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We have 111...11 a hundred times.

We begin with 1 in our quotient. We have 93 ones remaining. We add six zeros according to the division algorithm to get 1000000 in our quotient. We add another one to get 10000001 in our quotient and we have 86 ones remaining. We see that we get the following answer by the same logic : 1000000100000010000001...1000001 (14 ones because 7*14 = 98 ones taken out so we have 11 left as a remainder)

My question is not on the remainder but on the way the quotient is presented : Why do we say : 1000000100000010000001...100000100 is the quotient and not 1000000100000010000001...10000010 with only one zero at the end ? According to the division algorithm shouldn't I need only one zero to make "11" go down before hitting the decimals ?

Thanks

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  • $\begingroup$ 1000000100000010000001...1000000100 contains 101 digits $\endgroup$
    – MAN-MADE
    Jul 5 '17 at 4:56
  • $\begingroup$ you must see that here 1000000100000010000001..."100000"100, in the quoted part you are missing a 0 as you explained $\endgroup$
    – MAN-MADE
    Jul 5 '17 at 5:04
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We have that, letting $N$ be $111...1$ (100 ones), M be $1111111$, $Q$ be the quotient, and $R$ be the remainder,

$$N=M\cdot Q+R$$

However, if $M$ had only one zero at the end, $M\cdot Q$ would be $111...10$, while we need it to be $111...100$.

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  • $\begingroup$ I'm not sure of understanding why it needs to be two zeros and not only one. If we divided 11/1111111 we would need only one zero before hitting the decimals... I already know that there needs to be two of them, I'm asking an explanation for this $\endgroup$ Jul 5 '17 at 3:44
  • $\begingroup$ If $M\cdot Q$ only had one zero at the end, it would end up ending in $...111121$ when we added $11$. Do you understand why the answer is $11$? $\endgroup$ Jul 5 '17 at 20:57
  • $\begingroup$ @CarlSchildkraut Sorry to bump an old thread, but I came across this and don't understand the explanation. I also have no math background. $\endgroup$
    – soju
    Feb 24 '18 at 20:45
  • $\begingroup$ @soju Are you familiar with modular arithmetic at all? It makes some of the ideas here easier to understand. $\endgroup$ Feb 25 '18 at 19:08
  • $\begingroup$ @CarlSchildkraut Hey thanks for getting back to me. Not at all, I am currently self studying Gelfands Algebra. I have no math background, just learned long division yesterday. $\endgroup$
    – soju
    Feb 26 '18 at 0:18
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$111\dots 11$ ($100$ times) is very big.

Lets take $111,111,111,11$, divide it by $111$.

Now $11111111111=111\times 10^8+111\times 10^5+111\times 10^2+11$.

So remainder is $11$.

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Each of your blocks of $1000000$ has seven digits. You want the whole quotient to have $100-7=93$ digits. Thirteen blocks of $1000000$ account for $91$ digits so you need two more, which are $10$. I think you didn't count that first $1$ as a digit when you thought you needed two more zeros.

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  • $\begingroup$ The quotient does end in $10$. Why do you think we say it ends in $100$? $\endgroup$ Jul 6 '17 at 4:02
  • $\begingroup$ Forget it, I solved my problem. It was just a false manipulation from my part. And no, the quotient does indeed finish with 100. Just go try it in a online calculator they all give this answer $\endgroup$ Jul 6 '17 at 5:45
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HINT:

We know $$\underbrace{11\cdots11}_{n \text{ digits}}=\dfrac{10^n-1}{10-1}$$

Now use : Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$

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