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I came across the following:

$\begin{align} 1 \times 8 + 1 &= 9 \\ 12 \times 8 + 2 & = 98 \\ 123 \times 8 + 3 & = 987 \\ 1234 \times 8 + 4 & = 9876 \\ 12345 \times 8 + 5 & = 98765 \\ 123456 \times 8 + 6 & = 987654 \\ 1234567 \times 8 + 7 & = 9876543 \\ 12345678 \times 8 + 8 & = 98765432 \\ 123456789 \times 8 + 9 & = 987654321. \\ \end{align}$

I'm looking for an explanation for this pattern. I suspect that there is some connection to the series $\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \cdots$.

This post asks the same question but has no answers posted.

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    $\begingroup$ @Zacky I did try searching. If you happen to find a post where an answer is given, do let me know. $\endgroup$ – Omnomnomnom Dec 31 '19 at 0:31
  • $\begingroup$ I've already answered a question about digit reversal how is this not flagged as a duplicate? $\endgroup$ – Adam 2 days ago
  • $\begingroup$ @Adam I was curious about what you were referring too and having looked through your answers I see you must be referring to your answers here and here. There is no clear way to use those formulas to reverse digits to answer this question, nor is there a way to apply these answers to those questions. The thought that this question is "the same" as those others because it involves numbers with reversed digits is laughable. $\endgroup$ – Omnomnomnom 2 days ago
  • $\begingroup$ neat well I'm glad you find it funny if you really think there is not a common underlying principle it's best you laugh $\endgroup$ – Adam 2 days ago
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If I consider the equations you provide with your "ideas so far":

\begin{align} 1 \times 9 + 1 &= 10 \\ 12 \times 9 + 2 & = 110 \\ 123 \times 9 + 3 & = 1110 \\ \vdots\\ 123456789 \times 9 + 9 & = 1111111110, \\ \end{align}

The first equation being true, this system is equivalent to the system composed of their successive differences all of them having the common pattern :

$$\underbrace{11...1}_{k \ \text{digits}} \times 9 + 1 = 10^k$$

which is an (almost) evident fact.

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    $\begingroup$ A lot of great justifications, but this one is my favorite $\endgroup$ – Omnomnomnom Dec 31 '19 at 3:32
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Rewritten in sum form, your equations become:

$$\bigg(\sum_{r=1}^n8r\cdot10^{n-r}\bigg)+n=\bigg(\sum_{r=1}^n(10-r)\cdot10^{n-r}\bigg)$$ for $n\in\Bbb N\cap[1,9]$

Subtracting the RHS gives:

$$n=\sum_{r=1}^n\bigg[(10-9r)\cdot10^{n-r}\bigg]$$

We prove this via induction:

$$\text{Assume } k=\sum_{r=1}^k\bigg[(10-9r)\cdot10^{k-r}\bigg]$$ $$\text{Then } 10k=\sum_{r=1}^k\bigg[(10-9r)\cdot10^{k+1-r}\bigg]$$ $$\text{So } \sum_{r=1}^{k+1}\bigg[(10-9r)\cdot10^{k+1-r}\bigg]=10k+(10-(9k+9))\cdot10^{(k+1)-(k+1)}$$ $$=10k+(1-9k)\cdot1=k+1 \text{ a.r.}$$

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Ideas so far:

