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

LucyWhat
Apr 2, 2017

#1**+1 **

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

#4**+1 **

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**+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

heureka
Apr 3, 2017