0
$\begingroup$

I was looking around on the internet until I stumbled upon this equation.

$$111111111\times111111111 = 12345678987654321$$ How does this actually work? It is quite amazing how the number ascend and then descend by ones.

$\endgroup$
  • $\begingroup$ It is just a coincidence. You can yourself multiply out the numbers on pen and paper to see the reason $\endgroup$ – Darkrai Feb 16 '18 at 2:07
  • $\begingroup$ This looks like a trivial application of the Central Limit Theorem. $\endgroup$ – CopyPasteIt Feb 16 '18 at 3:43
8
$\begingroup$

Just multiply it out with the grade school pen-and-paper algorithm:

111111111 x 111111111
---------------------
            111111111
           111111111
          111111111
         111111111
        111111111
       111111111
      111111111
     111111111
    111111111
   ------------------
    12345678987654321

Each digit of the result comes from summing the digits in one column -- that is, counting how many ones there are. Since this is at most 9, there are no carries between columns.

$\endgroup$
  • 1
    $\begingroup$ +1 - nice picture! To the OP, note that we can use this same picture to cook up versions of this in arbitrary bases. For example, in base $5$ we'll want the end result to be "$1234321_5$"; thinking about the picture above, this will happen when we square "$1111_5$." $\endgroup$ – Noah Schweber Feb 16 '18 at 2:13
3
$\begingroup$

If you write one of the $111111111$ values up the side and the other across the top of a table, and multiply each digit selection individually, you get a table of $1$s. But the actual value behind those $1$s is a power of ten, so you collect like powers of ten as shown: enter image description here

$\endgroup$
2
$\begingroup$

$$\left(\sum_{k=0}^n 10^k\right)^2 = \sum_{i=0}^n \sum_{j=0}^n 10^{i+j} $$ There are $k+1$ occurrences of $10^k$, namely $(i,j)=(0,k),(1,k-1),\ldots, (k,0)$, if $k \le n$, and $2n+1-k$, namely $(i,j) = (k-n,n),(k-n+1,n-1),\ldots,(n,k-n)$, if $n < k \le 2n$. If $n \le 8$ this means the decimal representation of $(1\ldots1)\times(1\ldots1)$ is $12 \ldots n (n+1) n \ldots 21$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.