the last two-digit number in modulo 7 is the remainder of N divided by 7.

N = 389,613; ( 3 . 3 + 8 ) mod 7 ≡ 3 → 39,613; ( 3 . 3 + 9 ) mod 7 ≡ 4 → 4,613;

( 4 . 3 + 6 ) mod 7 ≡ 4 → 413; ( 4 . 3 + 1 ) mod 7 ≡ 6 → 63; 7|63 and 7|N

This procedure is not a rule of divisibility by the same reasons appointed regarding the previous procedure.

Better and quicker than this procedure is to deduce the digit that forms a two-digit number multiple of 7 with the first digit and get a new first digit by adding the deduced digit to the first and second digit in modulo 7. The procedure must be repeated until the penultimate digit of N is reached. The last result mod 7 determines if 7|N or the remainder if 7ƗN.

N = 389,613; ( 6 + 3 + 8) mod 7 ≡ 3 → 39,613; ( 6 + 3 + 9 ) mod 7 ≡ 4 → 4,613;

( 1 + 4 + 6) mod 7 ≡ 4 → 413; ( 1 + 4 + 1) mod 7 ≡ 6 → 63; 7|63 and 7|N

There is a procedure that consists of multiplying the first digit of N by 2 and adding the result to the number formed by the next two digits. The first digit is eliminated and the new first digit must be submitted to the same procedure until the antepenultimate digit is submitted to the procedure. Its application is very slow and so it cannot be considered a rule of divisibility according to the definition of divisibility rule.

N = 389,613; 2 . 3 + 89 = 95 → 95,613; 2 . 9 + 56 = 74 → 7,413; 2 . 7 + 41 = 55 → 553;

2 . 5 + 53 = 63; 7|63 and 7|N

The use of Modular Arithmetic turns quicker the application of the procedure:

( 2 . 3 + 89 ) mod 7 ≡ 4 → 4,613; ( 2 . 4 + 61 ) mod 7 ≡ 6 → 63; 7|63 and 7|N

Perhaps this version, created by me, turns the procedure into a rule of divisibility because its applications is very quick. It may be performed entirely through mental calculation.

Why did not anybody thought of this before?

This procedure is superior to the experts’ preferred procedure to test divisibility by 7 because its result in modulo 7 is equivalent to the remainder of N divided by 7 if 7ƗN.

See my post “Applying Pascal’s multipliers” to know a better and quicker variation of this procedure.

The last procedure I will comment consists of multiplying the last digit of N by 5 and adding it to the two-digit number formed by the two following digits to the left. It must be repeated until the two leftmost digits are reached. Its application is very slow and cannot be considered a rule of divisibility according to the Wikipedia definition.

N = 389,613; 38961 + 15 = 38976; 3,897 + 30 = 3,927; 392 + 35 = 427; 42 + 35 = 77; 7|77 and 7|N

                              ( 3 . 5)                           ( 6 . 5)                  ( 7 . 5 )            ( 7 . 5)

Better than applying this procedure is to perform this way: Deduce the ones of a two-digit number multiple of 7 that is formed with the rightmost digit of N. Add the deduced digit to the two rightmost digits of N and represent the sum in modulo 7; the result is the new rightmost digit. Repeat the procedure until N is reduced to a two-digit number. If this last number is a multiple of 7 then N is also a multiple of 7.

389,613(5); ( 1 + 3 + 5 ) mod 7 ≡ 2 → 38962(1); ( 6 + 2 + 1 ) mod 7 ≡ 2 → 3892(1);

( 9 + 2 + 1 ) mod 7 ≡ 5 → 385(6); ( 8 + 5 + 6 ) mod 7 ≡ 5 → 35; 7|35 and 7|N

I do not remember of any other procedure used to determine if a number is divisible by 7. If any person wants to contest my claim presenting any procedure that was not examined in this post, please, let me know. Especially if the procedure may fit into the definition of divisibility rule.

