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## 'Double Digit' printed from http://nrich.maths.org/

Everyone said correctly that the answer will always be $11$. Many
of you also managed to justify this using algebra. The first to do
sowas Stephen from Singapore International School. His solution
went along the following lines:

Let single numbers $x$ and $y$ represent our digits. Then our two
digit numbers will be $10x + y$ and $10y + x$.

For example, if $x=1$ and $y=2$, our two digit numbers are $12 =
10\times 1 + 2$ and $21 = 10\times 2 + 1$. Now $12 + 21 = 10\times
1 + 2 + 10\times 2 + 1 = 11\times 1 + 11\times 2$. So
$$\frac{12+21}{1+2} = \frac{11\times 1 + 11\times 2}{1+2} =
\frac{11\times 3}{3} = 11$$ In general, the sum of our two digit
numbers will be $(10x + y) + (10y + x) = 11x +11y$ and this sum
divided by $x+y$ will be 11.

Patrick from Woodbridge School extended the problem to three digits
in the following way:

Suppose we pick three digits, say $2$, $3$ and $7$. Then we
construct the six three digit numbers from our digits - in our case
$237$, $273$, $327$, $372$, $723$ and $732$. If we add these
numbers together and divide by $2+3+7$ we get $222$. Try this with
your own set of digits. Can you explain what is happening?

Patrick then extended the problem in the same way to four digits.
Investigate further.