The { Odd-On-Odd Format } for { 'odd' numbers }

In this Section , we attempt to determine whether a { number } is divisible by [ 3 ] , approaching from :

• the { Base-10 Format } ,
• the { Binary Format } , &
• the { Odd-On-Odd Format } .

In the last of these , we were able to move-in on a { Binary Code } from-left-to-right .

Whenever we wish to determine if a { number } , in { Base-10 Format } , is divisible by [ 3 ] ,

• we can always add up all the digits and determine if the [ sum ] is divisible by [ 3 ] ,

  and if it is , then the { number } is divisible by [ 3 ] .

A quick example here for the number [ 246,813,579 ] :

• { Sum of the digits } = 2 + 4 + 6 + 8 + 1 + 3 + 5 + 7 + 9 = 45 ,

  And [ 45 ] is divisible by [ 3 ] , so that [ 246,813, 579 ] is also divisible by [ 3 ] .

Let us briefly explain why this is so .

But let us first mark-off the digits as [ Digit #1 ] , [ Digit #2 ] , [ Digit #3 ] and so-on-and-so-forth ,

• counting from the [ Unit-Digit ] on-the-very-right and moving left-wards ,

  as per this diagram below :

And we note that the { unit-value } for each of the digits is always congruent to [ 1 mod 3 ] ,

• as per this table below :

Let us take { Digit #4 } as an example , and this { Digit } carries a [ unit-value ] of [ 1,000 ] :

• if this { Digit } is [ 1 ] , then the associated value is [ 1,000 ] , which is congruent to [ 1 mod 3 ] ,
• if this { Digit } is [ 2 ] , then the associated value is [ 2,000 ] , which is congruent to [ 2 mod 3 ] ,
• if this { Digit } is [ 3 ] , then the associated value is [ 3,000 ] , which is congruent to [ 3 mod 3 ] ,
• if this { Digit } is [ 4 ] , then the associated value is [ 4,000 ] , which is congruent to [ 4 mod 3 ] ,
• and so-on-and-so-forth ,

so that :

• adding up all the digits and finding out whether the { sum } is divisible by [ 3 ] is the same as :
  • adding up all the component parts of the { number } and finding out whether that { sum } , which is the { number } itself , is divisible by [ 3 ] .

  And that is to say :

  • { [ 200,000,000 + 40,000,000 + 6,000,000 + 800,000 + 10,000 + 3,000 + 500 + 70 + 9 ] mod 3 }

    is congruent to { [ 2 + 4 + 6 + 8 + 1 + 3 + 5 + 7 + 9 ] mod 3 } .

Let us now attempt to develop a procedure to determine whether a { number } , in { binary format } , is divisible by [ 3 ] .

And we start-off , again , by marking-off the digits as [ Digit #1 ] , [ Digit #2 ] , [ Digit #3 ] and so-on-and-so-forth ,

• counting from the [ Unit-Digit ] on-the-very-right and moving left-wards ,

  as per this diagram below :

And we note that the { unit-value } for each of the digits is always :

• alternately congruent to [ 1 mod 3 ] and [ 2 mod 3 ] ,

  as per this table below :

Our proposal here for the procedure , based on the same logical sequence above for { Base-10 } , is then :

• Set up the value [ S1 ] as the sum of all the { 'odd'-position digits } ,

• Set up the value [ S2 ] as the sum of all the { 'even'-position digits } ,

• Set up the value [ Adjusted-Sum ] such that :
  • [ Adjusted-Sum ] = 1 x [ S1 ] + 2 x [ S2 ]

  If the [ Adjusted-Sum ] is divisible by [ 3 ] , then the { number } is also divisible by [ 3 ] ;

  • otherwise the { number } is not divisible by [ 3 ] .

And the 'adjustment' here is based on applying different 'weights' :

• a 'weight' of [ 1 ] for all the { 'odd'-position digits } whose { unit-values } are always congruent to [ 1 mod 3 ] , &

• a 'weight' of [ 2 ] for all the { 'even'-position digits } whose { unit-values } are always congruent to [ 2 mod 3 ] .

