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8 dezenas impares fixas e 7 pares moveis

sorel

By sorel
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sorel Grand Master

sorel

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grande  COELHO,na planilia preciso das impares e pares das repetidas e das ausentes

 um exemplo repetiu 8 dezenas dificilmente seram 8 pares ou  8 impares

vamos supor que repetiu 5 impares difilmente vai repetir todas as 5 impares

e das ausentes tambem

objetivo usar das repetidas e das ausentes pares e impares e nao pelo todos as 25 dezenas

 consegue?

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gilberto miranda Experienced

gilberto miranda

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14 horas atrás, Crazy Rabbit disse:

Essa deu trabalho, tive que mudar o botão giratório, não sei

onde o ganhado colocou.

 

Todos os créditos ao GANHADO

 

La vai,

 

http://www.mediafire.com/file/ohg46a0z9x9o6b2

 

Crazy Rabbit

Muito bom mesmo , o coelho eu tenho uma outra aqui que gostaria que você desse uma olhada nela  caso me permita vou te enviar ela esta com uns problemas mas parece ser boa .

 

Saúde e sorte sempre.

Gilberto.

 

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sorel Grand Master

sorel

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ola Miranda a baixo a matriz das 7 moveis pares

 ou seja 3,4 das ausentes pares, somente bom podemos usar a matri permutável nas 8 impares fixas

 so precisa ser mais fixas

Talvez a matriz abaixo poderia ser de alguma utilidade para você. Assim, 3 números em uma linha, 4 números em uma coluna. E desde 7 * 4 é mais do que 7 * 3, você sacrifize colunas 2 e 6, desde que você menos como estes dois números em uma fileira.

X0xx000

X0x0x00

X00x00x

X000x0x

00xxx00

00xx00x

0000xxx

Eu acredito que esta matriz e seu espelho juntos cobrem todas as 10 combinações de números diferentes de 2 e 6 em suas primeiras 6 linhas. Ou seja, 12 linhas com 8 combinações de espelhos e 4 linhas contendo 2 repetidas combinações únicas (simétricas). A última linha contém o número 6 (espelho 2). Espero que isso ajude com o seu problema.

Para cobrir todos os 35, eu não sei se você pode compor 5 tais matrizes. Seria muito trabalho para tentar (a menos que você sabe alguma fórmula para ajudá-lo, o que eu não). Mas não pode certamente ser com matrizes do espelho, desde que há 5 delas, assim que pelo menos uma delas deve espelhar-se. Mas existem ony 3 combinações únicas, não 7.
 
 
 
 
 
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perhaps the matrix below could be of some use to you. So, 3 numbers in a row, 4 numbers in a column. And since 7*4 is more than 7*3, you sacrifize columns 2 and 6, since you least like these two numbers in a row. x0xx000 x0x0x00 x00x00x x000x0x 00xxx00 00xx00x 0000xxx I believe that this matrix and it's mirror together cover all 10 combinations of numbers other than 2 and 6 in their first 6 rows. That is, 12 rows with 8 mirror combinations and 4 rows containing 2 repeated unique (symetric) combinations. The last row contains number 6 (mirror 2). I hope this helps with your problem. To cover all 35, I don't know if you can compose 5 such matrices. It would be a lot of work to try (unless you know some formula to help you, which I don't). But it certainly can't be with mirror matrices, since there is 5 of them, so at least one of them should mirror itself. But there are ony 3 unique combinatios, not 7.
 
 
 
 
 
 
 
 
Definições de perhaps the matrix below could be of some use to you. So, 3 numbers in a row, 4 numbers in a column. And since 7*4 is more than 7*3, you sacrifize columns 2 and 6, since you least like these two numbers in a row. x0xx000 x0x0x00 x00x00x x000x0x 00xxx00 00xx00x 0000xxx I believe that this matrix and it's mirror together cover all 10 combinations of numbers other than 2 and 6 in their first 6 rows. That is, 12 rows with 8 mirror combinations and 4 rows containing 2 repeated unique (symetric) combinations. The last row contains number 6 (mirror 2). I hope this helps with your problem. To cover all 35, I don't know if you can compose 5 such matrices. It would be a lot of work to try (unless you know some formula to help you, which I don't). But it certainly can't be with mirror matrices, since there is 5 of them, so at least one of them should mirror itself. But there are ony 3 unique combinatios, not 7.
 
 
 
Sinônimos de perhaps the matrix below could be of some use to you. So, 3 numbers in a row, 4 numbers in a column. And since 7*4 is more than 7*3, you sacrifize columns 2 and 6, since you least like these two numbers in a row. x0xx000 x0x0x00 x00x00x x000x0x 00xxx00 00xx00x 0000xxx I believe that this matrix and it's mirror together cover all 10 combinations of numbers other than 2 and 6 in their first 6 rows. That is, 12 rows with 8 mirror combinations and 4 rows containing 2 repeated unique (symetric) combinations. The last row contains number 6 (mirror 2). I hope this helps with your problem. To cover all 35, I don't know if you can compose 5 such matrices. It would be a lot of work to try (unless you know some formula to help you, which I don't). But it certainly can't be with mirror matrices, since there is 5 of them, so at least one of them should mirror itself. But there are ony 3 unique combinatios, not 7.
 
 
 
Exemplos de perhaps the matrix below could be of some use to you. So, 3 numbers in a row, 4 numbers in a column. And since 7*4 is more than 7*3, you sacrifize columns 2 and 6, since you least like these two numbers in a row. x0xx000 x0x0x00 x00x00x x000x0x 00xxx00 00xx00x 0000xxx I believe that this matrix and it's mirror together cover all 10 combinations of numbers other than 2 and 6 in their first 6 rows. That is, 12 rows with 8 mirror combinations and 4 rows containing 2 repeated unique (symetric) combinations. The last row contains number 6 (mirror 2). I hope this helps with your problem. To cover all 35, I don't know if you can compose 5 such matrices. It would be a lot of work to try (unless you know some formula to help you, which I don't). But it certainly can't be with mirror matrices, since there is 5 of them, so at least one of them should mirror itself. But there are ony 3 unique combinatios, not 7.
 
 
 
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Traduções de perhaps the matrix below could be of some use to you. So, 3 numbers in a row, 4 numbers in a column. And since 7*4 is more than 7*3, you sacrifize columns 2 and 6, since you least like these two numbers in a row. x0xx000 x0x0x00 x00x00x x000x0x 00xxx00 00xx00x 0000xxx I believe that this matrix and it's mirror together cover all 10 combinations of numbers other than 2 and 6 in their first 6 rows. That is, 12 rows with 8 mirror combinations and 4 rows containing 2 repeated unique (symetric) combinations. The last row contains number 6 (mirror 2). I hope this helps with your problem. To cover all 35, I don't know if you can compose 5 such matrices. It would be a lot of work to try (unless you know some formula to help you, which I don't). But it certainly can't be with mirror matrices, since there is 5 of them, so at least one of them should mirror itself. But there are ony 3 unique combinatios, not 7.
 
 
 
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