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Many years ago, I noticed that $987654321/123456789 = 8.0000000729\ldots$.

I sent it in to Martin Gardner at Scientific American and he published it in his column!!!

My life has gone downhill since then:)

My questions are:

  • Why is this so?

  • What happens beyond the "$729$"?

  • What happens in bases other than $10$?

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    $\begingroup$ 8.000000072900000663390006036849054935326399911470239194379176... $\endgroup$ – copper.hat May 19 '13 at 7:10
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    $\begingroup$ I don't see it, why is this number interesting? $\endgroup$ – Matsemann May 19 '13 at 9:43
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    $\begingroup$ 8.0000000729000006633900060368490549353263999114702391943791766688505076865396199475105415223459278533479434654662855357431983752631052148942574555377428453934598930804850270324137459949650885541823058430589831718367468637143964598010077841891708361214546087052369392176561468806709366141055231883602610140783752281132145758302526400552990245032211229793122191117411939168448646432882682539232411107014941073835963771907270324356159951641055559933605595395810918101879354727102128016629364951327221057077711619407175736605299203108222748284827009391925785466524647745374294482906079794445326129452467 $\endgroup$ – Double AA May 19 '13 at 10:29
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    $\begingroup$ Unrelated, but ever notice that Sqrt(9.87654321) approximates pi? The number system is full of this kind of stuff. $\endgroup$ – cobaltduck May 19 '13 at 11:44
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    $\begingroup$ @cobaltduck: It is a terrible approximation given how many digits you're putting in. It has more than $10$ times the error of $\sqrt{9.87}$ $\endgroup$ – Jonas Meyer May 19 '13 at 18:28
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In base $n$ the numerator is $$p = n^{n-1} - \frac{n^{n-1}-1}{(n-1)^2}$$ and the denominator is $$q = \frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$

Note that $p = (n-2)q + n-1$ and for the quotient we get

\begin{align} \frac{p}{q} &= n-2 + \frac{(n-1)^3}{n^n} \frac{1}{1 - \frac{n^2-n+1}{n^n}} \\ &= n-2 + \frac{(n-1)^3}{n^n} \sum_{k=0}^{\infty} \left(\frac{n^2-n+1}{n^n}\right)^k. \end{align}

Indeed for $n=10$ this is

$$\frac{987654321}{123456789} = 8 + \frac{729}{10^{10}}\sum_{k=0}^{\infty}\left(\frac{91}{10^{10}}\right)^k $$

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    $\begingroup$ so methodical. Neat. $\endgroup$ – obataku May 19 '13 at 13:52
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    $\begingroup$ I almost don't want to change the upvote from being "123." $\endgroup$ – BarocliniCplusplus Jun 2 '20 at 23:51
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$$729=9^3$$ $$66339=9^3\cdot 91$$ $$6036849=9^3\cdot 91^2$$ $$...$$ $$987654321/123456789=8+9^3\cdot 10^{-10}\cdot\displaystyle\sum_{n=0}^{\infty}(91\cdot 10^{-10})^n$$

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    $\begingroup$ @FedericaMaggioni Please tell me you didn't just "find" that...! $\endgroup$ – Jonathan H May 19 '13 at 7:40
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    $\begingroup$ Could you add a bit more detail? $\endgroup$ – nbubis May 19 '13 at 8:08
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    $\begingroup$ Is it just me who is too stupid to understand what the pattern is? $\endgroup$ – Parth Kohli May 20 '13 at 10:11
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    $\begingroup$ @ΠάρτηΚοχλί look at Double AA's expansion in the comments above.. $\endgroup$ – Federica Maggioni May 20 '13 at 10:26
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    $\begingroup$ Ah, gotcha. Nice answer. $\endgroup$ – Parth Kohli May 20 '13 at 11:27
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Let $$S_n(a)=1 +2a+\ldots +na^{n-1}=\frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$ $$T_n(a)=a^{n-1}+2a^{n-2}+\ldots +n=a^{n-1}S_n(a^{-1}).$$

Then $$ \frac{S_n(a)}{T_n(a)}=\frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$ For $a=10,n=9$ we have $$ \frac{S_n(a)}{T_n(a)}\approx\frac{8\cdot 10^{10}+1}{10^{10}}. $$

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Just to add to the excellent answers above, some examples:

${987654321\,/\,123456789}\approx 8.00000007290000066339$

${{87654321}_9\,/\,{12345678}_9}\approx {7.000000628000056238}_9$

${{7654321}_8\,/\,{1234567}_8}\approx {6.0000052700046137}_8$

${{654321}_7\,/\,{123456}_7}\approx {5.00004260036036}_7$

${{\mathrm{fedcba987654321}}_{16}\,/\,{\mathrm{123456789abcdef}}_{16}}\approx {\mathrm{e.0000000000000d2f00000000000c693f}}_{16}$

