Multiplying $11$ by a positive whole number $a$ that's less than $10$ is easy. The result is simply $a$ repeated. For example, $$11 \times 2 = 22$$ $$11 \times 3 = 33$$ $$11 \times 4 = 44,$$ and so on. Most people know this, but what's less well-known is that there's also a neat trick to multiply larger numbers by $11.$ Suppose $a$ is a whole number with two digits. To work out $11 \times a,$ simply work out the sum of the digits of $a$ and drop that sum in-between the digits. For example, let $a = 23.$ The sum of its digits is $2+3=5.$ Dropping that sum between the digits gives $253,$ which is indeed equal to $11 \times 23.$ There's just one little caveat. If the sum of the digits of $a$ is $10$ or larger, you need to carry a digit. In other words, you stick the right-most digit of the sum between the original digits and add the left-most digit of the sum to the original left-most digit. For example, if $a = 75,$ then the sum of the digits is $12.$ We therefore stick a $2$ between the original digits $7$ and $5$ and add a $1$ to the $7$ to get $825,$ which again is the correct result. You can convince yourself that this trick always works using long multiplication. Suppose the digits of $a$ are $x$ and $y,$ so $a=xy.$ Long multiplication now tells us that \begin{tabular}{ c c c c c c} &$x$& $y$&$\times$&$1$&$1$& \\ \hline $=$&&& $x$ & $y$ & $0$ \\ & $+$&&&$x$&$y$ \\ \end{tabular} from which our result easily follows. Can you work out the trick for numbers $a$ with more than two digits? As a hint, here is the long multiplication when the digits of $a$ are $x_1,$ $x_2,$ $x_3,$ up to $x_n$: \begin{tabular}{ c c c c c c c c} &$x_1$ & $x_2$ & $x_3$ & $\dots$ & $x_n$ & $\times$&$1$&$1$& \\ \hline =&& & $x_1$ & $x_2$ & $x_3$ & $\dots$ & $x_n$ & $0$ \\ & +&&&$x_1$&$x_2$& $\dots$ & $x_{n-1}$ & $x_n$ \\ \end{tabular}