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If $4$ is written $4444$ times side by side, we shall get a number of $4444$ digits. What is the remainder when $7$ divides this large number?

Answer of this question is $1$ but I did not understand how?

Using this rule I tried solving the question, If P is a prime number>5 then any digit written (N-1) times will be divisible by P where N is a recurring decimal. Number of times digit gets repeated in $1/7$ is $6$

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  • $\begingroup$ Your numbers are of the form 4(10^n - 1)/9. This should help you. Think " periodic ". $\endgroup$
    – mick
    Mar 19, 2017 at 19:48

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Hint: consider $4,44,444,4444,44444,444444,\cdots $ modulo $7$. This is $4,2,3,6,1,0,\; 4,2,3,6,1,0,\cdots $. Do you see a pattern?

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  • $\begingroup$ Yes, I saw the pattern. But what I have to do of it now? @DietrichBurde $\endgroup$
    – Shobhit Gupta
    Mar 20, 2017 at 14:09
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The answer is not $1$.

$$ \frac{4}{9}(10^{4444}-1)\equiv 2(10^{4444}-1) \equiv 2(3^{4444}-1) \equiv 2(3^4-1) \equiv \color{red}{-1} \pmod{7} $$ by Fermat's little theorem.

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  • $\begingroup$ How did you get $2(10^{4444} - 1)$? $\endgroup$
    – Toby Mak
    Jun 14, 2017 at 13:23

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