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  • $\begingroup$ And what do these numbers show?? $\endgroup$ – Matsemann May 21 '13 at 19:04
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    $\begingroup$ They show empirically that the behavior seen in base 10 is present for all bases. $\endgroup$ – Mark Adler May 21 '13 at 20:13
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    $\begingroup$ And what behavior is that? I asked in a comment to the question what's so interesting about this, no one has answered. $\endgroup$ – Matsemann Jun 13 '13 at 8:59
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    $\begingroup$ The behavior is that the result is extremely close to the base minus two. Like a joke, if it has to be explained to you, it won't be funny. If you weren't intrigued by this when you first saw it, then you probably never will be. $\endgroup$ – Mark Adler Jun 13 '13 at 15:00
  • $\begingroup$ No, they show empirically that the behavior seen in base 10 is present for bases 9, 8, 7 and 16. $\endgroup$ – jwg Jun 28 '17 at 10:09
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$98765432 / 12345679 = 8$, exactly. You can see how the pattern works by multiplying out $12345679 * 8$ starting at either end.

This explains why your fraction is close to an integer. If you think the $729$ is interesting (I don't), it can be explained by some of the other answers here.

Edit:

What can we say about the fact that $12345679 * 8 = 98765432$? I have been aware of this 'factlet' for about 20 years, and remember it being used to 'demonstrate' calculators (which often had 8 digit displays back in the day).

I just recently realised that:

$$ \frac{1}{81} = \left(\frac{1}{9}\right)^2 = \left(\sum_{k=1}^{\infty}\frac{1}{10^k}\right)^2 = \sum_{k=1}^{\infty} \sum_{m=1}^{k-1} \frac{1}{10^m} \frac{1}{10^{k-m}} = \sum_{k=1}^{\infty} \frac{k-1}{10^k} $$

In other words, while $\frac{1}{9} = 0.1111111\ldots$ $$ \frac{1}{81} = 0.01 + 0.002 + 0.0003 + 0.0004 + 0.00005 \ldots $$

It is pretty easy to see that this infinite sum is going to converge to something starting $0.012345\ldots$. If you keep on adding, or work out $\frac{1}{81}$ by division, you get $$ 0.012345679012345679012345679\ldots $$ When you get to the point where you add $\frac{10}{10^{11}}$, the first carry happens, which leads to the 9 where you might expect an 8. After that every addition carries and the decimal expansion repeats every 9 digits (not every ten - because the amount we carry keeps on getting bigger and bigger).

Now, $\frac{8}{81} = \frac{9}{81} - \frac{1}{81}$, or $$ \frac{8}{81} = 0.11111111\ldots - 0.012345679012345\ldots $$ Think of each '1' digit in $0.111\ldots$ as being a '10' in the next column. This means that we can work out $\frac{8}{81}$ as the "10's complement" of $\frac{1}{81}$, since we are subtracting a digit between $1$ and $9$ from $10$, to get another single digit which appears in the same place. So $\frac{8}{81}$ starts $0.098765\ldots$. The only break in the pattern is when you get to the digit '0' - subtracting 0 from 10 leaves you with 10, or a '1' in the next digit on the left, changing the 1 to a 2.

So $$ \frac{8}{81} = 0.098765432098765432098765\ldots $$

and therefore $$ 0.0123456790123456790\ldots * 8 = 0.0987654320987654320\ldots $$ and clearly this gets you that $$ 12345679 * 8 = 98765432 $$

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    $\begingroup$ A bit late but note \times ($\times$) instead of *. $\endgroup$ – TheSimpliFire Apr 4 '18 at 18:01
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I think one should also think about how this is related to 9 being before 10. Notice the two following striking facts:

$12345679*8=98765432$ and $12345679*9=111111111$

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  • $\begingroup$ See @jwg's answer. $\endgroup$ – Jonathan H Jan 1 '19 at 14:07
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I think @BorisNovikov's answer is the best here, but I had to do some work to understand it; this is my attempt at clarifying his answer, hopefully this can help others too:

In the following, we adopt the notation $123456789 = [ 1, 2, ..., B-1 ]_B$ in base $B=10$.

Boris introduced the following sum: $$ S_d(B) = [d, d-1, ..., 1]_B = 1 + 2B + ... + dB^{d-1} = \sum_{k=1}^d kB^{k-1} = \frac{dB^{d+1} - (d+1)B^d + 1}{(B-1)^2} $$ where $d$ is the number of digits, $B$ the base, and the final formula is obtained by derivation of the usual geometric sum.

From there, it is easy to see that: $$ [1, 2, ..., d]_B = d + (d-1)B + ... + 1B^{d-1} = B^{d-1} S_d(1/B) $$ and putting both results together, we have: $$ \frac{ [B-1, B-2, ..., 1]_B }{ [1, 2, ..., B-1]_B } = \frac{ S_{B-1}(B) }{ B^{B-2} S_{B-1}(1/B) } = \frac{1 + (B-2)B^B}{B^B - B(B-1) - 1} $$

Clearly both the numerator and denominator are dominated by the terms in $B^B$, so this fraction is equivalent to $B-2$ as $B$ tends to infinity, and we can see that the exact value is very close already for $B=10$ (see also @Mark Adler's post for other examples). Beautiful, isn't it? :)

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