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Let $n$ be the smallest integer:$7n$ has $1984$ digits

What is the units digit of $n$?

My attempt:

I have deduced that the number $7n$ must begin with a $1$ and have $1982$ $0s$ after that, ending in the digit $c$ which I need to find.

Next I used long division to determine that:

$\frac{7n}{7} = \frac{10000000...00c}{7} =$ the $6$ digits $142857$ repeated many times till the number terminates on $c$ = one of the digits $1,4,2,8,5,7$

So to determine $c$, We calculate $1984mod6 \equiv 4mod6$

Therefore $c =$ the fourth digit in the pattern of 6 repeated digits = $8$

Is my attempt correct, and if yes, is it efficient?

EDIT:

Having looked at the inputs from members on the site, I have identified my errors. I should have used $1983\equiv 3mod6$ and then used the remainders corresponding to each recurring digit i.e. $1 \rightarrow 1, 4 \rightarrow 1, 2\rightarrow 3, 8 \rightarrow 2, 5\rightarrow 6, 7 \rightarrow 4$ Then the remainder corresponding to the third digit in $142857$ gives us $c=3$

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    well, if $10^{1083}\equiv 6 \pmod 7$ then $10^{1083}+1\equiv 0 \pmod 7$. – 2017-02-24
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    I don't understand what you're getting at – 2017-02-24
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    "What is the units digit of $n$?" Did you mean $7n$? – 2017-02-24
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    So we know that $10^{1083}+1$ is an integer with $1084$ digits AND it is divisible by $7$. Clearly it is the smallest such, yes? – 2017-02-24
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    @mathlove I did mean $n$ – 2017-02-24
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    @lulu But the number needs to be 1984 digits long – 2017-02-24
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    It is $1984$ digits long. Remember, $10^1$ is two digits long...$10^2$ is three digits long...$10^{1083}$ is $1084$ digits long. Adding $1$ doesn't change the length. – 2017-02-24
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    Sorry if I'm being dense, but could you make your comment into a more idiot proof one since I still don't understand what you're trying to say in relation to the question – 2017-02-24
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    @mrnovice: OK. I think that lulu has typos. He should mean that $10^{1\color{red}{9}83}\equiv 6\pmod 7$. So, $7n=100\cdots 01$. Now consider multiplication instead of division. Hint : $7\times 3=21$. – 2017-02-24
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    @mathlove I don't think there are any typos in what I wrote. True, I didn't do the division by $7$, but neither did I claim to. – 2017-02-24
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    @lulu: So, you actually meant $10^{1\color{red}{0}83}\equiv 6\pmod 7$ instead of $10^{1\color{red}{9}83}\equiv 6\pmod 7$? – 2017-02-24
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    @mathlove Thanks! For some reason, I am systematically mixing $1084$ and $1984$. One of those days. I'll post something below which (hopefully) omits the blunders. – 2017-02-24

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Note: in the comments I, for whatever reason, systematically confused $1984$ with $1084$ (and $1983$ with $1083$. In a weird way, I was consistent there). Let me post something which, hopefully, avoids the blunders.

First, let's produce the smallest multiple of $7$ which has $1984$ digits, call it $M$. Since every seventh integer is divisible by $7$ we know there is one near $10^{1983}$.

First: note that $10^6\equiv 1\pmod 7$ and $1983\equiv 3 \pmod 6$. It follows that $$10^{1983}\equiv 10^3\equiv 6 \pmod 7$$

This in turn implies that $$10^{1983}+1\equiv 0 \pmod 7$$ We see we have found our $M$.

Now, to answer the question, we need to compute the units place of $\frac M7$. But $7a\equiv 1\pmod {10} \implies a\equiv 3\pmod {10}$ so the units place we seek is $\fbox 3$.

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    Ok I think I understand my mistake now, I should have used $1983\equiv 3mod6$ and then used the remainders corresponding to each recurring digit i.e. $1 \rightarrow 1, 4 \rightarrow 1, 2\rightarrow 3, 8 \rightarrow 2, 5\rightarrow 6, 7 \rightarrow 4$ Then the remainder corresponding to the third digit in $142857$ gives us $c=3$ as you arrived at – 2017-02-24
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    That sounds right. Of course, you don't really need to compute the other digits (though you certainly can). If all you want is the result $\pmod {10}$ then you can just work $\pmod {10}$. We have $M\equiv 1 \pmod {10}$ so we want $n$ such that $7n\equiv M\equiv 1 \pmod {10}$ and that immediately gets us $n\equiv 3 \pmod {10}$. – 2017-02-24
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    Yes I see this method is more efficient. Was making sure I understood where I went wrong in my initial method. – 2017-02-24