tag:blogger.com,1999:blog-198934422024-03-07T00:55:36.377-08:00primeasdivisorSilvio Moura Velhohttp://www.blogger.com/profile/18322243925876063572noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-19893442.post-52957187257639729632012-10-06T07:13:00.001-07:002013-03-10T14:50:32.717-07:00RECIPE TO CREATE REAL RULES FOR DIVISIBILITY BY 7<div class="separator" style="clear: both; text-align: center;">
<iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/ZUozMuPE1RA?feature=player_embedded' frameborder='0'></iframe></div>
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BEFORE YOU READ THE TEXT I ASK YOU TO WATCH THIS VIDEO.</span><br />
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<span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;">
FOLLOW THESE </span></span><span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;">STEPS<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">1 – Choose a sequence of 3 to 6 of Pascal’s
multipliers: 546231546…<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">2 – Arrange the best way to perform the
operations.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">3 – Make use of the additive inverse mod 7 when
needed<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">4 – Create the algorithm<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">5 – Apply it.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">If necessary look at my last post.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">First Example:<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">1 - Chosen sequence: 231<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">2 – N = abc; 2 . a + bc<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">3 – In this case the additive inverse mod 7
will be used.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">5 – Application<o:p></o:p></span></span></span></div>
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<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">N = 154; - ( (2) + 54 ) mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">Ø</span><span lang="EN-US" style="mso-ansi-language: EN-US;">; 7|</span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">Ø</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">→</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 7|N<o:p></o:p></span></span></span></div>
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<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">N = 4.641; - 4 mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 3; ( 3 + (5) +
41 ) mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">Ø; 7|Ø<span style="mso-spacerun: yes;"> </span>→ 7|N<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 823.424; - ( (2) + 23 ) mod 7 ≡ 3;<span style="mso-spacerun: yes;">
</span>( 3 + (1) + 24 ) mod 7 ≡ Ø; 7|Ø → 7|N<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 6.453.243; - 6 mod 7 ≡ 1; - ( 1 + (1) + 53 ) mod 7 ≡ 1; ( 1 + (4) +
43 ) mod 7 ≡ 6; 7ł6; 7łN*<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">*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.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
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<b style="mso-bidi-font-weight: normal;"><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><o:p><span style="font-family: Calibri; font-size: large;"> </span></o:p></span></b></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Second example:<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">1 – Chosen sequence: 1546<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">2 – N = a.bcd; (6 . cd + a) mod 7; N is
reduced to a’b<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">3 – Additive inverse mod 7 is not
necessary as a link.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">4 – The algorithm: - (cd<span style="mso-spacerun: yes;"> </span>mod 7 + a ) mod 7 ≡ a’; it is applied from
right to left.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">5 – Application:<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 154 → Ø154; - (54 + Ø) mod 7 ≡ 2 = a’
→ a’b = 21; 7|21 and 7|N<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 4.641; - ( 41 mod 7 + 4 ) mod 7 ≡ 5 =
a’ → a’b = 56; 7|56 and 7|N<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 823.424; ( - 24 mod 7 + 3 ) mod 7 ≡ Ø
= a’ → a’b = Ø4; ( - 4 mod 7 + 8) mod 7 ≡ 4 = a’<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">→ a’b = 42; 7|42 and 7 |N<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 6.453.243; ( - 43 mod 8 + 3) mod 7 ≡ 2
= a’ → a’b = 22; ( - 22 mod 7 + 4 ) mod 7 ≡ 3 = a’<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">→ a’b = 35; ( - 35 mod 7 + Ø ) mod 7 ≡ Ø =
a’ → a’b = Ø6; 7ł6 and 7łN*<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">*Coincidentally the result was the same of
the anterior rule but generally this rule does not result in the remainder.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 38.453.952; ( - 52 mod 7 + 3 ) mod 7 ≡
Ø = a’ → a’b = Ø9; ( - 9 mod 7 + 4 ) mod 7 ≡ 2 = a’<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">→ a’b = 25; ( - 25 mod 7 + 3 ) mod 7 ≡ 6 =
a’ → a’b = 68; 68 mod 7 ≡ 5; 7ł5 and 7łN<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Changing what must be changed this rule
may be applied to divisibility by 11 and 13.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Third example:<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">1 – Chosen sequence: 462<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">2 – N = abc; 6 . a<sub>1</sub>b<sub>1</sub>
+ (x) + a<sub>2</sub>b<sub>2 …<o:p></o:p></sub></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">3 – Additive inverse mod 7 is not
