Proving a mathematical curiosity.

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 b, we want b-1 base-b digits, all ones. To standardize the base, we multiply each digit by an increasing power of b and sum. Since each digit is one, we get a nice geometric series which can easily be solved.

    \[\sum\limits_{i=1}^{b-1} b^{i-1} = \frac{b-b^b}{b-b^2}\]

When we multiply this number by itself, we are squaring it, so we end up with \left((b-b^b)/(b-b^2)\right)^2.

The hard part was writing a general form for the 1234 \dots (b-1) \dots 4321 number. To deal with this, I broke it down into two parts, as illustrated below.

Digit value 1 2 \dots (b-2) (b-1) (b-2) \dots 2 1
Place multiplier b^{2b-3} b^{2b-4} b^{2b-(b+1)} b^{2b-(b+2)}
b^{b-2} b^{b-3} b^1 b^0

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.

    \[\left( \sum\limits_{i=1}^{b-1} ib^{2b-(i+3)} \right) + \left( \sum\limits_{i=1}^{b-1} ib^{i-1} \right) - b^{b-2}\]

With a little thinking (or the help of a computer algebra system), we can get a neat closed form.

    \[\left( \sum\limits_{i=1}^{b-1} ib^{2b-(i+3)} + ib^{i-1} \right) - b^{b-2} = \left(\frac{b-b^b}{b^2-b}\right)^2\]

We can see that this is quite similar to the expression we got for the square above; the only difference is that the b-b^2 denominator has changed to b^2-b. 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.

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