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Tim from Gravesend Grammar School and Mohammad Afzaal Butt both sent us similar solutions to the problem. Well done Tim and Mohammad. Here is Mohammad's solution:
Let the three digit number be $xyz$. Hence the six digit number will be $xyzxyz$. Now
$$\eqalign { xyzxyz &= 100000x + 10000y + 1000z + 100x + 10y + z \cr
&= 100100x + 10010y + 1001z \cr
&= 1001 (100x + 10y + z) \cr
&= 7 \times 11 \times 13 (100x + 10y + z)} $$ Hence the number $xyzxyz$ is always divisible by $7$, $11$ and $13$.
Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. Try lots of examples. What happens? Can you explain it?
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
Take any four digit number. Move the first digit to the end and move the rest along. Now add your two numbers. Did you get a multiple of 11?