necessary as a link<o:p></o:p></span></span></span></div>
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<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">4 – The algorithm: ( - a</span><sub><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">1</span></sub><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">b<sub>1</sub> mod 7 + (x) ) mod 7 + a<sub>2</sub>b<sub>2
….<o:p></o:p></sub></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">5 – Application:<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 154; ( - 15 mod 7 + (1) ) mod 7 ≡ Ø; 7|Ø and 7|N<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 4.641;<span style="mso-spacerun: yes;"> </span><span style="mso-spacerun: yes;"> </span>- [ ( (1) + 64 mod 7 ) + (2) ] mod 7 ≡ Ø; 7|Ø
and 7|N<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 823.424; [ ( - 82 mod 7 + (6) ] mod 7 ≡ 1; - [ ( 1 + 42 ) mod 7 +
(1) ] mod 7 ≡ Ø; 7|Ø and 7|N<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 6.453.243; - [ ( (5) + 45 ) mod 7 + (6) ] mod 7 ≡ 5; - [ ( 5 + 24 ) mod
7 ) + 6 ] mod 7 ≡ 5; 7ł5 and 7łN<o:p></o:p></span></span></span></div>
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</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 38.453.952; ( - 3 mod 7 + 2 ) mod 7 ≡ 6; [ -( 6 + 45 mod 7 ) + (6) ]
mod 7 ≡ 4;<span style="mso-spacerun: yes;"> </span>[ - ( 4 + 95 mod 7) + (4) ]
mod 7 ≡ 3; 7ł3 and 7łN<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Changing what must be changed this rule may be applied to divisibility
by 11 and 13.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">The remainder is the value of the digit that forms with the final result
a two-digit number multiple of 7. <o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div align="center" class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: center;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Fourth example:<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">1 - Chosen sequence: 315<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">2 – N = abc; ab + c + (y)<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">3 – Additive inverse mod 7 is necessary<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">4 – The algorithm: - ( ab + c + (y) ) mod
7 …<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">5 – Application:<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 154; - ( 15 + 4 + (2) ) mod 7 ≡ Ø; 7|Ø
and <span style="mso-spacerun: yes;"> </span>7|N<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 4.641; - ( 4 + (2) ) mod 7 ≡ 1; - ( 1
+ 64 + 1 + (4) ) mod 7 ≡ Ø; 7|Ø and 7|N<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 823.424; - ( 82 + 3 + (5) ) mod 7 ≡ 4;
- ( 4 + 42 + 4 + (2) ) mod 7 ≡ 3; 7ł3 and 7łN<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 6.453.243; - ( 6 + (3) ) mod 7 ≡ 5; -
( 5 + 45 + 3 + (5) ) mod 7 ≡ 5;<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">- ( 5 + 24 + 3 + (5) ) mod 7 ≡ 5; 7ł5 and
7łN<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 38.453.952; - ( 3 + 8 + (4) ) mod 7 ≡
6; - ( 6 + 45 + 3 + (5) ) mod 7 ≡ 4; - ( 4 + 95 + 2 + (1) ) ≡ 3<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Changing what must be changed this rule
may be applied to divisibility by 13.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">The remainder is the value of the digit
that forms with the final result a two-digit number multiple of 7.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div align="center" class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: center;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Fifth Example:<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">1 - Chosen sequence: 546231<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">2 – N = abcdef; 5 . a + 6 . bc + 2 . d +
ef<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">3 – Additive inverse mod 7 is not necessary<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">4 – The algorithm: ( - bc mod 7) + a + (y)
+ (x) + ef<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">5 – Application:<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 154; - 54 mod 7 + 1 + 4 = 7; 7|7 and
7|N<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 4.641; - 4 mod 7 + (5) + 41 = 49; 7|49
and 7|N<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 823.424; - 23 mod 7 + 8 + (4) + (1) +
24 = 42; 7|42 and 7|N<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 6.453.243; 6 + (- 53 mod 7) + 4 + (2)
+ (4) + 43 = 62; 7ł62 and 7łN; <o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">remainder = 62 mod 7 ≡ 6<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 38.453.952; 38 + (- 53 mod 7) + 4 +
(2) + (4) + 52 = 103; 7ł103 and 7łN<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">remainder = 103 mod 7 ≡ 5<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">This rule demands more attention and I
included it because it uses the whole sequence of Pascal’s multipliers.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></div>
<div class="separator" style="clear: both; text-align: center;">
<iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/ZUozMuPE1RA?feature=player_embedded' frameborder='0'></iframe></div>
<span style="font-size: large;">
</span>Silvio Moura Velhohttp://www.blogger.com/profile/18322243925876063572noreply@blogger.com0tag:blogger.com,1999:blog-19893442.post-74039267466528409632012-09-18T15:56:00.000-07:002013-03-23T14:05:37.322-07:00THE MOST DIFFICULT ELEMENTARY PROBLEM<div class="separator" style="clear: both; text-align: center;">
<iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/ZUozMuPE1RA?feature=player_embedded' frameborder='0'></iframe></div>
<div style="text-align: center;">
<span style="font-size: large;">