Adding another $123 \cdots$ to both sides yields the following equivalent series equations: \begin{align} 1 \times 9 + 1 &= 10 \\ 12 \times 9 + 2 & = 110 \\ 123 \times 9 + 3 & = 1110 \\ \vdots\\ 123456789 \times 9 + 9 & = 1111111110, \\ \end{align} so it suffices to prove that the above pattern holds. We note that for $n = 1,\dots,9$, we can write the first number on the LHS of each equation as $$ 10^{n-1} \cdot (1 + 2 \cdot 10^{-1} + \cdots + n \cdot 10^{-(n-1)}). $$ Let $M = 1 + 2 \cdot 10^{-1} + \cdots + n \cdot 10^{-(n-1)}$. We have $$ \begin{align} M &= 1 + 2 \cdot 10^{-1} + \cdots + n \cdot 10^{-(n-1)} \\ & = (1 + 2 \cdot 10^{-1} + \cdots + n \cdot 10^{-(n-1)} + \cdots) - ((n+1) \cdot 10^{-n} + (n+2) \cdot 10^{-(n+1)} + \cdots) \\ & = \frac{1}{(1 - 10^{-1})^2} - ((n+1) \cdot 10^{-n} + (n+2) \cdot 10^{-(n+1)} + \cdots). \end{align} $$ Let $N = (n+1) \cdot 10^{-n} + (n+2) \cdot 10^{-(n+1)} + \cdots$. We can rewrite this as $$ \begin{align} M &= \sum_{k=n+1}^\infty k\cdot 10^{-(k-1)} = \sum_{k=1}^\infty (k+n)\cdot 10^{-(k+n-1)} \\ & = \sum_{k=1}^\infty k \cdot 10^{-(k+n-1)} + n \cdot \sum_{k=1}^\infty \cdot 10^{-(k+n-1)} \\ & = 10^{-n} \cdot \sum_{k=1}^\infty k \cdot 10^{-(k-1)} + n \cdot 10^{-n} \cdot \sum_{k=1}^\infty \cdot 10^{-(k-1)} \\ & = 10^{-n} \frac{1}{(1 - 10^{-1})^2} + n \cdot 10^{-n} \cdot \frac{1}{1 - 10^{-1}} \\ & = 10^{-n} \cdot \frac{1 + n \cdot(1 - 10^{-n})}{(1 - 10^{-1})^2} \end{align} $$ That is, we have $$ M = \frac{1}{(1 - 10^{-1})^2} - N = \frac{1 - 10^{-n}(1 + n \cdot(1 - 10^{-n}))}{(1 - 10^{-1})^2}. $$ With that, we can rewrite the LHS of the equation as $$ \begin{align} 10^{n-1}M + n &= \frac{10^{n-1} - 10^{-1}(1 + n \cdot(1 - 10^{-n}))}{(1 - 10^{-1})^2} + n \\ & = \frac{10^{n-1} - 10^{-1}(1 + n \cdot(1 - 10^{-n})) + n\cdot (1 - 10^{-1})^2}{(1 - 10^{-1})^2} \end{align} $$

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Well, it basically boils down to

$1111.....110 - 12345...(k-1)k = 987.....(10-k+1)(10-k)$.

This isn't too surprising. The last digit, derived from $0-k$ is $10-k$. We must borrow a $1$ so the next digit is from $0 - (k-1)$, and so on.

Thus if $1234....k\times 9 + k = 1111.....10$ then it follows that $1234...k\times 8 + k = 987.....(10-k)$.

But why should $1234...k\times 9 + k = 1111....10$?

Well, it stands to reason that $1234...k\times 9 = 1234...k(10 -1) = 12345....k0-12345...k$

Subtracting $0 - k$ we get that the last digit is $10-k$. Now we has to borrow $1$ for the previous column, and the next digits were $k- (k-1)$ but as we had to borrow we have $k-(k-1) -1 = 0$. Now we didn't borrow and the next column after that is $(k-1) -(k-2) =1$ and we don't borrow. All remaining columns are $(k-j) - (k-(j-1)=1$ and thus all remaining columns result in $1$.

So we can conclude that $12345...k0 - 12345...k = 111111.....10(10-k)$.

And if we add $k$ to that we have $(10-k) +k= 10$ and we carry the $1$ to the next column which goes from $0$ to $1$.

So $1234....k*9 + k = 11111.....1110$.

And that's it.

$12345...k*8 + k =$

$12345...k*9 +k - 12345....k =$

$12345...k*10 - 12345....k + k -12345...k =$

$11111....0(10-k) + k -12345...k =$

$11111.....10 - 12345...k =$

$987.....(10-k)$.

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........

Well, if we do an example it will be very clear:

$123456 = 123456$

$123456\times 10 = 1234560$

$123456\times 9 = 1234560 - 123456$

now doing subtraction and borrowing:

$\require{cancel}$

$\ \ \ 12345\cancel6^50$

$-\ \ 12345\ \ \ 6$

$\_\_\_\_\_\_\_\_$

$111104$

$123456\times 9 = 111104$

$123456\times 9 + 6 = 111104+6 = 111110$.

