I was playing with my calculator and found out $987654321$ $÷$ $123456789$ is very close to $8$. Trying out some more numbers i saw $998877665544332211$ $÷$ $112233445566778899$ gives $8.9$. When when you repeat the digits thrice, it gives $8.99$. Can it be proved that this value tends to $9$ if the digits are repeated forever.
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Note: 0.11111111.... times 9 is 0.99999999... – J. W. Tanner Jul 21 '19 at 20:52
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Yeah i guess if put any number ahead of even those 99 i mentioned above the result would still be pretty close to 8.9 – Manav Goyal Jul 21 '19 at 20:54
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Possible duplicate of Why is $\frac{987654321}{123456789} = 8.0000000729?!$ – Jam Jul 21 '19 at 20:56
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@Jam, I think that title is misleading, as he is asking for the limiting case, and so this is not really a duplicate of that. – Isaac Browne Jul 21 '19 at 20:57
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@IsaacBrowne It's not perfect and feel free to improve it. But it's definitely better than "interesting result". – Jam Jul 21 '19 at 20:59
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@Jam Sure, but I think the possible duplicate vote is unwarranted, as this isn't really a possible duplicate in anything but that title. – Isaac Browne Jul 21 '19 at 21:01
4 Answers
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The division becomes essentially $999999 \div 111111$, which is $9$. The difference between your dividend and the same number with all $9$s becomes a smaller fraction as the number of digits increases, as does the difference between the dividend and the number with all $1$s.
We can make this more precise. Define your ratio as $\frac {b_n}{c_n}$ where there are $n$ of each digit in the numerator and the denominator. The numerator then starts with $n\ 9$s and has $9n$ digits. We have $|b_n - 10^{9n}| \lt 2\cdot 10^{8n}$ and $|c_n-\frac {1}9\cdot 10^{9n}|\lt 2\cdot 10^{8n}$, so $|\frac {b_n}{c_n}-9| \lt \frac {20}{10^n}\to 0$
Ross Millikan
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Write $x_n, y_n$ for the versions with $n$ many repetitions - so e.g. $$x_4=111122223333444455556666777788889999$$ and $$y_4=999988887777666655554444333322221111.$$ We can get upper and lower bounds for $y_n\over x_n$ as follows:
For an upper bound, replace $y_n$ by $y^+_n$, which consists of $n+1$ many $9$s followed by $8n-1$ many $0$s; this is indeed $>y_n$. For example, $$y_2^+=999000000000000000.$$ Similarly, let $x_n^+$ be $n+1$ many $1$s followed by $8n-1$ many $0$s. Then $9={y^+_n\over x^+_n}>{y_n\over x_n}$.
For a ower bound, we do the opposite trick: replace $y_n$ with $y^-_n$ consisting of $n$ $9$s and then $8n$ $0$s, and replace $x_n$ with $x_n^-$ consisting of $n-1$ many $1$s, one $2$, and then $8n$ many $0$s. Then ${y^-_n\over x^-_n}<{y^n\over x_n}$, but it's easy to check that $\lim_{n\rightarrow\infty}{y^-_n\over x^-_n}=9$.
So we've sandwiched the actual limit above and below by $9$.
Noah Schweber
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As you increase the same digits your numbers are approaching to $$\frac {99999999999999...}{11111111111111....}=9$$
Mohammad Riazi-Kermani
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Putting $a=10^n$ we have $$\frac{(9..9)a^8+(8..8)a^7+(7..7)a^6+\cdots+(3..3)a^2+(2..2)a+(1..1)}{(1..1)a^8+(2..2)a^7+(3..3)a^6+\cdots+(7..7)a^2+(8..8)a+(9..9)}$$ so dividing by $a^8$ in both numerator and denominator one has in the limit $$\frac{99....99}{11....11}=9$$
Isaac Browne
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Piquito
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