Today, a thread full of cool math facts appeared on Reddit. As usual, someone mentioned the fact that 111111111 × 111111111 = 12345678987654321. In another reply, someone pointed out that this also works in other bases. For some reason, I decided that I needed to prove that it works in all bases.
To begin, I needed a general formula for values of the 111… terms. This was fairly straightforward: for a base 
, we want 
base- digits, all ones. To standardize the base, we multiply each digit by an increasing power of and sum. Since each digit is one, we get a nice geometric series which can easily be solved.
![\[\sum\limits_{i=1}^{b-1} b^{i-1} = \frac{b-b^b}{b-b^2}\]](https://lo.calho.st/wp-content/ql-cache/quicklatex.com-03480c9bd2192a952d324f5bdc4f2c95_l3.png "Rendered by QuickLaTeX.com")
When we multiply this number by itself, we are squaring it, so we end up with 
.
The hard part was writing a general form for the 
number. To deal with this, I broke it down into two parts, as illustrated below.
| Digit value | 1 | 2 |  |
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2 | 1 |
| Place multiplier |  |
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I calculated the values of the most-significant digits starting at the left, and the values of the least-significant digits starting at a right. To make the math come out nicely, I actually included the center digit in both formulas. That’s okay, since we can subtract it off once to make up for the duplicate. Now we have a summation formula for the value of the square.
![\[\left( \sum\limits_{i=1}^{b-1} ib^{2b-(i+3)} \right) + \left( \sum\limits_{i=1}^{b-1} ib^{i-1} \right) - b^{b-2}\]](https://lo.calho.st/wp-content/ql-cache/quicklatex.com-e2293358b55ea177f7faf960d4def316_l3.png "Rendered by QuickLaTeX.com")
With a little thinking (or the help of a computer algebra system), we can get a neat closed form.
![\[\left( \sum\limits_{i=1}^{b-1} ib^{2b-(i+3)} + ib^{i-1} \right) - b^{b-2} = \left(\frac{b-b^b}{b^2-b}\right)^2\]](https://lo.calho.st/wp-content/ql-cache/quicklatex.com-c52d8ce61f393a679032157cb2d35367_l3.png "Rendered by QuickLaTeX.com")
We can see that this is quite similar to the expression we got for the square above; the only difference is that the 
denominator has changed to 
. Fortunately, this negation goes away when squaring, so we can trivially prove that the two expressions are equal.
And there we have it: proof that this curiosity is true in any base of at least two.