$123456\times 8+ 6 = 111110-123456$ and

$\cancel 1 \cancel 1^{10}\cancel 1^{10}\cancel 1^{10}\cancel 1^{1}0$

$-1\ \ \ \ 2\ \ \ \ 3\ \ \ \ 4\ \ \ \ 5\ \ \ \ 6$

$\_\_\_\_\_\_\_\_$

$\ \ \ 9\ \ \ 8\ \ \ 7\ \ \ 6\ \ \ 5\ \ \ 4$

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More generally.

Well.....

$123....k = 123....k$

$123....k\times 10 = 123.....k0$

$123.....k\times 9 = 1234....k0 - 1234....k$

Now subtracting and borrowing we get...

$1234....k0 -1234....k = (1-0)(2-1)....([k-1]-[k-2])(k-(k-1)-1)(10-k)=1111....10(10-k)$

(example: $12340 -1234 = 11106$)

So $1234...k\times 9 = 111....10(10-k)$

$1234....k\times 9 + k = 11111.....10$.

And finally that means

$1324....k\times 8 + k = 11111....10- 1234....k$

And.... well, we'd better use sumation notation to figure that out.

$\sum_{i=1}^k 10^k - \sum_{i=1}^k i*10^{k-1}=$

$\sum_{i=1}^k(10-k)*10^{k-1}=$

$987....(10-k)$.

And that is that.

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I'll leave a formal prove with summation notation for an exercise for the reader.

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Oh....

I was wondering if their as an intuive we to see that $1234...k \times 9 + k = 1111...10$. I mean, my reasoning above worked but it didn't have the gut "well, of course".

But if $12345...k = $

$1111.... + 111111.... + ...... +1111 + 111 + 11 + 1$ then

$1234...k\times 9 = 99999.... + 9999.... + ...+ 999 + 99 + 9=$

$(10^k - 1) + (10^{k-1} -1) + ..... + (10^3-1) + (10^2 -1) + (10-1) =$

$111111....10 - k$.

I guess that .. fits.

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An example might help explain the pattern:

$$\begin{align} 12{\color\red3}\times8+{\color\red3}=987 &\implies12{\color\red3}0\times8+{\color\red3}\times10=9870\\ &\implies12{\color\red3}{\color\green4}\times8+{\color\green4}=9870+{\color\green4}\times8+{\color\green4}-{\color\red3}\times10=9870+({\color\green4}-{\color\red3})\times10-{\color\green4}=9870+{\color\yellow6} \end{align}$$

(My apologies if the colors, in particular the yellow $6$ at the very end, are hard to see.)

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$$\left\lfloor {10^n\over (1-x)^2} \right\rfloor \cdot 8+n= 9\cdot \left\lfloor {10^n\over (1-x)^2} \right\rfloor -\left\lfloor {10^n\over(1-x)^2} \right\rfloor +n$$

With $x=1$ is what you've observed ( yes I realize the division by 0, just don't know a better way yet to present what the OP sees). The real question though is what makes it work.

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    $\begingroup$ That's a neat way of putting it $\endgroup$ – Omnomnomnom Dec 31 '19 at 3:07
  • $\begingroup$ Yeah now if only it could easily be transformed to a different $x$ $\endgroup$ – Roddy MacPhee Dec 31 '19 at 4:05
  • $\begingroup$ @Omnomnonom serchop() is Your PARI GP friend it seems. $\endgroup$ – Roddy MacPhee Dec 31 '19 at 17:54
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This is something I noticed but I'm still thinking about if it means anything: $$\boxed{1\cdot8+1=9}\\\downarrow$$

$$10\cdot8+10=90$$ $$10\cdot8+18=98$$ $$(10+2)\cdot8+2=98$$ $$\boxed{12\cdot8+2=98}\\\downarrow$$

$$120\cdot8+20=980$$ $$120\cdot8+27=987$$ $$(120+3)\cdot8+3=987$$ $$\boxed{123\cdot8+3=987}\\\downarrow$$

$$1230\cdot8+30=9870$$ $$1230\cdot8+36=9876$$ $$(1230+4)\cdot8+4=9876$$ $$\boxed{1234\cdot8+4=9876}\\\downarrow\\\cdot\\\cdot\\\cdot$$

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