quarta-feira, 11 de setembro de 2013
OPPOSITION TO MY CLAIM
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.
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.
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
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
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.
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.
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