PLEASE, WATCH THIS VIDEO BEFORE YOU READ THE TEXT</span></div>
<div style="text-align: center;">
</div>
<div align="center" class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: center;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">CHALLENGE TO THE MATHEMATICAL ESTABLISHMENT<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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!!).<span style="mso-spacerun: yes;"> </span><o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">I learned and can teach <b style="mso-bidi-font-weight: normal;">anyone</b> how to create various real rules for divisibility by 7. Now
you can create your own rule!!!<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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!!!<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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. <span style="mso-spacerun: yes;"> </span>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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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 .<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">I challenge anyone to prove that:<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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;<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">2) My real rules (the first and the new ones)
were not created by me;<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">3) My real rules don’t work properly.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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. <o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">From now on I will present why my first real
rule for divisibility by 7 works.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div align="center" class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: center;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">Starting with the digit one in the center<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div align="center" class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: center;">
<b style="mso-bidi-font-weight: normal;"><span lang="EN-US" style="line-height: 115%; mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">(2) 1 (4)*<o:p></o:p></span></span></span></b></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="line-height: 115%; mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">(2) and (4)
were inferred because both (2) and 1 and 1 and (4) form a two-digit number
multiple of 7.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">If you want the product of 1 . 1 just keep the
1 the way it is.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">If you want the product of 6 . 1 just apply <span style="mso-spacerun: yes;"> </span>– 1 mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 6.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">If you substitute the one in the center by any
other digit you will observe that <b style="mso-bidi-font-weight: normal;">always</b>
the value of the digit on the left is the double of the value of the central
digit mod 7 and that <b style="mso-bidi-font-weight: normal;">always </b>the
value of the digit on the right is the quadruple of the central digit mod 7.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div align="center" class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: center;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">Pascal’s general criterion of divisibility<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">Regarding 7 the sequence of remainders is:<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">10<sup>0</sup>/7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">→</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> r = 1; 10<sup>1</sup>/7
</span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">→</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> r = 3; 10<sup>2</sup>/7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">→</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> r = 2; 10<sup>3</sup>/7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">→ r = 6; 10<sup>4</sup>/7 → r = 4; 10<sup>5</sup>/7 →
r = 5;<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">10<sup>6</sup>/7 → r = 1; 10<sup>7</sup>/7 → r = 3; …<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">From 10<sup>6</sup>/7 on the sequence repeats itself infinitely.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">To simplify communication I will call Pascal’s <b style="mso-bidi-font-weight: normal;">remainders as multipliers</b>.</span><span lang="EN-US" style="mso-ansi-language: EN-US;"><o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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 <b style="mso-bidi-font-weight: normal;">I think I
discovered something new!</b><o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><b style="mso-bidi-font-weight: normal;"><span lang="EN-US" style="mso-ansi-language: EN-US;">When N is a multiple
of 7</span></b><span lang="EN-US" style="mso-ansi-language: EN-US;">, if the order
of the sequence is maintained it is possible to apply any other multiplier to
the ones; <b style="mso-bidi-font-weight: normal;">the sum of the products will
be a multiple of 7</b>, 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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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:<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 154; 1 . 4 + 3 . 5 + 2 . 1 = 4 + 15 + 2 =
21; 7|21 and N; multipliers 1, 3 and 2<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;"><span style="mso-tab-count: 1;"> </span><span style="mso-spacerun: yes;"> </span>3 . 4 + 2 . 5 + 6 . 1 = 12 + 10 + 3 = 28; 7|28