A quick example here for the number [ 441 ] , as shown in the diagram , above :

• [ S1 ] = [ Digit #9 ] + [ Digit #7 ] + [ Digit #5 ] + [ Digit #3 ] + [ Digit #1 ]

  = 1 + 0 + 1 + 0 + 1 = 3

• [ S2 ] = [ Digit #8 ] + [ Digit #6 ] + [ Digit #4 ] + [ Digit #2 ]

  = 1 + 1 + 1 + 0 = 3

• [ Adjusted-Sum ] = 1 x [ S1 ] + 2 x [ S2 ] = 1 x 3 + 2 x 3 = 9 .

  Since [ 9 ] is divisible by [ 3 ] , the number [ 441 ] is also divisible by [ 3 ] .

Let us now attempt to develop a procedure to determine whether a { number } , in the { Odd-On-Odd Format } , is divisible by [ 3 ] .

But let us first set up three (3) { base numbers } to start-off our investigation here :

• the number [ 15 ] , which is congruent to [ 0 mod 3 ] ,
• the number [ 13 ] , which is congruent to [ 1 mod 3 ] , &
• the number [ 11 ] , which is congruent to [ 2 mod 3 ] ;

  as per this set of 3 diagrams below :

Let us now add a set of { 5 red-color balls } to-the-right of each of these 3 { base numbers } ,

• as per these 3 diagrams below :

Let us now make 3 observations on this { Adding-Red-After-Red System } :

• If we start-off with the value of [ 0 mod 3 ] and keep on adding { red-color balls } ,
  • the { resultant value } alternates bewteen [ 1 mod 3 ] & [ 0 mod 3 ] ;

• If we start-off with the value of [ 1 mod 3 ] and keep on adding { red-color balls } ,
  • the { resultant value } alternates bewteen [ 0 mod 3 ] & [ 1 mod 3 ] ;

• If we start-off with the value of [ 2 mod 3 ] and keep on adding { red-color balls } ,
  • the { resultant value } stays constant at [ 2 mod 3 ] .

Let us now add a set of { 5 blue-color balls } to-the-right of each of these 3 { base numbers } ,

• as per these 3 diagrams below :

Let us now make 3 observations on this { Adding-Blue-After-Blue System } :

• If we start-off with the value of [ 0 mod 3 ] and keep on adding { blue-color balls } ,
  • the { resultant value } alternates bewteen [ 2 mod 3 ] & [ 0 mod 3 ] ;

• If we start-off with the value of [ 1 mod 3 ] and keep on adding { blue-color balls } ,
  • the { resultant value } stays constant at [ 1 mod 3 ] ;

• If we start-off with the value of [ 2 mod 3 ] and keep on adding { blue-color balls } ,
  • the { resultant value } alternates bewteen [ 0 mod 3 ] & [ 2 mod 3 ] .

Let us take a quick application of these resultant rules on the example { N = 441 } , as shown below :

We always start-off with the very-first { red-color ball } on-the-left , which always gives us the number [ 3 ] ,

• and the starting value of [ 0 mod 3 ] .

• the next single { red-color ball } then brings the value from [ 0 mod 3 ] to [ 1 mod 3 ] ,
• the next single { blue-color ball } then let the value stay constant at [ 1 mod 3 ] ,
• the next set of { 3 red-color balls } then brings the value from [ 1 mod 3 ] to [ 0 mod 3 ] ,
• and the final set of { 2 blue-color balls } let the value stay-the-same at [ 0 mod 3 ] .

Thus , the number [ 441 ] is divisible by [ 3 ] .

But before we leave here , let us also make this observation , in preparartion for the next { Topic } below :

• If we add a set of { 2 red-color balls } or { 2 blue-color balls } , the value always stays the same ,
  • regardless of whether we started-off with a [ 0 mod 3 ] , a [ 1 mod 3 ] , or a [ 2 mod 3 ] .

  as per this set of 2 tables below .