 

 

 

 

A FALSE DIVISIBILITY RULE

THE EXPERTS’ PREFERRED PROCEDURE

Among the existing procedures to verify divisibility by 7 of any integer the one most cited by experts may be described this way:

 “To determine if a number is divisible by 7, take the last digit off the number, double it and subtract the doubled number from the remaining number. If the result is evenly divisible by 7 (e.g. 14, 7, 0, -7, etc.), then the number is divisible by seven. This may need to be repeated several times.” http://www.aaamath.com/div66_x7.htm

According to the above-mentioned description, its application demands two calculations (one multiplication and one subtraction) in order to take off successively each final digit of the tested number. It is applied very slowly especially to large numbers. For example, a ten-digit number requires sixteen calculations.

Sometimes this procedure is referred to as a rule, sometimes as a trick. It is not a rule of divisibility in the strict sense and much less a trick.

It is not a rule in the strict sense because the definition of “divisibility rule” includes the adjectives “shorthand”, “shortcut” or the equivalent. Although it is a mathematical rule because it is efficacious, it is not a rule of divisibility by definition because it is not efficient.

It is not a trick because Mathematics is not magic. Some experts and tutors present this procedure as a trick and, in order to keep the mystery, they do not explain why it works; and the “why it works” of this procedure is extremely primary.

To illustrate what I said let me show the practice of that procedure and a better alternative:

 

                                                     N = 696,816                                       696,816

                                                                 ─ 126                                              ─ 56          

                                                            696,690                                       696,760

                                             ─ 1,890                                           ─ 560

                                                          694,800                                       696,200

                                                         ─ 16,800                                        ─ 4,200

                                                          678,000                                       692,000

                                                       ─ 168,000                                     ─ 42,000

                                                          510,000                                      650,000

 

Excluding the zero digits, that took the places of the final digits removed in the process, the results are 51 and 65 that are not multiples of 7 and are equivalent in modulo 7; 7Ɨ51 and 7Ɨ 65 then 7ƗN.

I showed two alternatives (none of them is, by definition, a rule of divisibility by 7) side by side to put in evidence some facts:

1) In a two-digit multiple of 7 the tens are always the double of the ones.

Examples: 6 . 2 = 12 → 126 and 6 . 2 = 12; 12 mod 7 ≡ 5 → 56; 126 mod 7 ≡ 56 mod 7

                     9 . 2 = 18 → 189 and 9 . 2 = 18; 18 mod 7 ≡ 4 → 49; 189 mod 7 ≡ 49 mod 7

Why to do the multiplication 6 . 2, for example, if the resulting 12 mod 7 is equivalent to 5 mod 7 that may be deduced directly and quickly by the elementary 7 times table?

2) Although it puts in evidence the primarity of the procedure, it is simpler and quicker to use the second alternative: it is based directly on the 7 times table.

3) I included the zeros to demonstrate that, opposing some experts’ texts, the process preserves the value mod 7 of the original number in each step of the process.

4) It is easier and quicker to figure out the digit of the tens of a two-digit number multiple of 7 instead of multiplying one digit by 2 before each subtraction. The second alternative involves only subtractions of one-digit numbers easily deduced.

In sum this rule whose application is extremely slow consists of successive subtractions of multiples of 7; very rudimental indeed. It does not deserve the huge amount of papers written by such a great number of experts. None of these papers ever mentioned the successive subtractions of multiples of 7! Why not?

I think it is embarrassing that a few professional mathematicians try to impose and defend mathematical rules that are not, by definition, rules of divisibility by 7 just because they do not want to accept the real rules created by me.

My next post will be about another “rule” mentioned by some experts.

 

 

FEEDBACKS

FEEDBACKS TO MY VIDEO

After I posted my video about the first rule of divisibility by 7 of the History of Number Theory, created by me in 2005, I received a few feedbacks from experts on this matter.

I transcribe the following feedbacks I received from two reputable professionals who occupy important positions in traditional organizations of the mathematical world that together, congregate more than 800,000 math teachers around the world:

“Thank you!

Interesting indeed.

I have posted the link on the Facebook Site for …”

and

“Silvio,

This video is very interesting & I enjoyed watching it. What would be   even more interesting is a video of you explaining WHY this rule works   and HOW you developed it. Also, you used several examples, but that does not prove your rule works. Perhaps with more abstract mathematics, you could convince more people.

