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Double Digit

Stage: 3 Challenge Level: Challenge Level:1

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.