  Start-Off Value	Pair of color-balls added	Resultant Value
  0 mod 3		0 mod 3
  1 mod 3		1 mod 3
  2 mod 3		2 mod 3

  Start-Off Value	Pair of color-balls added	Resultant Value
  0 mod 3		0 mod 3
  1 mod 3		1 mod 3
  2 mod 3		2 mod 3

Let us now investigate what happens where we add a set of { 2 color-balls of different colors } .

For the { Red-color-ball / Blue-color-ball Combination } :

• the applicable equation is :
  • { Resultant Value } = [ 2* ( 2* { Start-off Value } + 1 ) - 1 ] mod 3

  or ,

  • { Resultant Value } = [ 4 * { Start-off Value } + 2 - 1 ] mod 3

  yielding :

  • { Resultant Value } = [ 4 * { Start-off Value } + 1 ] mod 3

  Since ( [ 4 * { Start-Off Value } ] mod 3 ) is always congruent to ( [ 1 * { Start-Off Value } ] mod 3 ) ,

  • ( because [ 4 ] is congruent to [ 1 mod 3 ] )

  the above equation then becomes :

  • { Resultant Value } = [ { Start-off Value } + 1 ] mod 3

  We then have this summary table below for the { Red-Blue Combination } :

  Start-Off Value	Pair of color-balls added	Resultant Value
  0 mod 3		1 mod 3
  1 mod 3		2 mod 3
  2 mod 3		0 mod 3

For the { Blue-color-ball / Red-color-ball Combination } :

• the applicable equation is :
  • { Resultant Value } = [ 2* ( 2* { Start-off Value } - 1 ) + 1 ] mod 3

  or ,

  • { Resultant Value } = [ 4 * { Start-off Value } - 2 + 1 ] mod 3

  yielding :

  • { Resultant Value } = [ 4 * { Start-off Value } - 1 ] mod 3

  Again , since ( [ 4 * { Start-Off Value } ] mod 3 ) is always congruent to ( [ 1 * { Start-Off Value } ] mod 3 ) ,

  the above equation then becomes :

  • { Resultant Value } = [ { Start-off Value } - 1 ] mod 3

  We then have this summary table below for the { Blue-Red Combination } :

  Start-Off Value	Pair of color-balls added	Resultant Value
  0 mod 3		2 mod 3
  1 mod 3		0 mod 3
  2 mod 3		1 mod 3

Let us now summarize the actions for adding a set of { 2 color-balls } :

• : the value remains the same for the { red-red combination } ,

• : add [ 1 ] to the value for the { red-blue combination } ,

• : the value remains the same for the { blue-blue combination } ,

• : add [ 2 ] to the value for the { blue-red combination } .

Let us now look at a quick application to the number [ 441 ] , as follows :

Again , we always start-off with the very-first { red-color ball } on-the-left , which always gives us the number [ 3 ] ,

• and the starting value of [ 0 mod 3 ] .

• the next 3 pairs of { 2 color-balls } then brings the value from [ 0 mod 3 ] to [ 2 mod 3 ] ,
  • the 2 { red-blue combinations } adds [ 2 ] to the value and the single [ red-red combination ] let the value remain the same .

• the final { blue-color ball } then brings the value from [ 2 mod 3 ] to [ 0 mod 3 ] .

  And [ 441 ] is therefore divisble by [ 3 ] .

It is then recognized that this { adding pairs of color-balls } system does move us from a [ binary / base-2 ] evaluation scheme to a [ quad / base-4 ] evaluation scheme .

While the reader may now wish to try divisibility by [ 5 ] , [ 7 ] , [ 11 ] , [ 13 ] , etc. ,

• and posssible move towards the { Chinese Remainder Theorem } ,

let us now take a look at the number [ 233 ] , arising from the { Fermat Little Theorem } ,