Sincerely,”

I do not mention their names because it seems they forgot the experts’ first commandment: “Do not appreciate the work of an outsider even if you cannot prove he is incorrect.”

My video explaining why my rule works will be posted soon.

There are also a few shy feedbacks like “Mathematically interesting” or the like.

It seems that the few experts that tried to deny my claim (“the first rule”) do not know that Mathematics is based on definitions. The rules they defend do not fit into the definition of “divisibility rule”:  “A divisibility rule is a SHORTHAND WAY of determining whether a given number is divisible by a fixed divisor without performing the division, usually by examining its digits” (Wikipedia).

The expert who apparently is the best in mental calculation alleged that it takes time to figure out "what digit should I put here to make a multiple of 7”. Anyone who knows the 7 times table is able to do this very quickly; like the Brazilian student of the primary school who tested my rule in 2006.

 Another expert cited these addresses to exemplify the existing rules of divisibility by 7 before 2005:

 (www.aaamath.com/div66_x7.htm,

www.maa.org/mathland/mathtrek_05_23_05.html,

www.math.hmc.edu/funfacts/ffiles/10005.5.shtml

Anyone can access the addresses mentioned above: THEY DO NOT PRESENT A SINGLE REAL RULE OF DIVISIBILITY BY 7!!!! The respective procedures do not fit into the definition of “divisibility rule”.

Of course, when I stated my claim I had already accessed the mentioned sites (as many others) and concluded that, according to the definition of “divisibility ruIe”, no real rule of divisibility by 7 had been created before 2005. The procedures presented in those sites, especially when applied to larger numbers, are very slow and do not fit into the definition of “divisibility rule”.

Professional mathematicians know that mathematical rules used to verify if a number is divisible by other are not rules of divisibility if they do not fit the respective definition.

There is a paper written by Prof. Bruce Ikenaga entitled “Divisibility Tests and Factoring” in which the author referred to the expert’s favorite rule this way:

“It is difficult to factor a large, arbitrary integer in a reasonable amount of time. You can use simple divisibility tests like those above to deal with "obvious" cases, but the general problem is the object of current research.”

I agree with Prof. Ikenaga.

My rules can be used quickly and precisely to test the divisibility by 7 of numbers of any magnitude.

In my next post I will comment the existing false rules for divisibility by 7 before 2005; I will explain why they work and demonstrate why their application is very slow and so they are not SHORTHAND WAYS according to the definition of divisibility rule.

 

 

 

 

 

 

OTHER QUICK RULES

OTHER QUICK RULES

I will present these rules just to show how my research provided various possibilities to the creation of numerous quick rules for divisibility by 7.

Practice turns quick and easy the application of my rules. At first, as any new knowledge, it may look somehow complicated. Knowing the multiplication table of seven and having some skill with mental calculations are the requirements to apply my rules.

Remember that, without success, researchers have been trying to create a real rule of divisibility by 7 since the beginning of the first millennium (Talmud) according to the History Of Number Theory. When I say real rule I refer to a rule that fits into the definition of divisibility rule according to Wikipedia:

“A divisibility rule is a shorthand way of determining whether a given number is divisible by a fixed divisor without performing the division, usually by examining its digits.”

I think that I presented my best rules in my previous posts and that the explanations of why they work are enough to understand the “why they work” of the next rules. Therefore, I will present only the respective algorithm and some numerical examples. If necessary, I will highlight important details.

Rule 1

N = abc,def → abxc, def

─ ab mod 7 + x + de = de’→ de'f; de'xf; R = ─  de' mod 7 + x

 R = ─ de’ mod 7 + x; if 7|R then 7|N

N = 946,132 → 9456, 132 → 222 → 2242 → 10

─ 94 mod 7 + 5 + 13 = 22; R = ─ 22 mod 7 + 4 = 10; 7Ɨ10 and 7ƗN

Remainder: (R mod 7 . 4) mod 7; (10 . 4) mod 7 ≡ 5 = r (remainder of the division N/7)

The algorithm must be repeated until the last period is reached. It uses repetitively the multipliers 462 to each period of N. It is easy to extend its application to larger numbers.

