Maths in a minute: Easy 11

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.

Eleven

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

 

$x$

$y$

$\times $

$1$

$1$

 

$=$

   

$x$

$y$

$0$

 

$+$

   

$x$

$y$

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 $x_ n$:

 

$x_1$

$x_2$

$x_3$

$\dots $

$x_ n$

$\times $

$1$

$1$

 

=

   

$x_1$

$x_2$

$x_3$

$\dots $

$x_ n$

$0$

 

+

   

$x_1$

$x_2$

$\dots $

$x_{n-1}$

$x_ n$