DOTS Division
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
Age
14 to 16
Challenge level
Problem
Take any pair of two digit numbers $ ab $ and $ cd $ where, without loss of generality, $ ab> cd $. Form two 4 digit numbers $ abcd $ and $ cdab $ and calculate: \[\frac{abcd^2-cdab^2}{ab^2-cd^2}\] Repeat this with other choices of $ab$ and $cd$. There is a common feature of all the answers. What is it? Why does this occur? Generalise this to $n$ digits for other values of $n$.
Student Solutions
We have received a very clearly explained solution to this, but unfortunately whoever sent it did not include their name. If it was you, let us know!
Let $ab=x$ and $cd=y$ (where $ab$ means $a$ as the tens digit, and $b$ as the ones digit, not $a$ times $b$).
\[\frac{abcd^2-cdab^2}{ab^2-cd^2} = \frac{(100x+y)^2 - (100y+x)^2}{x^2-y^2}\] \[= \frac{(10000x^2+200xy+y^2) - (10000y^2+200xy+x^2)}{x^2-y^2}\] \[= \frac{9999x^2 - 9999y^2}{x^2-y^2}\] \[= \frac{9999(x^2-y^2)}{x^2-y^2}\] \[ = 9999 \]
(since $x> y$ we are not dividing by zero)