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 $\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \cdots$.
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$\endgroup$ 57 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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:
$$\text{Assume } k=\sum_{r=1}^k\bigg[(10-9r)\cdot10^{k-r}\bigg]$$$$\text{Then } 10k=\sum_{r=1}^k\bigg[(10-9r)\cdot10^{k+1-r}\bigg]$$$$\text{So } \sum_{r=1}^{k+1}\bigg[(10-9r)\cdot10^{k+1-r}\bigg]=10k+(10-(9k+9))\cdot10^{(k+1)-(k+1)}$$$$=10k+(1-9k)\cdot1=k+1 \text{ a.r.}$$
$\endgroup$ $\begingroup$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$$ \begin{align} 10^{n-1}M + n &= \frac{10^{n-1} - 10^{-1}(1 + n \cdot(1 - 10^{-n}))}{(1 - 10^{-1})^2} + n \\ & = \frac{10^{n-1} - 10^{-1}(1 + n \cdot(1 - 10^{-n})) + n\cdot (1 - 10^{-1})^2}{(1 - 10^{-1})^2} \end{align} $$
$\endgroup$ $\begingroup$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
$\cancel 1 \cancel 1^{10}\cancel 1^{10}\cancel 1^{10}\cancel 1^{1}0$
$-1\ \ \ \ 2\ \ \ \ 3\ \ \ \ 4\ \ \ \ 5\ \ \ \ 6$
$\_\_\_\_\_\_\_\_$
$\ \ \ 9\ \ \ 8\ \ \ 7\ \ \ 6\ \ \ 5\ \ \ 4$
====
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.
+++++++
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.
$\endgroup$ $\begingroup$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.)
$\endgroup$ $\begingroup$$$\left\lfloor {10^n\over (1-x)^2} \right\rfloor \cdot 8+n= 9\cdot \left\lfloor {10^n\over (1-x)^2} \right\rfloor -\left\lfloor {10^n\over(1-x)^2} \right\rfloor +n$$
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.
$\endgroup$ 3 $\begingroup$This is something I noticed but I'm still thinking about if it means anything:$$\boxed{1\cdot8+1=9}\\\downarrow$$
$$10\cdot8+10=90$$$$10\cdot8+18=98$$$$(10+2)\cdot8+2=98$$$$\boxed{12\cdot8+2=98}\\\downarrow$$
$$120\cdot8+20=980$$$$120\cdot8+27=987$$$$(120+3)\cdot8+3=987$$$$\boxed{123\cdot8+3=987}\\\downarrow$$
$$1230\cdot8+30=9870$$$$1230\cdot8+36=9876$$$$(1230+4)\cdot8+4=9876$$$$\boxed{1234\cdot8+4=9876}\\\downarrow\\\cdot\\\cdot\\\cdot$$
$\endgroup$