and N; multipliers 3, 2 and 6<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;"><span style="mso-tab-count: 1;"> </span><span style="mso-spacerun: yes;"> </span>2 . 4 + 6 . 5 + 4 . 1 = 8 + 30 + 4 = 42; 7|42
and N; multipliers 2, 6 and 4<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;"><span style="mso-tab-count: 1;"> </span><b style="mso-bidi-font-weight: normal;">5 . 4 + 1 . 5 + 3 . 1 = 20 + 5 + 3 = 28;
7|28 and N; 5, 1 and 3 *</b><o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<b style="mso-bidi-font-weight: normal;"><span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">*This is the sequence
of multipliers applied to my first real rule for divisibility by 7: 5, 1 and 3
from right to left.<o:p></o:p></span></span></span></b></div>
<span style="font-size: large;">
</span><br />
<div align="center" class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: center;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">THE ALGORITHM<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">Let N = abc;<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;"><span style="mso-spacerun: yes;"> </span>- ( (x)
+ a + b + c + (y) ) mod 7<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">Remember: ( (x) + a ) mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 3 . a mod 7; b = 1 . b and ( c + (y) ) mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 5 . c mod 7<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 382.473<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">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. <b style="mso-bidi-font-weight: normal;">It is
indifferent if the inverse additive mod 7 is applied or not to the last period.<o:p></o:p></b></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">It must be applied successively to each period
of N. I prefer to apply the algorithm from left to right.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">– ( (6) + 3 + 8 + 2 + (1) ) mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 1; 1 + (1) + 4 + 7 + 3 + 5 = 21; 7|21 and N<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">N = 32.473; – ( 3 + 2 + (1) ) mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 1; 1 + (1) + 4 + 7 + 3 + 5 = 21; 7|21 and N<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">N = 4.473; - ( 4 + (2) ) mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 1; 1 + (1) + 4 + 7 + 3 + 5 = 21; 7|21 and N<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">Please note that you only infer (x)’s before
the hundreds digits.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div align="center" class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: center;">
<span lang="EN-US" style="mso-ansi-language: EN-US;"><span style="font-family: Calibri;"><span style="font-size: large;">Application to a really large number<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">N = 695.246.594.226 – ( (5) + 6 + 9 + 5 + (6) )
</span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 4; - ( 4 + (4) + 2 + 6 + (3) ) mod 7 </span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">≡</span><span lang="EN-US" style="mso-ansi-language: EN-US;"> 2; <o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span style="font-family: Calibri;"><span style="font-size: large;"><span lang="EN-US" style="mso-ansi-language: EN-US;">- ( 2 + (3) + 5 + 9 + 4 + (2) ) mod 7 = 3; 3 +
(4) + 2 + 2 + 6 + (3) = 20; 7</span><span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;">ł20 and 7łN<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div align="center" class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: center;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">The reason for the use of the inverse additive mod 7<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Observe that:<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 155.554; 554 – 155 = 399; 7|399 and 7|N<o:p></o:p></span></span></span></div>
<span style="font-size: large;">
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<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 155.555.554; 554 + 155 – 555 = 154; 7|154 and 7|N<o:p></o:p></span></span></span></div>
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</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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 <b style="mso-bidi-font-weight: normal;">will be always equivalents mod 7</b>.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">N = 155; (2) + 1 + 5 + 5 + (6) = 19; 19 mod 7 ≡ 5; - 5 mod 7 ≡ 2<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">4 . 1 + 6 . 5 + 2 . 5 = 4 + 30 + 10 = 44; 44 mod 7 ≡ 2<o:p></o:p></span></span></span></div>
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</span><br />
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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:<o:p></o:p></span></span></span></div>
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</span><br />
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Examples:<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">n = 54; ( 3 . 5 + 1 . 4) mod 7 ≡ 54 mod 7 ≡ 5<o:p></o:p></span></span></span></div>
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</span><br />
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">n = 83; ( 3 . 8 + 1 . 3) mod 7 ≡ 83 mod 7 ≡ 6<o:p></o:p></span></span></span></div>
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</span><br />
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">n = 23; ( 3 . 2 + 1 . 3) mod 7 ≡ 23 mod 7 ≡ 2<o:p></o:p></span></span></span></div>
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</span><br />