This rule may be simplified even more:

N = abc,def 

( ─ ab mod 7 + x + de ) mod 7 = e' → e'f; if 7|e'f then 7|N 

N = 946,132 → 9456,132

( ─ 94 mod 7 + 5 + 13 ) mod 7 ≡ 1 → 12; 7Ɨ12 and r = 12 mod 7 ≡ 5 (remainder of N/7)

In this case the multipliers applied to the 5 initial digits are: 46231.

Rule 2

N = abc.def → aybc, dyef

─ bc mod 7 + a + y + ef; R = ─ ef’ mod 7 + d + y. if 7|R then 7|N

N = 946,132 → 9146, 1432 → 145 → 1445

─ 46 mod 7 + 9 + 1 + 32 = 45; ─ 45 mod 7 + 1 + 4 = 9; 7Ɨ9 and 7ƗN

Remainder: ─ R mod 7 ≡ r; ─ 9 mod 7 ≡ 5 = r (remainder of N/7)

The algorithm must be repeated until the last period is reached. It uses repetitively the multipliers 546 to each period of N. It is easy to extend its application to more extensive numbers.

Rule 3

N = a,bcd

(─ cd mod 7) . 3 + b → ab’; if 7|ab’ then 7|N

N = 946,132

[(─ 32 mod 7) . 3 + 1 ] mod 7 = b’ = 3 → 9463

[(─ 63 mod 7) . 3 +4 = b‘ = 4 → 94; 7Ɨ94 and 7ƗN

r = (94 mod 7 . 4) mod 7 ≡ 5 (remainder of N/7)

Regarding larger numbers the algorithm must be applied until the leftmost digit is reached.

The procedure to determine the remainder is the same of a previous rule, already presented, that works with pairs of digits.

Rule 4

N = a,bcd → a,bycd

─ (a + b + y) mod 7 + cd

N = 946,132; 9426132 → 6732 → 67032

─ (9 + 4 + 2) mod 7 + 61 = 67

─ ( 6 + 7 + 0) mod 7 + 32 = 33; 7Ɨ33 and 7ƗN

r = 33 mod 7 = 5 (remainder of the division N/7)

The algorithm must be applied repetitively to four digits each time until the last digit is reached.

The multipliers applied are: 6231. To reach this conclusion it is necessary some reasoning.

Rule 5

N = a,bcd → a,bxcd

[─ (x + c + d) mod 7 + a] mod 7

This algorithm must be applied from right to left. In each application, the last two digits must be eliminated.

N = 946,132; 61632

[─ ( 6 + 3 + 2 ) mod 7 + 6] mod 7 = 2 → 9421 → 94421

[─ (4 + 2 + 1 ) mod 7 + 9] mod 7 ≡ 2 → 24; 7Ɨ24 and 7ƗN

r = (24 mod 7 . 4) mod 7 ≡ 5 = r (remainder of N/7)

The procedure to determine the remainder is the same of a previous rule, already presented, that works with pairs of digits.

CONCLUSION

After I deciphered Pascal's criterion of divisibility by seven it became very easy to create quick rules of divisibility by seven. Of course, there are many other rules that I might present, but I think that the rules already presented are enough to illustrate my discoveries.

In my next post I will comment some feedbacks I received from professional experts.

MY SIMPLEST RULE

THE SIMPLEST RULE

My rule is different of the known procedures based on the same principle.

This rule is very simple. It derives from the fact that if 7|N then the alternating sum of blocks of three digits from right to left are equivalent mod 7.

It is also versatile and quick. It works for divisibility by 7, 11 and 13.

My rule works alternating the signs plus and minus applied to the numbers formed by the digits of the hundreds and the tens of each period of N. The resultant algebraic sum mod 7 produces the tens of a reduced number derived from N. Regarding the tens, if the first period of N is incomplete, the procedure starts with the single tens of the first period or with the hundreds and tens of the second period.

