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My proffesor gave me this to problem to resolve it, but I really do not know how to do this 

8444...4...455= divisible to 19

Between the 8 and the two 5 are 2017 fours

 

The concept of the divisibility of 19 is:

We have to take the digit of the units  and multiply it by 2, this number we add the rest of the numbers

 

If someone knows how to do this let me know pls or if u just know a little bit or have an idea of how to do it tell me cheeky

LucyWhat  Apr 2, 2017
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 #1
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I will give it a go!

8444...4...455= divisible to 19

 

8444...4...45+(2x5) =

8444...4...55 /19 =444,445

P.S. Any number of ODD fours between 8 and 55 should be divisible by 19. Since 2017 fours is an odd number of fours, it should also be divisible by 19.

That is my take on it!.

Guest Apr 2, 2017
 #2
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Thank u so much!! You save me haha laugh

LucyWhat  Apr 2, 2017
 #3
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I just have a question, why you divided 8444...4...455/19?

LucyWhat  Apr 2, 2017
 #4
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To show you that any ODD number of fours between 8 and 55 is EVENLY divisible by 19.

If you have 7 fours between 8 and 55 you have this:

844444445+(2x5)=

844444455/19 =44,444,445.

Guest Apr 2, 2017
 #5
avatar+18358 
+2

My proffesor gave me this to problem to resolve it, but I really do not know how to do this 

8444...4...455= divisible to 19

Between the 8 and the two 5 are 2017 fours

 

8444...4...455 is divisible to 19, if 8444...4...455 modulo 19 = 0

 

  • We calculate 8444...4...455 modulo 19:
    We divide 8444...4...455 in parts of 3-digits. ( The partition is arbitrary ).
    But the first part is 8444.

So the Number is:

\(\begin{array}{lcll} \underbrace{8444}_{\text{ Fist Part }}\ \underbrace{444}_{\text{ Second Part }}\ 444\ 444\ 444\ 444\ 444\ 444\ 444\ \ldots \underbrace{444}_{\text{ Part 672 }} \ \underbrace{455}_{\text{ Last Part }} \qquad \text{2017 fours} \\ \end{array} \)

 

\(\begin{array}{|rcll|} \hline && \text{first part} \pmod {19} \\ &\equiv& 8444\pmod {19}\\ &=& \color{red}8 \color{black}\qquad (8444=444\cdot 19 + \color{red}8 \color{black}) \\\\ && \text{The last remainder } = \color{red}8 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline && \text{Second part} \pmod {19} \\ &\equiv& \underbrace{8}_{\text{ The last remainder }}\underbrace{444}_{\text{ Second part }}\pmod {19}\\ &=& \color{red}8 \color{black}\qquad (8444=444\cdot 19 + \color{red}8 \color{black}) \\\\ && \text{The new last remainder } = 8 \\ \hline \end{array} \)

 

\(\cdots \)

 

\(\begin{array}{|rcll|} \hline && \text{Part 672} \pmod {19} \\ &\equiv& \underbrace{8}_{\text{ The last remainder }}\underbrace{444}_{\text{ Part 672 }}\pmod {19}\\ &=& \color{red}8 \color{black}\qquad (8444=444\cdot 19 + \color{red}8 \color{black}) \\\\ && \text{The new last remainder } = 8 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \text{Last part} \pmod {19} \\ &\equiv& \underbrace{8}_{\text{ The last remainder }}\underbrace{455}_{\text{ Last part }}\pmod {19}\\ &=& \color{red}0 \color{black}\qquad (8455=445\cdot 19 + \color{red}0 \color{black}) \\\\ && \text{The new last remainder } = 0 \\ \hline \end{array} \)

 

8444...4...455 is divisible to 19, because 8444...4...455 modulo 19 = 0

 

laugh

heureka  Apr 3, 2017

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