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">Examples:<o:p></o:p></span></span></span></div>
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</span><br />
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">n = 54; ( 4 . 5 + 6 . 4) mod 7 ≡ - 54 mod 7 ≡ 2 ≡ (6 . 54) mod 7<o:p></o:p></span></span></span></div>
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</span><br />
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">n = 83; ( 4 . 8 + 6 . 3 ) mod 7 ≡ - 83 mod 7 ≡ 1 ≡ ( 6 . 83 ) mod 7<o:p></o:p></span></span></span></div>
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</span><br />
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">n = 23; ( 4 . 2 + 6 . 3) mod 7 ≡ - 23 mod 7 ≡ 5 ≡ ( 6 . 23 ) mod 7<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">These last examples are based on the fact that for any number this equation
is valid:<o:p></o:p></span></span></span></div>
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<span style="mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">- N mod x ≡ [( x – 1) N ] mod x<o:p></o:p></span></span></span></div>
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<span style="mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;"><span style="mso-spacerun: yes;"> </span>x . N mod x – N mod x<o:p></o:p></span></span></span></div>
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<span style="mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">- N mod x <span style="mso-tab-count: 1;"> </span>≡ Ø<span style="mso-spacerun: yes;"> </span>– N mod
x<o:p></o:p></span></span></span></div>
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</span><br />
<div class="MsoNormal" style="margin: 0cm 0cm 10pt; text-align: justify;">
<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><span style="font-family: Calibri;"><span style="font-size: large;">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.<o:p></o:p></span></span></span></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US; mso-bidi-font-family: Calibri; mso-bidi-theme-font: minor-latin;"><o:p><span style="font-family: Calibri; font-size: large;"> </span></o:p></span></div>
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<b style="mso-bidi-font-weight: normal;"><span lang="EN-US" style="mso-ansi-language: EN-US;"><o:p><span style="font-family: Calibri; font-size: large;"> </span></o:p></span></b></div>
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<span lang="EN-US" style="mso-ansi-language: EN-US;"><o:p><span style="font-family: Calibri; font-size: large;"> </span></o:p></span></div>
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</span>Silvio Moura Velhohttp://www.blogger.com/profile/18322243925876063572noreply@blogger.com0tag:blogger.com,1999:blog-19893442.post-1134662891612817652005-12-15T06:11:00.000-08:002013-03-23T14:07:57.612-07:00<div class="separator" style="clear: both; text-align: center;">
<iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/ZUozMuPE1RA?feature=player_embedded' frameborder='0'></iframe></div>
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PLEASE WATCH THIS VIDEO BEFORE YOU READ THE TEXT</div>
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<strong>DIVISIBILITY TESTS</strong></div>
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<strong>To find a quick method for divisibility by 7 is the most difficult problem related to Elementary Mathematics.</strong></div>
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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.</div>
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I have created the site: <a href="http://www.7and13divisibility.com/">www.7and13divisibility.com</a> (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.</div>
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<strong>"MOURA VELHO METHOD"1</strong></div>
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<strong>divisibility by 7</strong></div>
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<strong>This method is based entirely on mental calculation. Forget paper and pencil!<br />Look at this practical example and follow the explanation.<br /><br />Let n = 36,554<br /><br />36(3),(3)554(2)<br /><br />3+6+3 = 12; 12 to 14 = "2" ; "2" +3+5+5+4+2 = 21<br /><br />then 7 divides n exactly.<br /><br />The digits between parenthesis and their neighbors form a multiple of 7; they are mentally inserted while the addition is processed.<br />No insertion is needed regarding the tens.<br />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".<br /><br />If the last sum is a multiple of 7 then the same happens to the given number.<br /><br />Other examples:<br /><br />1) Let n = 266,453<br /><br />(4)266(3),(1)453(5)<br /><br />4+2+6+6+3 = 21; "0"+1+4+5+3+5 = 18<br /><br />then 266,453 is not a multiple of 7.<br /><br />2) Let n = 3,126,495<br /><br />3(5),(2)126(3),(1)495(6)<br /><br />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<br /><br />then 3,126,495 is not a multiple of 7.<br /><br />THE REMAINDER<br /><br />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.<br />If you would like to know the remainder when the given number is not a multiple of 7 then there is a final step:<br /><br />Create a last link with the last sum; the digit that forms a multiple of 7 with this last link is the remainder.<br /><br />The last sum of the example 1 is 18; 18 to 21 = 3; 3(5); then the remainder is 5.<br />Regarding example 2 we have 26; 26 to 28 = 2; 2(1); then the remainder is 1.<br /><br /><br />Copyright 2005 by Silvio Moura Velho, all rights reserved</strong></div>
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Silvio Moura Velhohttp://www.blogger.com/profile/18322243925876063572noreply@blogger.com0