The same procedure regarding the digits of the ones place produces the ones place of a reduced number derived from N.

If the resultant two-digit number is a multiple of seven then N is also a multiple of seven. Otherwise, the resultant number mod 7 is the remainder of the division of N by 7.

The procedure uses Modular Arithmetic this way:

N = abc,def,ghi

T1 =  ─ ab mod 7 + de; T2 = ( ─ T1 mod 7 + gh ) mod 7 ≡ tens

O1 = ─ c mod 7 + f; O2 = ( ─ O1 mod 7 + i) mod 7 ≡ ones

Example: N = 681,439,654

T1 = ─ 68 mod 7 + 43 = 45; T2 = ( ─ 45 mod 7 + 65 ) mod 7 ≡ 6 = the tens

Common language: 68 to 70 = 2; 2 + 43 = 45; 45 to 49 = 4;

4 + 65 = 69; 69 ─ 63 = 6 (the tens)

O1 = ─ 1 mod 7 + 9 = 15; O2 = (─ 15 mod 7 + 4 ) mod 7 ≡ 3 (tens)

Reduced number: 63; 7|63 and 7|N

N = 96,461

9 to 14 = 5; 5 + 46 = 51; 51 mod 7 ≡ 2 (tens)

6 to 7 = 1; 1 + 1 = 2 (ones)

7Ɨ22 and 7ƗN; 22 mod 7 ≡ 1 (remainder of N/7)

I wonder why nobody has thought in this solution before.

My next post will be about, mostly as a curiosity, the vast possibilities of creation of quick rules of divisibility by 7 proportionated by my research.

 

 

MY BEST RULE

MY BEST RULE OF DIVISIBILITY BY 7 

This rule is versatile because, changing what must be changed, it also works for divisibility by 11 and 13. Its application is very quick.

 

 

Its application involves four digits each time. The two final digits are eliminated and, the number formed by them, is subtracted of the fourth digit from right to left using modular arithmetic. The third digit remains unchanged.

THE ALGORITHM: N = a,bcd

N is reduced to a’b

a’ = (─ cd mod 7 + a ) mod 7 → if 7|a’b then 7|N

N = 6.832; a’ = ( ─ 32 mod 7 + 6) mod 7 ≡ 2; a’b = 28

Common language: 32 to 35 = 3, 3 + 6 = 9; cast out the sevens = 2

28 is the result.

7|28 and 7|N

For larger number it is necessary to apply repetitively the algorithm.

N = 39,948,412

( ─ 12 mod 7 + 8) mod 7 ≡ 3 → 39,943,4

( ─ 34 mod 7 + 9) mod 7 ≡ 3 → 39,34

( ─ 34 mod 7 + 3) mod 7 ≡ 4 → 49; 7|49 and 7|N

This rule works because it applies repeatedly these four multipliers determined by Pascal’s theorem 2.5: 1546

─ cd mod 7 ≡ 6 cd mod 7 ≡ (4c + 6d) mod 7

(1 . a’ + 5 . b) mod 7 ≡ (1 . a + 5 . b + 4 . c + 6 . d) mod7 

If N = 6,832 → a’b = 28; (1 . 2 + 5 . 8) mod 7 ≡ (1 . 6 + 5 . 8 + 4 . 3 + 6 . 2) mod 7 ≡ 0; 7|0 and 7|N

There is a practical way to determine the remainder.

Order the pairs of digits of N from right to left considering as a pair the eventual leftmost isolated digit.

From right to left alternate repetitively the digits 421 starting with 1 to each pair. Multiply the leftmost result by 1, 2 or 4 mod 7 according to the position of the pair. The product is the remainder of the division of N by 7.

N = 624.599.183

       2  1   4   2   1; the result must multiplied by 2 mod 7.

(─ 83 mod 7 + 9) mod 7 ≡ 3

(─ 31 mod 7 + 5) mod 7 ≡ 2

(─ 29 mod 7 + 2) mod 7 ≡ 1

(─ 14 mod 7 + 0) mod 7 ≡ 0 → 06; (6 . 2) mod 7 ≡ 5 = r

In this case 5 is the remainder of the division of N by 7.

This procedure to determine the remainder is also based on Pascal’s theorem 2.5.

 

A VARIATION OF MY FIRST RULE

This variation is a little quicker than my first rule.

N = abc,def

Insert “y” after “c” in a way that 7|cy

Algorithm: S₁ = ab + c + y

ab mod 7 ≡ (3 . a + b) mod 7; c + y = 5c

S₁ = 3a + b + 5c

─ S₁ mod 7 ≡ 4a + 6 b + 2c

Insert “y” after “f” in a way that 7|fy

S₂ = 3d + e + 5f

SP = ─ S₁ mod 7 + S₂ = 4a + 6b + 2c + 3d + e + 5f

Without performing any multiplication, it is obtained the equivalent to a sum of products that apply the multipliers:

462315 just like my first rule.

N = 946,134
 ─ ( 94 + 6 + 3 ) mod 7 + 13 = 15 → 154
─ ( 15 + 4 + 2 ) mod ≡ 0; 7|0 and 7|N

To turn this rule even quicker use this algorithm:
N = abc,def
[ ─ ( ab + c + y) mod 7 + de ] mod 7 ≡ e'→ e'f; if 7|e'f then 7|N
N = 946,134;
 [ ─ ( 94 + 6 + 3) mod 7 + 13] mod 7 ≡ 1 → 14; 7|14 and 7|N

I already demonstrated in a previous post that this order of multipliers is equivalent to the application of Pascal’s theorem 2.5.

The repetitive use of the additive inverse mod 7 to each sum obtained results in the alternation of the multipliers (462) (315) applied to the periods that form N.

In my next post I will present a simple rule of divisibility by 7 that is not based on Pascal’s theorem.

 

 

APPLYING PASCAL'S MULTIPLIERS

PASCAL MULTIPLIERS ACCORDING TO A “MOURA VELHO RULE”

 

N = abc

l. Insert an “x” before “a” in a way that 7|xa → x ≡ 2a mod 7

2. Eliminate “a” and perform the addition S = x + bc; if 7|N then 7|S; if 7ƗN then S mod 7 ≡ r (remainder of N/7)

bc mod 7 ≡ 3b + c → S = 2a + 3b + c

If N = abc,def

3. Calculate ─ S mod 7 ≡ 5a + 4b + 6c = S1

4. Repeat the initial procedure to “def” → S2 = 2d + 3e + f

5. Perform SP = S1 + S2 = 5a + 4b + 6c + 2d + 3e + f; if 7|N then 7|SP and if 7ƗN then SP mod 7 ≡ r

Without performing any multiplication, the result is equivalent to a sum of products in which the series of multipliers applied to each digit is: 546231, that is the same series of multipliers determined by Pascal’s theorem 2.5.

Regarding large numbers, the repetitive and cumulative application of the additive inverse modulo 7 to each sum, in the passage of one period to another, results in the alternation of the multipliers (546) and (231) exactly as prescribed by Pascal’s theorem 2.5.

Example:

N = 94,652,392

─ 94 mod 7 ≡ 4; “4” must be added to the next sum

Observe that 94 mod 7 ≡ 24 mod 7 → ─ 24 mod 7 ≡ 4

5652; “4” + 5 + 52 = 61; ─ 61 mod 7 ≡ 2; “2” must be added to the next sum

6392; “2” + 6 + 92 = 100; 7Ɨ100 and 100 mod 7 ≡ 2 = r (remainder of N/7)

Observe that 92 mod 7 ≡ 22 mod 7 → “2” + 6 + 22 = 30 and 30 mod 7 ≡ 2

With practice this rule may be applied entirely through mental calculation.

A quicker version of this rule:

N = 94,652,392 → (4)946,523,920

( 4 + 46 ) mod 7 ≡ 1 → (2)152,392; ( 2 + 52 ) mod 7 ≡ 5 → (3)5,392;

( 3 + 39 ) mod 7 ≡ 0 → 02 = r (remainder of N/7)

In my next post, I will present my best rule of divisibility by 7 and a variation of my first rule.